📚 Question Bank 📝 Mock Tests⚡ Live Quiz🎓 Teacher Dashboard ← All Grades ℹ️ About Us
Log In
25 objectives • 250 MCQs • 100 short answers

Student Revision Notes

Term 3 Physics Quick Review

250 MCQs • 100 Short Answers
Adapted Question Credit
50 MCQs are adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank. Adapted questions show a small citation inside their answer box.

Grade 10 Advanced Physics - Term 3 EOT Revision Notes

Page mapping used: actual PDF page = textbook page + 5. Example: textbook page 90 -> PDF page 95.

Formula Sheet

  • Refraction: n1 sin θ1 = n2 sin θ2; angles are measured from the normal.
  • Refractive index: n = c/v, where c = 3.00 × 108 m/s.
  • Boundary wave relation: v=fλ; frequency remains constant during refraction.
  • Critical angle: For denser to less dense: sin c = n2/n1.
  • Thin lens: 1/f = 1/do + 1/di; convex f positive, concave f negative.
  • Double slit: λ =xd/L and mλ =xm d/L for bright fringes.
  • Single slit: 2x1=2λ L/w; central maximum width is 2x1.
  • Diffraction grating: mλ =d sinθ; m=0 central, m=1,2,3... higher orders.
  • Relative velocity 1D: va/b + vb/c = va/c.
  • Relative velocity 2D: Use vector components, Pythagoras, and tanθ =vy/vx.

Objective-wise Quick Notes

LO01 - Refraction ray diagrams

  • Book page(s): 90 | actual PDF page(s): 95
  • Exam focus: Describe refraction of light (or a wave) as it crosses the boundary between two different mediums and represent that in a ray diagram.

LO02 - Wave quantities during refraction

  • Book page(s): 92-93 | actual PDF page(s): 97-98
  • Exam focus: Identify that during refraction, the wavelength and speed of light (or a wave) change but frequency remains the same.

LO03 - Snell's law

  • Book page(s): 91-93 | actual PDF page(s): 96-98
  • Exam focus: State and apply Snell's law of refraction.

LO04 - Refractive index

  • Book page(s): 93, 96 | actual PDF page(s): 98, 101
  • Exam focus: Calculate the refractive index of a medium using n=c/v.

LO05 - Critical angle

  • Book page(s): 94 | actual PDF page(s): 99
  • Exam focus: Define the critical angle as the angle of incidence at which the refracted light ray lies along the boundary of the two mediums.

LO06 - Refraction critical angle and TIR calculations

  • Book page(s): 92-94, 96 | actual PDF page(s): 97-99, 101
  • Exam focus: Apply the concepts of refraction, critical angle, and total internal reflection to solve numerical problems.

LO07 - Lens parts

  • Book page(s): 98-99 | actual PDF page(s): 103-104
  • Exam focus: Identify the principal axis, focal points, and focal length of convex or concave lenses.

LO08 - Convex lens ray diagrams

  • Book page(s): 98-99, 101 | actual PDF page(s): 103-104, 106
  • Exam focus: Draw a ray diagram to find the image of an object at different distances from a convex lens and determine image location and properties.

LO09 - Convex lens magnifier

  • Book page(s): 99 | actual PDF page(s): 104
  • Exam focus: Describe how a convex lens acts as a magnifier.

LO10 - Thin lens equation

  • Book page(s): 100-101, 106 | actual PDF page(s): 105-106, 111
  • Exam focus: Apply the thin lens equation to calculate image distance, object distance, or focal length using appropriate algebraic signs.

LO11 - Coherent and incoherent light

  • Book page(s): 112-113 | actual PDF page(s): 117-118
  • Exam focus: Define coherent and incoherent light and explain how coherent light is generated by passing monochromatic light through slits.

LO12 - Double-slit bright and dark fringes

  • Book page(s): 113-114 | actual PDF page(s): 118-119
  • Exam focus: Explain how bright and dark interference fringes are created in a double-slit investigation with monochromatic light.

LO13 - Double-slit wavelength equation

  • Book page(s): 115 | actual PDF page(s): 120
  • Exam focus: Apply λ =xd/L to calculate wavelength or an unknown distance in a double-slit investigation.

LO14 - Constructive interference positions

  • Book page(s): 115 | actual PDF page(s): 120
  • Exam focus: Explain that constructive interference occurs at locations x_m such that mλ =xm d/L, where m=0,1,2...

LO15 - Interference calculations

  • Book page(s): 116, 120 | actual PDF page(s): 121, 125
  • Exam focus: Apply λ =xd/L and solve problems on interference of light.

LO16 - Thin-film interference

  • Book page(s): 117 | actual PDF page(s): 122
  • Exam focus: Define and explain thin-film interference.

LO17 - Diffraction of light

  • Book page(s): 121-122 | actual PDF page(s): 126-127
  • Exam focus: Define diffraction as bending of a wave as it passes the edge of a barrier and explain diffraction of light.

LO18 - Single-slit diffraction calculations

  • Book page(s): 123-124, 129 | actual PDF page(s): 128-129, 134
  • Exam focus: Apply 2x1=2λ L/w to solve problems on single-slit diffraction.

LO19 - Double-slit vs single-slit comparison

  • Book page(s): 121-122 | actual PDF page(s): 126-127
  • Exam focus: Compare Young's Double Slit investigation with Single Slit Diffraction regarding spacing, source, width, and intensity.

LO20 - Diffraction grating definitions

  • Book page(s): 124-126 | actual PDF page(s): 129-131
  • Exam focus: Define a diffraction grating, reflection grating, and grating spectroscope.

LO21 - Diffraction grating equation

  • Book page(s): 126 | actual PDF page(s): 131
  • Exam focus: Explain constructive interference from a diffraction grating using mλ =d sinθ, where m=1,2,3...

LO22 - Grating spectroscope applications

  • Book page(s): 126 | actual PDF page(s): 131
  • Exam focus: Explain how a grating spectroscope works and give applications of diffraction gratings such as gemstone analysis.

LO23 - Frame of reference

  • Book page(s): 148 | actual PDF page(s): 153
  • Exam focus: Define a frame of reference.

LO24 - Relative velocity in one dimension

  • Book page(s): 148-149, 152 | actual PDF page(s): 153-154, 157
  • Exam focus: Calculate relative velocity using vector addition and subtraction in one dimension: va/b + vb/c = va/c.

LO25 - Relative velocity in two dimensions

  • Book page(s): 150-152 | actual PDF page(s): 155-157
  • Exam focus: Calculate relative velocity in two dimensions using vector addition/subtraction graphically and arithmetically.

Common Mistakes

  • Measuring angles from the surface instead of the normal.
  • Forgetting mm, cm, nm, and μm conversions.
  • Using n1θ1=n2θ2 instead of Snell's law with sine.
  • Forgetting frequency stays constant during refraction.
  • Applying total internal reflection from air to glass.
  • Using positive focal length for a concave lens.
  • Confusing double-slit separation d with single-slit width w.
  • Forgetting m=0 is the central maximum.
  • Ignoring direction signs in 1D relative velocity.
  • Giving a 2D vector answer without direction.

Grade 10 Advanced • pages 90

LO01: Refraction ray diagrams

Describe refraction of light (or a wave) as it crosses the boundary between two different mediums and represent that in a ray diagram.

10 MCQs • 4 Short Answers
LO01-MCQ01 MCQ easy ray diagram • 4 marks

A ray of light enters glass from air at an angle to the normal. Which feature must appear in the correct ray diagram?

Medium 1Medium 2normalir
Refraction Boundary
AThe ray must travel along the boundary.
BThe ray bends at the boundary and angles are measured from the normal.
CThe ray bends before it reaches the boundary.
DAngles are measured from the surface.
Answer
B. The ray bends at the boundary and angles are measured from the normal.
Refraction occurs at a boundary and angles are measured from the normal.
LO01-MCQ02 MCQ easy definition • 4 marks

What is refraction?

AChange in direction of a wave as it crosses into a different medium because its speed changes.
BBouncing of a wave from a surface.
CSpreading of a wave around a barrier.
DSeparation of white light only.
Answer
A. Change in direction of a wave as it crosses into a different medium because its speed changes.
Refraction is due to speed change at a boundary.
LO01-MCQ03 MCQ easy diagram • 4 marks

The normal in a refraction diagram is drawn

Medium 1Medium 2normalir
Refraction Boundary
Aparallel to the boundary.
Balong the incident ray.
Conly inside the denser medium.
Dperpendicular to the boundary at the point of incidence.
Answer
D. perpendicular to the boundary at the point of incidence.
The normal is an imaginary perpendicular line.
LO01-MCQ04 MCQ medium interpretation • 4 marks

A ray bends towards the normal when entering a new medium. The new medium is probably

Medium 1Medium 2normalir
Refraction Boundary
Aidentical to the first medium.
Ba perfect mirror.
Cmore optically dense.
Dless optically dense.
Answer
C. more optically dense.
Bending towards the normal indicates lower speed and higher optical density.
LO01-MCQ05 MCQ medium interpretation • 4 marks

A ray bends away from the normal after crossing a boundary. Its speed has

Astayed the same while direction changed.
Bincreased.
Cdecreased.
Dbecome zero.
Answer
B. increased.
Bending away from the normal indicates speed increases.
LO01-MCQ06 MCQ medium misconception • 4 marks

A student measures the angle of incidence from the surface. What is wrong?

AAngles of incidence and refraction are measured from the normal.
BOnly reflected angles use the normal.
CThe angle of incidence is always 90 degrees.
DThe surface is the correct reference line.
Answer
A. Angles of incidence and refraction are measured from the normal.
All optical angles in Snell/refraction diagrams are measured from the normal.
LO01-MCQ07 MCQ medium special case • 4 marks

If a light ray enters a new medium along the normal, its direction

Medium 1Medium 2normalir
Refraction Boundary
Abends towards the surface.
Bbends away from the normal.
Cmust totally internally reflect.
Ddoes not change.
Answer
D. does not change.
At normal incidence, direction is unchanged although speed may change.
LO01-MCQ08 MCQ easy application • 4 marks

Which pair of media can produce refraction at a boundary?

Asame glass and same glass
Bvacuum and vacuum
Cair and water
Dair and air
Answer
C. air and water
Refraction needs a change of medium or wave speed.
LO01-MCQ09 MCQ medium Refraction concepts • 4 marks

When light passes from air into glass at an angle, which statement is correct?

Medium 1Medium 2normalir
Refraction Boundary
AIt bends away from the normal because its speed increases.
BIt bends toward the normal because its speed decreases.
CIts frequency becomes zero.
DIts color must become red.
Answer
B. It bends toward the normal because its speed decreases.
Entering a medium with larger refractive index reduces light speed, so the ray bends toward the normal.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO01-MCQ10 MCQ medium Refraction concepts • 4 marks

A mirage is mainly caused by:

Medium 1Medium 2normalir
Refraction Boundary
Apolarization by sunglasses
Bdiffraction from a single slit
Cgradual refraction of light through layers of air at different temperatures
Dcomplete absorption of sunlight by sand
Answer
C. gradual refraction of light through layers of air at different temperatures
Warm and cool air layers have slightly different refractive indexes, bending light paths.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO01-SA01 Short Answer • 4 marks

Draw and label a ray diagram for light entering glass from air at an angle.

Medium 1Medium 2normalir
Refraction Boundary
Expected Answer

Ray bends towards normal.

Mark Scheme
  • 1 mark: boundary shown
  • 1 mark: normal drawn perpendicular
  • 1 mark: ray bends towards normal in glass
  • 1 mark: i and r labelled from normal
LO01-SA02 Short Answer • 3 marks

Explain why a ray changes direction during refraction.

Expected Answer

Speed change causes bending.

Mark Scheme
  • 1 mark: speed changes at boundary
  • 1 mark: one side of wavefront changes speed first
  • 1 mark: direction bends
LO01-SA03 Short Answer • 3 marks

Correct this: refraction is the same as reflection.

Expected Answer

They are different boundary behaviours.

Mark Scheme
  • 1 mark: reflection stays in original medium
  • 1 mark: refraction crosses boundary
  • 1 mark: refraction involves speed change
LO01-SA04 Short Answer • 3 marks

Describe what happens when light enters a new medium along the normal.

Expected Answer

Direction unchanged; speed/wavelength may change.

Mark Scheme
  • 1 mark: direction unchanged
  • 1 mark: speed may change
  • 1 mark: frequency unchanged

Grade 10 Advanced • pages 92-93

LO02: Wave quantities during refraction

Identify that during refraction, the wavelength and speed of light (or a wave) change but frequency remains the same.

10 MCQs • 4 Short Answers
LO02-MCQ01 MCQ easy concept • 4 marks

During refraction of light, which quantity remains the same?

Afrequency
Bspeed
Cwavelength
Ddirection always
Answer
A. frequency
Frequency is determined by the source.
LO02-MCQ02 MCQ medium wave equation • 4 marks

Light slows down when it enters glass from air. Its wavelength

Aincreases.
Bbecomes zero.
Cstays the same.
Ddecreases.
Answer
D. decreases.
v=fλ; if f is constant and v decreases, λ decreases.
LO02-MCQ03 MCQ medium interpretation • 4 marks

Light travels from glass to air. Which change is correct?

Afrequency increases and wavelength increases.
Bfrequency decreases and speed increases.
Cspeed increases and wavelength increases.
Dspeed decreases and wavelength decreases.
Answer
C. speed increases and wavelength increases.
Frequency stays constant; speed and wavelength increase together.
LO02-MCQ04 MCQ medium reasoning • 4 marks

Why does frequency stay constant at a boundary?

AIt becomes zero at the boundary.
BIt is fixed by the source producing the wave.
CIt is fixed by the material only.
DIt is always equal to the angle of refraction.
Answer
B. It is fixed by the source producing the wave.
The source controls the oscillation rate.
LO02-MCQ05 MCQ medium interpretation • 4 marks

A ray bends towards the normal. Which set of changes is most likely?

Medium 1Medium 2normalir
Refraction Boundary
Aspeed decreases, wavelength decreases, frequency unchanged
Bspeed increases, wavelength increases, frequency unchanged
Cspeed decreases, wavelength increases, frequency decreases
Dspeed unchanged, wavelength decreases, frequency increases
Answer
A. speed decreases, wavelength decreases, frequency unchanged
Towards the normal means lower speed; frequency unchanged, so wavelength decreases.
LO02-MCQ06 MCQ easy formula • 4 marks

Which formula links speed, frequency, and wavelength?

Af = vλ
Bλ = vf
Cv = f/λ
Dv = fλ
Answer
D. v = fλ
The wave equation is v=fλ.
LO02-MCQ07 MCQ medium calculation • 4 marks

A wave has frequency 5 × 1014 Hz and speed 2 × 108 m/s in a medium. What is its wavelength?

A2.5 × 106 m
B5 × 1014 m
C4 × 10-7 m
D1 × 1023 m
Answer
C. 4 × 10-7 m
Use λ=v/f.
LO02-MCQ08 MCQ medium calculation • 4 marks

A wave has frequency 6 × 1014 Hz and speed 2.2 × 108 m/s in a medium. What is its wavelength?

A6 × 1014 m
B3.8 × 10-7 m
C1.3 × 1023 m
D2.7 × 106 m
Answer
B. 3.8 × 10-7 m
Use λ=v/f.
LO02-MCQ09 MCQ medium Refraction concepts • 4 marks

Which quantity remains unchanged when light refracts from one medium into another?

Medium 1Medium 2normalir
Refraction Boundary
AWavelength
BFrequency
CSpeed
DDirection
Answer
B. Frequency
The source fixes the frequency. Speed and wavelength change together in the new medium.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO02-MCQ10 MCQ medium Wavelength during refraction • 4 marks

Light with wavelength 600 nm in air enters a medium with refractive index n = 1.50. Approximately what is its wavelength in the medium?

Medium 1Medium 2normalir
Refraction Boundary
A900 nm
B600 nm
C200 nm
D400 nm
Answer
D. 400 nm
Frequency stays constant and v decreases by n, so wavelength becomes λ/n = 600/1.50 = 400 nm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO02-SA01 Short Answer • 3 marks

A light wave of frequency 6.0 × 1014 Hz travels in glass at 2.0 × 108 m/s. Calculate wavelength.

Expected Answer

3.3 × 10-7 m

Mark Scheme
  • 1 mark: λ=v/f
  • 1 mark: substitution correct
  • 1 mark: 3.3 × 10-7 m
LO02-SA02 Short Answer • 3 marks

State what happens to speed, wavelength, and frequency from air to denser medium.

Expected Answer

Speed and wavelength decrease; frequency constant.

Mark Scheme
  • 1 mark: speed decreases
  • 1 mark: wavelength decreases
  • 1 mark: frequency unchanged
LO02-SA03 Short Answer • 3 marks

Use v=fλ to explain why wavelength changes but frequency does not.

Expected Answer

Frequency stays fixed; λ follows speed.

Mark Scheme
  • 1 mark: state v=fλ
  • 1 mark: frequency set by source
  • 1 mark: λ changes when v changes
LO02-SA04 Short Answer • 2 marks

Water waves have f=4.0 Hz and λ=0.75 m in shallow water. Find speed.

Expected Answer

3.0 m/s

Mark Scheme
  • 1 mark: v=fλ
  • 1 mark: v=3.0 m/s

Grade 10 Advanced • pages 91-93

LO03: Snell's law

State and apply Snell's law of refraction.

10 MCQs • 4 Short Answers
LO03-MCQ01 MCQ medium calculation • 4 marks

Light travels from n1=1.00 to n2=1.50 with angle of incidence 30°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A30.0°
B60.0°
C0.0°
D19.5°
Answer
D. 19.5°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ02 MCQ medium calculation • 4 marks

Light travels from n1=1.00 to n2=1.33 with angle of incidence 40°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A50.0°
B0.0°
C28.9°
D40.0°
Answer
C. 28.9°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ03 MCQ medium calculation • 4 marks

Light travels from n1=1.50 to n2=1.00 with angle of incidence 30°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A0.0°
B48.6°
C30.0°
D60.0°
Answer
B. 48.6°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ04 MCQ medium calculation • 4 marks

Light travels from n1=1.33 to n2=1.00 with angle of incidence 35°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A49.7°
B35.0°
C55.0°
D0.0°
Answer
A. 49.7°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ05 MCQ medium calculation • 4 marks

Light travels from n1=1.00 to n2=1.52 with angle of incidence 45°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A35.0°
B17.5°
C0.0°
D27.7°
Answer
D. 27.7°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ06 MCQ medium calculation • 4 marks

Light travels from n1=1.50 to n2=1.33 with angle of incidence 25°. Find the angle of refraction.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A65.0°
B0.0°
C28.5°
D25.0°
Answer
C. 28.5°
Use n1 sinθ1 = n2 sinθ2.
LO03-MCQ07 MCQ easy formula • 4 marks

Which equation is Snell's law?

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
An1θ1 = n2θ2
Bn1 cos θ1 = n2 cos θ2
Cn1/θ1 = n2/θ2
Dn1 sin θ1 = n2 sin θ2
Answer
D. n1 sin θ1 = n2 sin θ2
Snell's law uses sine of angles measured from the normal.
LO03-MCQ08 MCQ easy definition • 4 marks

In Snell's law, the angles are measured from the

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
Aincident ray.
Bscreen.
Cnormal.
Dsurface.
Answer
C. normal.
Optical angles are measured from the normal.
LO03-MCQ09 MCQ hard Apply Snell’s law • 4 marks

A ray travels from medium 1 (n = 1.00) into medium 2 (n = 1.50) with θ1 = 30°. What is θ2?

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
ANo refraction; TIR occurs
B12.5°
C27.5°
D19.5°
Answer
D. 19.5°
n1 sinθ1 = n2 sinθ2, so sinθ2 = (1.00 sin 30°)/1.50; θ2 = 19.5°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO03-MCQ10 MCQ hard Apply Snell’s law • 4 marks

A ray travels from medium 1 (n = 1.00) into medium 2 (n = 1.33) with θ1 = 45°. What is θ2?

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
A32.1°
BNo refraction; TIR occurs
C25.1°
D40.1°
Answer
A. 32.1°
n1 sinθ1 = n2 sinθ2, so sinθ2 = (1.00 sin 45°)/1.33; θ2 = 32.1°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO03-SA01 Short Answer • 4 marks

State Snell's law and define symbols.

n1n2θ1θ2n1 sin θ1 = n2 sin θ2
Snells Angles
Expected Answer

n1 sinθ1=n2 sinθ2

Mark Scheme
  • 1 mark: n1 sinθ1=n2 sinθ2
  • 1 mark: n values are refractive indices
  • 1 mark: θ1 incident angle from normal
  • 1 mark: θ2 refracted angle from normal
LO03-SA02 Short Answer • 4 marks

Light enters glass n=1.50 from air at i=35.0°. Calculate r.

Expected Answer

22.5°

Mark Scheme
  • 1 mark: use Snell's law
  • 1 mark: 1.00 sin35 =1.50 sin r
  • 1 mark: sin r=0.382
  • 1 mark: r=22.5°
LO03-SA03 Short Answer • 3 marks

Explain why r in glass is smaller than i in air.

Expected Answer

Higher n bends ray towards normal.

Mark Scheme
  • 1 mark: glass has larger n
  • 1 mark: light slows down
  • 1 mark: bends towards normal
LO03-SA04 Short Answer • 4 marks

Light goes from water n=1.33 to air at i=30.0°. Calculate r.

Expected Answer

41.7°

Mark Scheme
  • 1 mark: use Snell's law
  • 1 mark: 1.33 sin30=sin r
  • 1 mark: sin r=0.665
  • 1 mark: r=41.7°

Grade 10 Advanced • pages 93, 96

LO04: Refractive index

Calculate the refractive index of a medium using n=c/v.

10 MCQs • 4 Short Answers
LO04-MCQ01 MCQ medium calculation • 4 marks

The speed of light in a material is 2.00 × 108 m/s. What is its refractive index?

A15.0
B5.0 × 108
C1.50
D0.67
Answer
C. 1.50
Use n=c/v.
LO04-MCQ02 MCQ medium calculation • 4 marks

The speed of light in a material is 2.25 × 108 m/s. What is its refractive index?

A5.2 × 108
B1.33
C0.75
D13.3
Answer
B. 1.33
Use n=c/v.
LO04-MCQ03 MCQ medium calculation • 4 marks

The speed of light in a material is 1.50 × 108 m/s. What is its refractive index?

A2.00
B0.50
C20.0
D4.5 × 108
Answer
A. 2.00
Use n=c/v.
LO04-MCQ04 MCQ medium calculation • 4 marks

The speed of light in a material is 1.24 × 108 m/s. What is its refractive index?

A0.41
B24.2
C4.2 × 108
D2.42
Answer
D. 2.42
Use n=c/v.
LO04-MCQ05 MCQ medium calculation • 4 marks

The speed of light in a material is 2.40 × 108 m/s. What is its refractive index?

A12.5
B5.4 × 108
C1.25
D0.80
Answer
C. 1.25
Use n=c/v.
LO04-MCQ06 MCQ medium calculation • 4 marks

The speed of light in a material is 1.80 × 108 m/s. What is its refractive index?

A4.8 × 108
B1.67
C0.60
D16.7
Answer
B. 1.67
Use n=c/v.
LO04-MCQ07 MCQ easy unit • 4 marks

Why has refractive index no unit?

ASpeed has no unit.
BIt is measured in degrees.
CIt is a ratio of two speeds.
DIt is always equal to zero.
Answer
C. It is a ratio of two speeds.
Units cancel in c/v.
LO04-MCQ08 MCQ medium calculation • 4 marks

A material has n=1.60. What is the speed of light in it?

A3.00 × 108 m/s
B1.88 × 108 m/s
C4.80 × 108 m/s
D1.60 × 108 m/s
Answer
B. 1.88 × 108 m/s
v=c/n=3.00 × 108/1.60.
LO04-MCQ09 MCQ easy Calculate refractive index • 4 marks

Light travels in a transparent medium at 2.00 × 10^8 m/s. What is the refractive index of the medium? Use c = 3.00 × 10^8 m/s.

A1.75
B0.67
C1.30
D1.50
Answer
D. 1.50
n = c/v = (3.00 × 10^8)/(2.00 × 10^8) = 1.50.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO04-MCQ10 MCQ easy Calculate refractive index • 4 marks

Light travels in a transparent medium at 2.25 × 10^8 m/s. What is the refractive index of the medium? Use c = 3.00 × 10^8 m/s.

A1.58
B1.33
C0.75
D1.13
Answer
B. 1.33
n = c/v = (3.00 × 10^8)/(2.25 × 10^8) = 1.33.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO04-SA01 Short Answer • 3 marks

Calculate n if light speed in medium is 2.05 × 108 m/s.

Expected Answer

1.46

Mark Scheme
  • 1 mark: n=c/v
  • 1 mark: substitution
  • 1 mark: n=1.46
LO04-SA02 Short Answer • 3 marks

A medium has n=1.75. Calculate light speed.

Expected Answer

1.71 × 108 m/s

Mark Scheme
  • 1 mark: v=c/n
  • 1 mark: substitution
  • 1 mark: 1.71 × 108 m/s
LO04-SA03 Short Answer • 3 marks

Explain why diamond n=2.42 bends light more than water n=1.33.

Expected Answer

Diamond is more optically dense.

Mark Scheme
  • 1 mark: larger n
  • 1 mark: lower speed
  • 1 mark: greater bending at boundary
LO04-SA04 Short Answer • 2 marks

State refractive index in terms of speed.

Expected Answer

n=c/v

Mark Scheme
  • 1 mark: ratio of speed in vacuum/air to speed in material
  • 1 mark: n=c/v

Grade 10 Advanced • pages 94

LO05: Critical angle

Define the critical angle as the angle of incidence at which the refracted light ray lies along the boundary of the two mediums.

10 MCQs • 4 Short Answers
LO05-MCQ01 MCQ easy definition • 4 marks

The critical angle is the angle of incidence for which the refracted ray

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Ahas no wavelength.
Btravels along the boundary.
Ctravels back along the incident path.
Dhas angle of refraction 0 degrees.
Answer
B. travels along the boundary.
At critical angle, r=90 degrees.
LO05-MCQ02 MCQ medium condition • 4 marks

Critical angle occurs when light travels from

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Amore optically dense to less optically dense medium.
Bless dense to more dense medium.
Cair to glass only.
Dvacuum to vacuum.
Answer
A. more optically dense to less optically dense medium.
Critical angle requires higher n to lower n.
LO05-MCQ03 MCQ easy concept • 4 marks

For i greater than the critical angle, the ray undergoes

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Aordinary refraction only.
Bdiffuse reflection only.
Cdispersion only.
Dtotal internal reflection.
Answer
D. total internal reflection.
TIR occurs for dense-to-less-dense and i>c.
LO05-MCQ04 MCQ easy definition • 4 marks

At the critical angle in glass-air, the refracted angle is

Aequal to the incident angle.
B45 degrees always.
C90 degrees.
D0 degrees.
Answer
C. 90 degrees.
The ray travels along the boundary.
LO05-MCQ05 MCQ medium definition • 4 marks

Which angle is the critical angle?

AAngle of reflection in air.
BAngle of incidence in the denser medium.
CAngle of refraction in the denser medium.
DAngle from the surface only.
Answer
B. Angle of incidence in the denser medium.
It is an incidence angle in the higher-n medium.
LO05-MCQ06 MCQ medium interpretation • 4 marks

A ray in water reaches water-air boundary at exactly c. It will

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Arefract along the boundary.
Bbend towards the normal into air.
Ctotally internally reflect only.
Dstop moving.
Answer
A. refract along the boundary.
At c, the refracted ray has r=90 degrees.
LO05-MCQ07 MCQ hard reasoning • 4 marks

Diamond-air has a smaller critical angle than glass-air because diamond has

Asmaller frequency.
Bzero light speed.
Cno reflection.
Dlarger refractive index.
Answer
D. larger refractive index.
sin c = n2/n1; larger n1 gives smaller c.
LO05-MCQ08 MCQ easy application • 4 marks

Which application uses total internal reflection?

Amagnetic induction
Bheating by friction
Coptical fibre communication
Dordinary shadow formation
Answer
C. optical fibre communication
Optical fibres guide light by TIR.
LO05-MCQ09 MCQ medium Refraction concepts • 4 marks

A smaller critical angle means that total internal reflection is:

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Aeasier to achieve because θi only needs to exceed a smaller angle
Bonly possible for red light
Cindependent of refractive index
Dimpossible
Answer
A. easier to achieve because θi only needs to exceed a smaller angle
Once θi is greater than θc, the ray reflects back into the higher-index medium.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO05-MCQ10 MCQ medium Calculate critical angle • 4 marks

Light travels from a medium with n = 1.50 into a medium with n = 1.00. What is the critical angle?

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
A51.8°
B41.8°
C48.2°
D31.8°
Answer
B. 41.8°
At the critical angle, θ2 = 90°, so sinθc = n2/n1 = 0.667; θc = 41.8°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO05-SA01 Short Answer • 2 marks

Define critical angle.

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Expected Answer

Incident angle giving r=90°

Mark Scheme
  • 1 mark: angle of incidence in denser medium
  • 1 mark: refracted ray is 90°/along boundary
LO05-SA02 Short Answer • 2 marks

State conditions for total internal reflection.

Expected Answer

Dense-to-less-dense and i>c.

Mark Scheme
  • 1 mark: higher n to lower n
  • 1 mark: i greater than critical angle
LO05-SA03 Short Answer • 3 marks

Why can TIR not occur from air to glass?

Expected Answer

TIR requires high n to low n.

Mark Scheme
  • 1 mark: air lower n
  • 1 mark: travels into higher n
  • 1 mark: does not bend away to r=90°
LO05-SA04 Short Answer • 2 marks

At critical angle in glass-air, describe refracted ray.

c normal less dense medium more dense medium refracted ray along boundary angle of refraction = 90° Critical Angle
Critical Angle
Expected Answer

Along the boundary.

Mark Scheme
  • 1 mark: travels along boundary
  • 1 mark: r=90°

Grade 10 Advanced • pages 92-94, 96

LO06: Refraction critical angle and TIR calculations

Apply the concepts of refraction, critical angle, and total internal reflection to solve numerical problems.

10 MCQs • 4 Short Answers
LO06-MCQ01 MCQ medium calculation • 4 marks

Light travels from water (n=1.33) to air (n=1.00). Calculate the critical angle.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A48.8°
B41.2°
C1.33°
D90.0°
Answer
A. 48.8°
Use sin c = n2/n1.
LO06-MCQ02 MCQ medium calculation • 4 marks

Light travels from glass (n=1.50) to air (n=1.00). Calculate the critical angle.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A48.2°
B1.50°
C90.0°
D41.8°
Answer
D. 41.8°
Use sin c = n2/n1.
LO06-MCQ03 MCQ medium calculation • 4 marks

Light travels from diamond (n=2.42) to air (n=1.00). Calculate the critical angle.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A2.42°
B90.0°
C24.4°
D65.6°
Answer
C. 24.4°
Use sin c = n2/n1.
LO06-MCQ04 MCQ medium calculation • 4 marks

Light travels from glass (n=1.50) to water (n=1.33). Calculate the critical angle.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A90.0°
B62.5°
C27.5°
D1.13°
Answer
B. 62.5°
Use sin c = n2/n1.
LO06-MCQ05 MCQ medium calculation • 4 marks

Light travels from flint glass (n=1.62) to air (n=1.00). Calculate the critical angle.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A38.1°
B51.9°
C1.62°
D90.0°
Answer
A. 38.1°
Use sin c = n2/n1.
LO06-MCQ06 MCQ medium application • 4 marks

Glass has critical angle 42°. A ray inside glass strikes the boundary at 50°. What happens?

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
Atotal internal reflection
Bordinary refraction only
Crefraction along boundary
Dbending towards normal in air
Answer
A. total internal reflection
Since i>c, TIR occurs.
LO06-MCQ07 MCQ medium application • 4 marks

Water-air critical angle is 48.8°. A ray in water strikes at 30°. What happens?

AIt totally internally reflects.
BIt travels along the boundary.
CIt stops at the surface.
DIt refracts out into air.
Answer
D. It refracts out into air.
Since i<c, refraction is possible.
LO06-MCQ08 MCQ medium comparison • 4 marks

For outside medium air, which material gives the smallest critical angle?

Aglass, n=1.50
Bice, n=1.31
Cdiamond, n=2.42
Dwater, n=1.33
Answer
C. diamond, n=2.42
Larger n gives smaller c.
LO06-MCQ09 MCQ medium Refraction concepts • 4 marks

Total internal reflection can occur only when light travels:

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
Afrom a higher-index medium to a lower-index medium and θi > θc
Bfrom air into glass at any angle
Cfrom a lower-index medium to a higher-index medium
Donly at normal incidence
Answer
A. from a higher-index medium to a lower-index medium and θi > θc
TIR needs n1 > n2 and an incidence angle greater than the critical angle.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO06-MCQ10 MCQ medium Calculate critical angle • 4 marks

Light travels from a medium with n = 1.33 into a medium with n = 1.00. What is the critical angle?

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
A38.8°
B41.2°
C48.8°
D58.8°
Answer
C. 48.8°
At the critical angle, θ2 = 90°, so sinθc = n2/n1 = 0.752; θc = 48.8°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO06-SA01 Short Answer • 4 marks

Calculate critical angle for glass n=1.50 to air.

Refraction, Critical Angle, and Total Internal Reflection Light travels from a more dense medium to a less dense medium. i < c The ray refracts out of the medium. Some bending occurs at the boundary. i = c The refracted ray travels along the boundary. The angle of refraction is 90°. i > c Total internal reflection occurs. No refracted ray leaves the medium. TIR condition: more dense to less dense medium, and i > c.
Tir Cases
Expected Answer

41.8°

Mark Scheme
  • 1 mark: sin c=n2/n1
  • 1 mark: sin c=1/1.50
  • 1 mark: sin c=0.667
  • 1 mark: c=41.8°
LO06-SA02 Short Answer • 3 marks

Water-air c=48.8°. A ray in water strikes at 55°. Explain.

Expected Answer

Total internal reflection.

Mark Scheme
  • 1 mark: from higher to lower n
  • 1 mark: 55°>48.8°
  • 1 mark: TIR occurs
LO06-SA03 Short Answer • 4 marks

Fibre core n=1.48, cladding n=1.44. Find c.

Expected Answer

76.7°

Mark Scheme
  • 1 mark: sin c=n2/n1
  • 1 mark: 1.44/1.48
  • 1 mark: 0.973
  • 1 mark: c≈76.7°
LO06-SA04 Short Answer • 3 marks

Why is TIR useful in optical fibres?

Expected Answer

It keeps signal in fibre.

Mark Scheme
  • 1 mark: light remains trapped
  • 1 mark: repeated TIR guides signal
  • 1 mark: low loss

Grade 10 Advanced • pages 98-99

LO07: Lens parts

Identify the principal axis, focal points, and focal length of convex or concave lenses.

10 MCQs • 4 Short Answers
LO07-MCQ01 MCQ easy labeling • 4 marks

The principal axis of a lens is

FFobjectimage
Convex Lens
Athe curved surface of the lens.
Bany line parallel to the object.
Cthe edge of the lens.
Dthe straight line through the optical centre and focal points.
Answer
D. the straight line through the optical centre and focal points.
Principal axis is the main reference line for lens diagrams.
LO07-MCQ02 MCQ easy definition • 4 marks

A convex lens is also called a

FFobjectimage
Convex Lens
Aplane mirror.
Breflection grating.
Cconverging lens.
Ddiverging lens.
Answer
C. converging lens.
Convex lenses converge parallel rays.
LO07-MCQ03 MCQ easy definition • 4 marks

A concave lens is also called a

FFobjectvirtual image
Concave Lens
Athin film.
Bdiverging lens.
Cconverging lens.
Dconvex mirror.
Answer
B. diverging lens.
Concave lenses diverge parallel rays.
LO07-MCQ04 MCQ easy definition • 4 marks

Focal length is the distance from optical centre to

Afocal point.
Bobject only.
Cscreen only.
Dedge of the lens.
Answer
A. focal point.
Focal length is measured to F along the principal axis.
LO07-MCQ05 MCQ medium ray rule • 4 marks

Parallel rays incident on a convex lens emerge

FFobjectimage
Convex Lens
Aparallel with no bending.
Bas if from the far focus.
Cback along the same path only.
Dthrough the focus on the far side.
Answer
D. through the focus on the far side.
Convex lenses focus parallel rays.
LO07-MCQ06 MCQ medium ray rule • 4 marks

Parallel rays incident on a concave lens emerge

FFobjectvirtual image
Concave Lens
Awithout refraction.
Bforming a real image at 2F.
Cdiverging as if from the focus on the incident side.
Dmeeting at the far focus.
Answer
C. diverging as if from the focus on the incident side.
Concave lenses make rays diverge.
LO07-MCQ07 MCQ easy identification • 4 marks

A lens thicker in the middle than at the edges is

Aopaque.
Bconvex.
Cconcave.
Dplane.
Answer
B. convex.
Convex lenses are thicker at the centre.
LO07-MCQ08 MCQ easy identification • 4 marks

A lens thinner in the middle than at the edges is

Aconcave.
Bconvex.
Cplane mirror.
Ddiffraction grating.
Answer
A. concave.
Concave lenses are thinner at the centre.
LO07-MCQ09 MCQ easy Lens concepts • 4 marks

A convex lens is also called a:

FFobjectimage
Convex Lens
Apolarizing filter
Bconverging lens
Cdiverging lens
Dplane mirror
Answer
B. converging lens
A convex lens makes parallel rays converge toward a focal point.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO07-MCQ10 MCQ medium Lens diagrams and aberrations • 4 marks

In a ray diagram for a convex lens, a ray parallel to the principal axis refracts:

FFobjectimage
Convex Lens
Aonly into a dark fringe
Bback through the source
Cparallel forever
Dthrough the far focal point
Answer
D. through the far focal point
This is one of the principal rays used to locate images.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO07-SA01 Short Answer • 4 marks

Label principal axis, optical centre, F and focal length for a convex lens.

FFobjectimage
Convex Lens
Expected Answer

Correct lens labels.

Mark Scheme
  • 1 mark: principal axis through centre
  • 1 mark: optical centre marked
  • 1 mark: F on both sides
  • 1 mark: f measured centre to F
LO07-SA02 Short Answer • 4 marks

Compare convex and concave lenses for parallel rays.

Expected Answer

Convex converges; concave diverges.

Mark Scheme
  • 1 mark: convex converges
  • 1 mark: convex focus far side
  • 1 mark: concave diverges
  • 1 mark: concave appears from near focus
LO07-SA03 Short Answer • 2 marks

Define focal length.

Expected Answer

Distance centre to focus.

Mark Scheme
  • 1 mark: distance from optical centre to F
  • 1 mark: along principal axis
LO07-SA04 Short Answer • 2 marks

Explain error if F is drawn above principal axis.

Expected Answer

F belongs on the principal axis.

Mark Scheme
  • 1 mark: F lies on principal axis
  • 1 mark: at focal distance from centre

Grade 10 Advanced • pages 98-99, 101

LO08: Convex lens ray diagrams

Draw a ray diagram to find the image of an object at different distances from a convex lens and determine image location and properties.

10 MCQs • 4 Short Answers
LO08-MCQ01 MCQ medium image properties • 4 marks

For a convex lens, an object is placed beyond 2F. What image is formed?

FFobjectimage
Convex Lens
Areal, upright, enlarged
Bno refraction occurs
Creal, inverted, diminished, between F and 2F
Dvirtual, upright, reduced
Answer
C. real, inverted, diminished, between F and 2F
Use standard convex lens image-position rules.
LO08-MCQ02 MCQ medium image properties • 4 marks

For a convex lens, an object is placed at 2F. What image is formed?

FFobjectimage
Convex Lens
Ano refraction occurs
Breal, inverted, same size, at 2F
Cvirtual, upright, reduced
Dreal, upright, enlarged
Answer
B. real, inverted, same size, at 2F
Use standard convex lens image-position rules.
LO08-MCQ03 MCQ medium image properties • 4 marks

For a convex lens, an object is placed between F and 2F. What image is formed?

FFobjectimage
Convex Lens
Areal, inverted, enlarged, beyond 2F
Bvirtual, upright, reduced
Creal, upright, enlarged
Dno refraction occurs
Answer
A. real, inverted, enlarged, beyond 2F
Use standard convex lens image-position rules.
LO08-MCQ04 MCQ medium image properties • 4 marks

For a convex lens, an object is placed at F. What image is formed?

FFobjectimage
Convex Lens
Avirtual, upright, reduced
Breal, upright, enlarged
Cno refraction occurs
Dimage at infinity/no image on nearby screen
Answer
D. image at infinity/no image on nearby screen
Use standard convex lens image-position rules.
LO08-MCQ05 MCQ medium image properties • 4 marks

For a convex lens, an object is placed inside F. What image is formed?

FFobjectimage
Convex Lens
Areal, upright, enlarged
Bno refraction occurs
Cvirtual, upright, enlarged, same side as object
Dvirtual, upright, reduced
Answer
C. virtual, upright, enlarged, same side as object
Use standard convex lens image-position rules.
LO08-MCQ06 MCQ medium ray rule • 4 marks

Which ray rule for a convex lens is correct?

FFobjectimage
Convex Lens
AA ray through focus becomes vertical.
BAll rays pass through 2F before the lens.
CA ray parallel to the principal axis refracts through the far focus.
DA ray through the optical centre reflects back.
Answer
C. A ray parallel to the principal axis refracts through the far focus.
This is a standard ray construction rule.
LO08-MCQ07 MCQ easy concept • 4 marks

An image that can be projected on a screen is

Aalways same size.
Breal.
Cvirtual.
Dalways upright.
Answer
B. real.
Real rays meet at real images.
LO08-MCQ08 MCQ medium image property • 4 marks

For a convex lens, an object inside the focal length produces

FFobjectimage
Convex Lens
Aa virtual upright enlarged image.
Ba real inverted image.
Ca real reduced image.
Dan image at infinity.
Answer
A. a virtual upright enlarged image.
This is the magnifier condition.
LO08-MCQ09 MCQ medium Convex lens image properties • 4 marks

For a convex lens, an object placed greater than 2f forms an image that is:

FFobjectimage
Convex Lens
Avirtual, upright, reduced
Bno image at any distance
Creal, upright, enlarged
Dreal, inverted, reduced
Answer
D. real, inverted, reduced
Ray diagrams for a convex lens show that when the object is greater than 2f, the image is real, inverted, reduced.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO08-MCQ10 MCQ medium Convex lens image properties • 4 marks

For a convex lens, an object placed at 2f forms an image that is:

FFobjectimage
Convex Lens
Ano image at any distance
Breal, upright, enlarged
Creal, inverted, same size
Dvirtual, upright, reduced
Answer
C. real, inverted, same size
Ray diagrams for a convex lens show that when the object is at 2f, the image is real, inverted, same size.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO08-SA01 Short Answer • 5 marks

Draw convex lens image for object beyond 2F and state properties.

FFobjectimage
Convex Lens
Expected Answer

Real, inverted, diminished.

Mark Scheme
  • 1 mark: lens/F/2F shown
  • 1 mark: parallel ray through F
  • 1 mark: central ray straight
  • 1 mark: image between F and 2F
  • 1 mark: real inverted diminished
LO08-SA02 Short Answer • 3 marks

State image for object between F and 2F.

Expected Answer

Real inverted enlarged.

Mark Scheme
  • 1 mark: real
  • 1 mark: inverted
  • 1 mark: enlarged beyond 2F
LO08-SA03 Short Answer • 3 marks

Differentiate real and virtual image.

Expected Answer

Real can be projected; virtual cannot.

Mark Scheme
  • 1 mark: real rays meet
  • 1 mark: virtual rays appear to meet
  • 1 mark: real can be projected; virtual cannot
LO08-SA04 Short Answer • 3 marks

Object at F for convex lens. Describe image.

Expected Answer

Image at infinity.

Mark Scheme
  • 1 mark: emerging rays parallel
  • 1 mark: image at infinity
  • 1 mark: no nearby screen image

Grade 10 Advanced • pages 99

LO09: Convex lens magnifier

Describe how a convex lens acts as a magnifier.

10 MCQs • 4 Short Answers
LO09-MCQ01 MCQ easy condition • 4 marks

A convex lens acts as a magnifier when the object is

FFobjectimage
Convex Lens
Aexactly on the screen.
Binside the focal length.
Cat 2F.
Dbeyond 2F.
Answer
B. inside the focal length.
Magnifier use requires object distance less than f.
LO09-MCQ02 MCQ easy image • 4 marks

The image formed by a simple magnifying glass is usually

Avirtual, upright, enlarged.
Breal, inverted, reduced.
Creal, upright, same size.
Dvirtual, inverted, diminished.
Answer
A. virtual, upright, enlarged.
A magnifier produces a virtual upright enlarged image.
LO09-MCQ03 MCQ medium reasoning • 4 marks

Why cannot the usual magnified image be projected onto a screen?

AIt is too bright.
BIt has no frequency.
CIt is behind the screen only.
DIt is virtual; rays only appear to come from it.
Answer
D. It is virtual; rays only appear to come from it.
Virtual images cannot be caught on a screen.
LO09-MCQ04 MCQ medium application • 4 marks

A convex lens of focal length 10 cm is used as a magnifier. Which object distance works?

A20 cm
B30 cm
C6 cm
D10 cm
Answer
C. 6 cm
Object distance must be less than f.
LO09-MCQ05 MCQ easy application • 4 marks

Which lens is used as a simple magnifier?

Adiffraction grating only
Bconvex lens
Cconcave lens
Dopaque lens
Answer
B. convex lens
Convex lenses can produce enlarged virtual images.
LO09-MCQ06 MCQ hard ray reasoning • 4 marks

When a convex lens acts as a magnifier, rays after the lens are usually

Adiverging and traced backward by the eye.
Bparallel and never traced backward.
Cconverging to a screen only.
Dnot refracted.
Answer
A. diverging and traced backward by the eye.
The eye sees backward extensions as a virtual image.
LO09-MCQ07 MCQ medium application • 4 marks

If an object moves outside the focal length of a convex magnifying lens, the image can become

Aalways virtual upright.
Bno image ever.
Csame as a plane mirror.
Dreal and inverted.
Answer
D. real and inverted.
Outside f, a convex lens can form real inverted images.
LO09-MCQ08 MCQ medium concept • 4 marks

The magnifier increases apparent size mainly by producing

Azero frequency light.
Btotal internal reflection only.
Ca larger angular image at the eye.
Da larger mass.
Answer
C. a larger angular image at the eye.
Magnification increases angular size.
LO09-MCQ09 MCQ medium Convex lens image properties • 4 marks

For a convex lens, an object placed less than f forms an image that is:

FFobjectimage
Convex Lens
Avirtual, upright, reduced
Bvirtual, upright, enlarged
Cno image at any distance
Dreal, upright, enlarged
Answer
B. virtual, upright, enlarged
Ray diagrams for a convex lens show that when the object is less than f, the image is virtual, upright, enlarged.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO09-MCQ10 MCQ medium Mixed exam review • 4 marks

A student wants to magnify a small insect with a single lens. Which lens should be used?

FFobjectimage
Convex Lens
AConcave lens far from the insect
BDiffraction grating only
CPlane glass sheet
DConvex lens with the insect inside the focal length
Answer
D. Convex lens with the insect inside the focal length
A convex lens acts as a magnifier when the object is closer than the focal length.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO09-SA01 Short Answer • 4 marks

Describe how a convex lens acts as magnifier.

Expected Answer

Object inside f gives virtual upright enlarged.

Mark Scheme
  • 1 mark: convex lens used
  • 1 mark: object inside focal length
  • 1 mark: rays diverge/apparent image
  • 1 mark: virtual upright enlarged
LO09-SA02 Short Answer • 3 marks

f=8.0 cm, object at 5.0 cm. State image type.

Expected Answer

Virtual upright enlarged.

Mark Scheme
  • 1 mark: do<f
  • 1 mark: virtual
  • 1 mark: upright enlarged
LO09-SA03 Short Answer • 2 marks

Why is concave lens not a simple magnifier?

Expected Answer

It reduces real objects.

Mark Scheme
  • 1 mark: concave diverges
  • 1 mark: forms reduced virtual image
LO09-SA04 Short Answer • 2 marks

Give one use and object position for magnifier.

Expected Answer

Reading small print; object inside f.

Mark Scheme
  • 1 mark: valid use
  • 1 mark: object inside focal length

Grade 10 Advanced • pages 100-101, 106

LO10: Thin lens equation

Apply the thin lens equation to calculate image distance, object distance, or focal length using appropriate algebraic signs.

10 MCQs • 4 Short Answers
LO10-MCQ01 MCQ medium calculation • 4 marks

A convex lens has focal length 10 cm. An object is 30 cm from the lens. Find image distance.

A40.0 cm
B15.0 cm
C30.0 cm
D10.0 cm
Answer
B. 15.0 cm
Use 1/f=1/do+1/di.
LO10-MCQ02 MCQ medium calculation • 4 marks

A convex lens has focal length 12 cm. An object is 18 cm from the lens. Find image distance.

A36.0 cm
B18.0 cm
C12.0 cm
D30.0 cm
Answer
A. 36.0 cm
Use 1/f=1/do+1/di.
LO10-MCQ03 MCQ medium calculation • 4 marks

A convex lens has focal length 20 cm. An object is 60 cm from the lens. Find image distance.

A60.0 cm
B20.0 cm
C80.0 cm
D30.0 cm
Answer
D. 30.0 cm
Use 1/f=1/do+1/di.
LO10-MCQ04 MCQ medium calculation • 4 marks

A convex lens has focal length 15 cm. An object is 45 cm from the lens. Find image distance.

A15.0 cm
B60.0 cm
C22.5 cm
D45.0 cm
Answer
C. 22.5 cm
Use 1/f=1/do+1/di.
LO10-MCQ05 MCQ medium calculation • 4 marks

A convex lens has focal length 8 cm. An object is 24 cm from the lens. Find image distance.

A32.0 cm
B12.0 cm
C24.0 cm
D8.0 cm
Answer
B. 12.0 cm
Use 1/f=1/do+1/di.
LO10-MCQ06 MCQ medium calculation • 4 marks

A convex lens has focal length 10 cm. An object is 15 cm from the lens. Find image distance.

A30.0 cm
B15.0 cm
C10.0 cm
D25.0 cm
Answer
A. 30.0 cm
Use 1/f=1/do+1/di.
LO10-MCQ07 MCQ hard calculation • 4 marks

A concave lens has f=-20 cm and object distance 30 cm. What is image distance?

FFobjectvirtual image
Concave Lens
A+12.0 cm
B+60.0 cm
C-60.0 cm
D-12.0 cm
Answer
D. -12.0 cm
Use 1/f=1/do+1/di, so di is negative for a virtual image.
LO10-MCQ08 MCQ easy formula • 4 marks

The thin lens equation is

A1/f = do + di
B1/f = 1/do + 1/di
Cf = do + di
Df = do/di
Answer
B. 1/f = 1/do + 1/di
The equation uses reciprocals.
LO10-MCQ09 MCQ hard Thin lens equation • 4 marks

A convex lens has f = 10 cm. An object is placed 30 cm from the lens. What is the image distance?

FFobjectimage
Convex Lens
A-15.0 cm
B25.0 cm
C15.0 cm
D5.0 cm
Answer
C. 15.0 cm
Using 1/f = 1/do + 1/di: 1/di = 1/10 - 1/30; di = 15.0 cm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO10-MCQ10 MCQ hard Concave lens equation • 4 marks

A concave lens has f = -10 cm and an object distance do = 20 cm. Which result is correct?

FFobjectimage
Convex Lens
A6.7 cm, m = -0.33
B20.0 cm, m = 1.00
C-6.7 cm, m = 0.33
DNo image forms
Answer
C. -6.7 cm, m = 0.33
1/di = 1/f - 1/do = 1/(-10) - 1/(20); di = -6.7 cm and m = -di/do = 0.33. The negative di means a virtual image.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO10-SA01 Short Answer • 4 marks

Convex lens f=15 cm, object=45 cm. Find image distance.

Expected Answer

22.5 cm

Mark Scheme
  • 1 mark: 1/f=1/do+1/di
  • 1 mark: 1/15=1/45+1/di
  • 1 mark: 1/di=0.0444
  • 1 mark: di=22.5 cm
LO10-SA02 Short Answer • 4 marks

Concave lens f=-12 cm, object=24 cm. Find image distance.

Expected Answer

-8.0 cm

Mark Scheme
  • 1 mark: use thin lens equation
  • 1 mark: 1/-12=1/24+1/di
  • 1 mark: 1/di=-0.125
  • 1 mark: di=-8.0 cm
LO10-SA03 Short Answer • 2 marks

State sign convention for focal length.

Expected Answer

Convex +, concave -.

Mark Scheme
  • 1 mark: convex positive
  • 1 mark: concave negative
LO10-SA04 Short Answer • 4 marks

Real image at 30 cm, object at 20 cm. Find f.

Expected Answer

12 cm

Mark Scheme
  • 1 mark: 1/f=1/20+1/30
  • 1 mark: 1/f=5/60
  • 1 mark: f=12 cm
  • 1 mark: unit included

Grade 10 Advanced • pages 112-113

LO11: Coherent and incoherent light

Define coherent and incoherent light and explain how coherent light is generated by passing monochromatic light through slits.

10 MCQs • 4 Short Answers
LO11-MCQ01 MCQ easy definition • 4 marks

Coherent light sources have

Aconstant phase difference and same frequency.
Brandom phase differences.
Cno wavelength.
Dzero amplitude.
Answer
A. constant phase difference and same frequency.
Coherence means fixed phase relationship.
LO11-MCQ02 MCQ easy definition • 4 marks

Incoherent light sources have

Aconstant phase difference.
Bno energy.
Conly one wavelength exactly.
Drandom or changing phase difference.
Answer
D. random or changing phase difference.
Incoherent waves do not keep fixed phase.
LO11-MCQ03 MCQ medium explanation • 4 marks

Why are two slits illuminated by one monochromatic source coherent?

double slitscreenm=0m=1dL
Double Slit
AEach slit uses a separate battery.
BThe light stops at the slits.
CThe slits act as secondary sources with a fixed phase relationship.
DThe slits change light to white.
Answer
C. The slits act as secondary sources with a fixed phase relationship.
One source feeds both slits.
LO11-MCQ04 MCQ easy design • 4 marks

Which setup is best for stable interference?

double slitscreenm=0m=1dL
Double Slit
Atwo random torches
Bone laser and two narrow slits
Ctwo independent lamps of different colors
Done white lamp with no slit
Answer
B. one laser and two narrow slits
Laser light is coherent and monochromatic.
LO11-MCQ05 MCQ easy definition • 4 marks

Monochromatic light means light with

Aone wavelength/frequency.
Bzero wavelength.
Cmany unrelated colors.
Dchanging speed in vacuum.
Answer
A. one wavelength/frequency.
Monochromatic means single color/wavelength.
LO11-MCQ06 MCQ medium reasoning • 4 marks

If two sources are incoherent, the interference pattern is

Aperfectly fixed.
Bonly a central dark band.
Cidentical to a lens image.
Dunstable or washed out.
Answer
D. unstable or washed out.
Random phase destroys stable fringes.
LO11-MCQ07 MCQ easy concept • 4 marks

Coherence is most directly linked to

Amagnetic charge.
Bfocal length only.
Cphase relationship.
Dmass density.
Answer
C. phase relationship.
Interference requires stable phase.
LO11-MCQ08 MCQ medium concept • 4 marks

The double-slit sources are coherent because they are

double slitscreenm=0m=1dL
Double Slit
Aseparated by a lens.
Bderived from the same incoming wave.
Cdifferent colors.
Dproduced by the screen.
Answer
B. derived from the same incoming wave.
Common origin creates fixed phase relation.
LO11-MCQ09 MCQ easy Interference concepts • 4 marks

Coherent light sources have:

double slitscreenm=0m=1dL
Double Slit
Aconstant phase relationship and same frequency
Bzero wavelength
Cno amplitude
Drandom phases and many unrelated frequencies
Answer
A. constant phase relationship and same frequency
Stable interference patterns require waves with a constant phase relationship.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO11-MCQ10 MCQ easy Interference concepts • 4 marks

Monochromatic light means light with:

double slitscreenm=0m=1dL
Double Slit
Ano frequency
Bone wavelength or a very narrow range of wavelengths
Call possible wavelengths equally
Donly sound energy
Answer
B. one wavelength or a very narrow range of wavelengths
Mono means one; monochromatic light is useful for clear interference fringes.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO11-SA01 Short Answer • 4 marks

Define coherent and incoherent light.

Expected Answer

Coherent: fixed phase; incoherent: random.

Mark Scheme
  • 1 mark: coherent same frequency/wavelength
  • 1 mark: coherent constant phase difference
  • 1 mark: incoherent random phase
  • 1 mark: no stable fringes for incoherent
LO11-SA02 Short Answer • 3 marks

Explain coherent light in double slit.

double slitscreenm=0m=1dL
Double Slit
Expected Answer

One source feeds both slits.

Mark Scheme
  • 1 mark: one monochromatic source
  • 1 mark: passes through two slits
  • 1 mark: slits act as coherent secondary sources
LO11-SA03 Short Answer • 3 marks

Why are two classroom lamps unsuitable?

Expected Answer

They are incoherent.

Mark Scheme
  • 1 mark: independent sources
  • 1 mark: random phase difference
  • 1 mark: fringes unstable/wash out
LO11-SA04 Short Answer • 2 marks

State two properties of good double-slit light source.

Expected Answer

Monochromatic and coherent.

Mark Scheme
  • 1 mark: monochromatic
  • 1 mark: coherent

Grade 10 Advanced • pages 113-114

LO12: Double-slit bright and dark fringes

Explain how bright and dark interference fringes are created in a double-slit investigation with monochromatic light.

10 MCQs • 4 Short Answers
LO12-MCQ01 MCQ easy concept • 4 marks

A bright fringe forms where waves arrive

double slitscreenm=0m=1dL
Double Slit
Aout of phase and cancel.
Bwith zero amplitude at both slits.
Conly after reflection.
Din phase and interfere constructively.
Answer
D. in phase and interfere constructively.
Bright fringe = constructive interference.
LO12-MCQ02 MCQ medium path difference • 4 marks

A dark fringe forms when path difference is

double slitscreenm=0m=1dL
Double Slit
Azero only.
B2mλ only.
C(m + 1/2)λ.
Dmλ.
Answer
C. (m + 1/2)λ.
Half-integer wavelength gives destructive interference.
LO12-MCQ03 MCQ easy central • 4 marks

The central fringe is bright because

double slitscreenm=0m=1dL
Double Slit
Awavelength is zero.
Bpath difference is zero.
Cone slit is closed.
Dlight is incoherent.
Answer
B. path difference is zero.
Zero path difference means in phase.
LO12-MCQ04 MCQ easy superposition • 4 marks

Crest meeting crest gives

Aconstructive interference.
Bdestructive interference.
CTIR.
Drefraction only.
Answer
A. constructive interference.
In-phase waves add.
LO12-MCQ05 MCQ easy superposition • 4 marks

Crest meeting trough gives

Aconstructive interference.
Bmagnification.
Cdispersion.
Ddestructive interference.
Answer
D. destructive interference.
Out-of-phase waves cancel.
LO12-MCQ06 MCQ easy evidence • 4 marks

Double-slit fringes show that light behaves as

Aa static object.
Ba magnetic pole.
Ca wave.
Donly a particle with no wave properties.
Answer
C. a wave.
Interference is wave behavior.
LO12-MCQ07 MCQ medium proportionality • 4 marks

Increasing slit separation d makes fringe spacing

double slitscreenm=0m=1dL
Double Slit
Ainfinite.
Bsmaller.
Clarger.
Dunchanged.
Answer
B. smaller.
x=λL/d.
LO12-MCQ08 MCQ medium proportionality • 4 marks

Increasing screen distance L makes fringe spacing

double slitscreenm=0m=1dL
Double Slit
Alarger.
Bsmaller.
Czero.
Dunchanged.
Answer
A. larger.
x is directly proportional to L.
LO12-MCQ09 MCQ easy Interference concepts • 4 marks

Constructive interference occurs when two waves meet:

double slitscreenm=0m=1dL
Double Slit
Aafter being absorbed
Bin phase, crest with crest
Cexactly out of phase, crest with trough
Dwith zero frequency
Answer
B. in phase, crest with crest
In-phase waves add amplitudes and produce a bright fringe for light.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO12-MCQ10 MCQ easy Interference concepts • 4 marks

Destructive interference in a double-slit pattern corresponds to:

double slitscreenm=0m=1dL
Double Slit
Alarger wavelength
Bmaximum brightness
Cno wave behavior
Ddark fringes
Answer
D. dark fringes
At dark fringes, waves arrive out of phase and cancel partly or completely.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO12-SA01 Short Answer • 3 marks

Explain bright fringe creation.

double slitscreenm=0m=1dL
Double Slit
Expected Answer

Constructive interference.

Mark Scheme
  • 1 mark: waves overlap
  • 1 mark: in phase/path difference mλ
  • 1 mark: constructive interference
LO12-SA02 Short Answer • 3 marks

Explain dark fringe creation.

Expected Answer

Destructive interference.

Mark Scheme
  • 1 mark: waves overlap
  • 1 mark: out of phase/path difference (m+1/2)λ
  • 1 mark: destructive interference
LO12-SA03 Short Answer • 2 marks

Why is central band bright?

double slitscreenm=0m=1dL
Double Slit
Expected Answer

Zero path difference.

Mark Scheme
  • 1 mark: equal path lengths
  • 1 mark: zero path difference/in phase
LO12-SA04 Short Answer • 2 marks

Give two ways to increase fringe spacing.

Expected Answer

Increase λ/L or reduce d.

Mark Scheme
  • 1 mark: increase λ or L
  • 1 mark: decrease d

Grade 10 Advanced • pages 115

LO13: Double-slit wavelength equation

Apply λ =xd/L to calculate wavelength or an unknown distance in a double-slit investigation.

10 MCQs • 4 Short Answers
LO13-MCQ01 MCQ medium calculation • 4 marks

Fringe spacing x=3.2 mm, slit separation d=0.40 mm, screen distance L=2.0 m. Find wavelength.

double slitscreenm=0m=1dL
Double Slit
A6.40 × 10-4 m
B2.50 × 10-1 m
C6.40 × 10-7 m
D1.60 × 101 m
Answer
C. 6.40 × 10-7 m
Use λ =xd/L after converting mm to m.
LO13-MCQ02 MCQ medium calculation • 4 marks

Fringe spacing x=4.0 mm, slit separation d=0.25 mm, screen distance L=2.0 m. Find wavelength.

double slitscreenm=0m=1dL
Double Slit
A1.25 × 10-1 m
B5.00 × 10-7 m
C3.20 × 101 m
D5.00 × 10-4 m
Answer
B. 5.00 × 10-7 m
Use λ =xd/L after converting mm to m.
LO13-MCQ03 MCQ medium calculation • 4 marks

Fringe spacing x=2.5 mm, slit separation d=0.18 mm, screen distance L=1.5 m. Find wavelength.

double slitscreenm=0m=1dL
Double Slit
A3.00 × 10-7 m
B2.08 × 101 m
C3.00 × 10-4 m
D1.08 × 10-1 m
Answer
A. 3.00 × 10-7 m
Use λ =xd/L after converting mm to m.
LO13-MCQ04 MCQ medium calculation • 4 marks

Fringe spacing x=5.0 mm, slit separation d=0.30 mm, screen distance L=2.5 m. Find wavelength.

double slitscreenm=0m=1dL
Double Slit
A4.17 × 101 m
B6.00 × 10-4 m
C1.50 × 10-1 m
D6.00 × 10-7 m
Answer
D. 6.00 × 10-7 m
Use λ =xd/L after converting mm to m.
LO13-MCQ05 MCQ medium calculation • 4 marks

Fringe spacing x=1.8 mm, slit separation d=0.15 mm, screen distance L=0.9 m. Find wavelength.

double slitscreenm=0m=1dL
Double Slit
A3.00 × 10-4 m
B7.50 × 10-2 m
C3.00 × 10-7 m
D1.08 × 101 m
Answer
C. 3.00 × 10-7 m
Use λ =xd/L after converting mm to m.
LO13-MCQ06 MCQ easy formula • 4 marks

The double-slit equation for wavelength is

double slitscreenm=0m=1dL
Double Slit
Aλ = L/(xd)
Bλ = d/(xL)
Cλ = xd/L
Dλ = xL/d
Answer
C. λ = xd/L
This is the coverage relation.
LO13-MCQ07 MCQ easy symbol • 4 marks

In λ =xd/L, x represents

double slitscreenm=0m=1dL
Double Slit
Acritical angle.
Bfringe spacing on the screen.
Cslit width.
Dspeed of light.
Answer
B. fringe spacing on the screen.
x is spacing between adjacent bright or dark fringes.
LO13-MCQ08 MCQ easy symbol • 4 marks

In λ =xd/L, d represents

double slitscreenm=0m=1dL
Double Slit
Aseparation between the two slits.
Bscreen distance.
Ccentral maximum width.
Dfocal length.
Answer
A. separation between the two slits.
d is double-slit separation.
LO13-MCQ09 MCQ hard Find wavelength from double slit • 4 marks

The first bright fringe is x = 2.0 mm from the center. The slit separation is d = 0.25 mm and L = 1.0 m. What is the wavelength?

double slitscreenm=0m=1dL
Double Slit
A250 nm
B750 nm
C700 nm
D500 nm
Answer
D. 500 nm
For m = 1, λ = xd/L = 500 nm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO13-MCQ10 MCQ hard Find wavelength from double slit • 4 marks

The first bright fringe is x = 3.0 mm from the center. The slit separation is d = 0.20 mm and L = 1.2 m. What is the wavelength?

double slitscreenm=0m=1dL
Double Slit
A750 nm
B250 nm
C700 nm
D500 nm
Answer
D. 500 nm
For m = 1, λ = xd/L = 500 nm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO13-SA01 Short Answer • 4 marks

x=3.2 mm, d=0.40 mm, L=2.0 m. Find λ.

Expected Answer

6.4 × 10-7 m

Mark Scheme
  • 1 mark: λ =xd/L
  • 1 mark: convert x,d to m
  • 1 mark: substitute
  • 1 mark: λ=6.4 × 10-7 m
LO13-SA02 Short Answer • 2 marks

Rearrange λ =xd/L for d.

Expected Answer

d=λL/x

Mark Scheme
  • 1 mark: λL=xd
  • 1 mark: d=λL/x
LO13-SA03 Short Answer • 4 marks

λ=5.0 × 10-7 m, d=0.25 mm, L=1.5 m. Find x.

Expected Answer

3.0 mm

Mark Scheme
  • 1 mark: x=λL/d
  • 1 mark: d=2.5 × 10-4 m
  • 1 mark: substitute
  • 1 mark: x=3.0 mm
LO13-SA04 Short Answer • 2 marks

How reduce uncertainty in fringe spacing?

Expected Answer

Measure many fringes and average.

Mark Scheme
  • 1 mark: measure several fringes
  • 1 mark: divide by number of spacings

Grade 10 Advanced • pages 115

LO14: Constructive interference positions

Explain that constructive interference occurs at locations x_m such that mλ =xm d/L, where m=0,1,2...

10 MCQs • 4 Short Answers
LO14-MCQ01 MCQ easy condition • 4 marks

Constructive interference in double slit occurs when path difference is

double slitscreenm=0m=1dL
Double Slit
Aalways zero only.
Bmλ.
C(m+1/2)λ.
Dλ/4 only.
Answer
B. mλ.
Whole-number wavelengths add in phase.
LO14-MCQ02 MCQ easy order • 4 marks

The central bright band has order

double slitscreenm=0m=1dL
Double Slit
Am=0.
Bm=1.
Cm=2.
Dm=1/2.
Answer
A. m=0.
Central maximum has zero path difference.
LO14-MCQ03 MCQ easy order • 4 marks

The first bright fringe has order

Am=0.
Bm=1/2.
Cm=3/2.
Dm=1.
Answer
D. m=1.
First bright is first order.
LO14-MCQ04 MCQ medium formula • 4 marks

The equation for bright fringe position is

double slitscreenm=0m=1dL
Double Slit
Amλ = x_m L/d.
Bλ = Ld/x_m^2.
Cmλ = xm d/L.
Dmλ = L/(x_m d).
Answer
C. mλ = xm d/L.
This is the constructive interference relation.
LO14-MCQ05 MCQ medium calculation • 4 marks

If the second bright fringe is 6.0 mm from the centre, fringe spacing is

A2.0 mm.
B3.0 mm.
C6.0 mm.
D12 mm.
Answer
B. 3.0 mm.
x_m=mx, so x=6.0/2.
LO14-MCQ06 MCQ medium calculation • 4 marks

A pattern has fringe spacing 2.5 mm. The third bright fringe is

A7.5 mm from the centre.
B2.5 mm from the centre.
C5.0 mm from the centre.
D1.25 mm from the centre.
Answer
A. 7.5 mm from the centre.
x_m=mx=3×2.5 mm.
LO14-MCQ07 MCQ medium calculation • 4 marks

For m=2, λ=600 nm, d=0.30 mm, L=1.5 m. x_m is

A3.0 mm.
B0.60 mm.
C60 mm.
D6.0 mm.
Answer
D. 6.0 mm.
x_m=mλL/d.
LO14-MCQ08 MCQ easy pattern • 4 marks

Bright fringes on the two sides of the central maximum are

double slitscreenm=0m=1dL
Double Slit
Aonly above the slit.
Brandom.
Csymmetric.
Donly on the left.
Answer
C. symmetric.
Orders +m and -m are symmetric.
LO14-MCQ09 MCQ hard Double-slit calculation • 4 marks

In a double-slit experiment, λ = 500 nm, d = 0.25 mm, L = 1.0 m. What is the distance x1 from the central bright fringe to order m = 1?

double slitscreenm=0m=1dL
Double Slit
A2.00 mm
B4.00 mm
C1.00 mm
D5.00 mm
Answer
A. 2.00 mm
Use mλ = xm d/L, so x_m = mλL/d = 2.00 mm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO14-MCQ10 MCQ hard Double-slit calculation • 4 marks

In a double-slit experiment, λ = 650 nm, d = 0.50 mm, L = 2.0 m. What is the distance x2 from the central bright fringe to order m = 2?

double slitscreenm=0m=1dL
Double Slit
A5.20 mm
B8.20 mm
C2.60 mm
D10.40 mm
Answer
A. 5.20 mm
Use mλ = xm d/L, so x_m = mλL/d = 5.20 mm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO14-SA01 Short Answer • 3 marks

State constructive interference condition and define m.

Expected Answer

Path difference=mλ.

Mark Scheme
  • 1 mark: path difference integer wavelength
  • 1 mark: mλ =xm d/L
  • 1 mark: m=0,1,2... order
LO14-SA02 Short Answer • 2 marks

Fringe spacing 4.0 mm. Find fifth bright distance.

Expected Answer

20 mm

Mark Scheme
  • 1 mark: x_m=mx
  • 1 mark: x5=20 mm
LO14-SA03 Short Answer • 2 marks

Explain why central bright has m=0.

Expected Answer

Zero path difference.

Mark Scheme
  • 1 mark: equal distances from slits
  • 1 mark: path difference zero
LO14-SA04 Short Answer • 4 marks

Second bright at 5.4 mm, d=0.20 mm, L=1.8 m. Find λ.

Expected Answer

3.0 × 10-7 m

Mark Scheme
  • 1 mark: mλ =xm d/L
  • 1 mark: λ=x_m d/(mL)
  • 1 mark: substitute m=2
  • 1 mark: λ=3.0 × 10-7 m

Grade 10 Advanced • pages 116, 120

LO15: Interference calculations

Apply λ =xd/L and solve problems on interference of light.

10 MCQs • 4 Short Answers
LO15-MCQ01 MCQ medium calculation • 4 marks

A double-slit setup uses λ=630 nm, d=0.35 mm, and L=2.2 m. Find fringe spacing.

double slitscreenm=0m=1dL
Double Slit
A3.96 mm
B3960.00 μm
C0.35 mm
D555.56 m
Answer
A. 3.96 mm
Use x=λL/d.
LO15-MCQ02 MCQ medium calculation • 4 marks

A double-slit setup uses λ=560 nm, d=0.28 mm, and L=2.0 m. Find fringe spacing.

double slitscreenm=0m=1dL
Double Slit
A4000.00 μm
B0.28 mm
C500.00 m
D4.00 mm
Answer
D. 4.00 mm
Use x=λL/d.
LO15-MCQ03 MCQ medium calculation • 4 marks

A double-slit setup uses λ=650 nm, d=0.40 mm, and L=1.6 m. Find fringe spacing.

double slitscreenm=0m=1dL
Double Slit
A0.40 mm
B615.38 m
C2.60 mm
D2600.00 μm
Answer
C. 2.60 mm
Use x=λL/d.
LO15-MCQ04 MCQ medium calculation • 4 marks

A double-slit setup uses λ=480 nm, d=0.24 mm, and L=1.5 m. Find fringe spacing.

double slitscreenm=0m=1dL
Double Slit
A500.00 m
B3.00 mm
C3000.00 μm
D0.24 mm
Answer
B. 3.00 mm
Use x=λL/d.
LO15-MCQ05 MCQ medium calculation • 4 marks

A double-slit setup uses λ=600 nm, d=0.50 mm, and L=2.5 m. Find fringe spacing.

double slitscreenm=0m=1dL
Double Slit
A3.00 mm
B3000.00 μm
C0.50 mm
D833.33 m
Answer
A. 3.00 mm
Use x=λL/d.
LO15-MCQ06 MCQ easy proportionality • 4 marks

If screen distance L is doubled, fringe spacing

Adoubles.
Bhalves.
Cbecomes four times.
Dstays unchanged.
Answer
A. doubles.
x is directly proportional to L.
LO15-MCQ07 MCQ easy proportionality • 4 marks

If slit separation d is doubled, fringe spacing

Adoubles.
Bbecomes four times.
Cbecomes zero.
Dhalves.
Answer
D. halves.
x is inversely proportional to d.
LO15-MCQ08 MCQ medium comparison • 4 marks

Which color gives largest fringe spacing in same setup?

Ablue, λ=450 nm
Bviolet, λ=400 nm
Cred, λ=650 nm
Dgreen, λ=530 nm
Answer
C. red, λ=650 nm
Larger wavelength gives larger spacing.
LO15-MCQ09 MCQ hard Find wavelength from double slit • 4 marks

The first bright fringe is x = 1.8 mm from the center. The slit separation is d = 0.30 mm and L = 1.5 m. What is the wavelength?

double slitscreenm=0m=1dL
Double Slit
A560 nm
B360 nm
C540 nm
D180 nm
Answer
B. 360 nm
For m = 1, λ = xd/L = 360 nm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO15-MCQ10 MCQ hard Find wavelength from double slit • 4 marks

The first bright fringe is x = 4.0 mm from the center. The slit separation is d = 0.50 mm and L = 2.0 m. What is the wavelength?

double slitscreenm=0m=1dL
Double Slit
A1200 nm
B500 nm
C1000 nm
D1500 nm
Answer
C. 1000 nm
For m = 1, λ = xd/L = 1000 nm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO15-SA01 Short Answer • 4 marks

λ=630 nm, d=0.35 mm, L=2.2 m. Find x.

Expected Answer

4.0 mm

Mark Scheme
  • 1 mark: x=λL/d
  • 1 mark: convert units
  • 1 mark: substitute
  • 1 mark: x≈4.0 mm
LO15-SA02 Short Answer • 4 marks

x=2.8 mm, L=1.6 m, d=0.32 mm. Find λ.

Expected Answer

5.6 × 10-7 m

Mark Scheme
  • 1 mark: λ =xd/L
  • 1 mark: convert units
  • 1 mark: substitute
  • 1 mark: λ=5.6 × 10-7 m
LO15-SA03 Short Answer • 2 marks

Why measure across 10 fringes?

Expected Answer

Improves accuracy.

Mark Scheme
  • 1 mark: larger distance reduces percentage uncertainty
  • 1 mark: averaging reduces random error
LO15-SA04 Short Answer • 2 marks

State two ways to increase fringe spacing.

Expected Answer

Increase λ/L or decrease d.

Mark Scheme
  • 1 mark: increase λ or L
  • 1 mark: decrease d

Grade 10 Advanced • pages 117

LO16: Thin-film interference

Define and explain thin-film interference.

10 MCQs • 4 Short Answers
LO16-MCQ01 MCQ easy definition • 4 marks

Thin-film interference is caused by interference of light reflected from

top surface reflectionbottom surface reflection
Thin Film
Aa black surface only.
Ba thick opaque block.
Ca single ray with no reflection.
Dtop and bottom surfaces of a thin layer.
Answer
D. top and bottom surfaces of a thin layer.
Two reflected rays can interfere.
LO16-MCQ02 MCQ easy example • 4 marks

A common example of thin-film interference is

top surface reflectionbottom surface reflection
Thin Film
Aa single plane-mirror image.
Ba falling stone.
Ccolors on a soap bubble.
Da shadow behind a wall.
Answer
C. colors on a soap bubble.
Soap films show interference colors.
LO16-MCQ03 MCQ medium explanation • 4 marks

Soap-bubble colors change because

top surface reflectionbottom surface reflection
Thin Film
Aair has no refractive index.
Bfilm thickness changes.
Cgravity changes light speed to zero.
Dthe bubble emits only red light.
Answer
B. film thickness changes.
Different thickness gives different path difference.
LO16-MCQ04 MCQ easy evidence • 4 marks

Thin-film interference is evidence that light

Ahas wave behavior.
Bhas no frequency.
Ccannot reflect.
Dis always incoherent.
Answer
A. has wave behavior.
Interference is a wave effect.
LO16-MCQ05 MCQ medium reasoning • 4 marks

In thin films, path difference depends mainly on

Amass of observer.
Bscreen shape only.
Cgravitational field only.
Dfilm thickness and refractive index.
Answer
D. film thickness and refractive index.
Optical path length controls interference.
LO16-MCQ06 MCQ medium application • 4 marks

Oil films appear colored in white light because

top surface reflectionbottom surface reflection
Thin Film
Aonly red reflects.
Boil has no thickness.
Cdifferent wavelengths are reinforced at different thicknesses.
Dall wavelengths are absorbed.
Answer
C. different wavelengths are reinforced at different thicknesses.
Constructive interference can select wavelengths.
LO16-MCQ07 MCQ medium concept • 4 marks

Destructive interference in a thin film happens when reflected waves arrive

Awith no wavelength.
Bout of phase.
Ccrest with crest.
Dwith zero path difference always.
Answer
B. out of phase.
Out-of-phase waves cancel.
LO16-MCQ08 MCQ medium scale • 4 marks

Thin-film effects are strongest when thickness is comparable to

Athe wavelength of light.
Bthe distance to the Sun.
Cthe mass of the film.
DEarth's diameter.
Answer
A. the wavelength of light.
Interference requires path differences of order wavelength.
LO16-MCQ09 MCQ easy Interference concepts • 4 marks

Thin-film interference can produce colors because:

top surface reflectionbottom surface reflection
Thin Film
Alight stops at the surface
Bthe film creates sound waves
Cdifferent wavelengths interfere differently after reflections in the film
Dall wavelengths are absorbed equally
Answer
C. different wavelengths interfere differently after reflections in the film
Reflected waves from different film boundaries can add or cancel depending on wavelength, thickness, and phase inversion.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO16-MCQ10 MCQ medium Thin-film interference • 4 marks

A thin soap film shows bright coloured bands under white light. What is the best explanation?

top surface reflectionbottom surface reflection
Thin Film
ADifferent wavelengths undergo constructive and destructive interference after reflection from the film surfaces.
BThe film converts all colours into red light.
CThe film blocks diffraction completely.
DThe colours appear because light frequency becomes zero inside the film.
Answer
A. Different wavelengths undergo constructive and destructive interference after reflection from the film surfaces.
Thin-film colours are produced because different wavelengths interfere differently after reflections within the film.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO16-SA01 Short Answer • 2 marks

Define thin-film interference.

top surface reflectionbottom surface reflection
Thin Film
Expected Answer

Interference from surfaces of a thin film.

Mark Scheme
  • 1 mark: interference from waves reflected/transmitted at thin layer surfaces
  • 1 mark: due to path difference/phase changes
LO16-SA02 Short Answer • 4 marks

Why soap bubble shows colors?

Expected Answer

Different colors reinforced at different thicknesses.

Mark Scheme
  • 1 mark: white light has many wavelengths
  • 1 mark: reflection from top and bottom surfaces
  • 1 mark: thickness varies
  • 1 mark: different wavelengths interfere
LO16-SA03 Short Answer • 2 marks

Give two examples of thin-film interference.

Expected Answer

Soap bubble and oil film.

Mark Scheme
  • 1 mark: soap bubble/oil film/anti-reflection coating
  • 1 mark: second valid example
LO16-SA04 Short Answer • 3 marks

How does anti-reflection coating reduce glare?

Expected Answer

Destructive interference reduces reflection.

Mark Scheme
  • 1 mark: two reflected rays
  • 1 mark: coating thickness gives out-of-phase reflection
  • 1 mark: destructive interference

Grade 10 Advanced • pages 121-122

LO17: Diffraction of light

Define diffraction as bending of a wave as it passes the edge of a barrier and explain diffraction of light.

10 MCQs • 4 Short Answers
LO17-MCQ01 MCQ easy definition • 4 marks

Diffraction is

single slit width wcentral maximum
Single Slit
Achange of direction due to medium change.
Bsplitting white light only.
Cbending/spreading of a wave around an edge or through an aperture.
Dbouncing from a mirror.
Answer
C. bending/spreading of a wave around an edge or through an aperture.
Diffraction is wave spreading.
LO17-MCQ02 MCQ medium condition • 4 marks

Diffraction is strongest when gap size is

single slit width wcentral maximum
Single Slit
Aunrelated to wavelength.
Bsimilar to the wavelength.
Cmillions of times larger than wavelength.
Dzero always.
Answer
B. similar to the wavelength.
Comparable size gives strong spreading.
LO17-MCQ03 MCQ medium pattern • 4 marks

Single-slit diffraction produces

single slit width wcentral maximum
Single Slit
Aa broad central bright band with weaker side bands.
Btwo equal bright lines only.
Cno central maximum.
Da real image only.
Answer
A. a broad central bright band with weaker side bands.
The central maximum is wide and bright.
LO17-MCQ04 MCQ easy evidence • 4 marks

Light diffraction supports the idea that light

Ahas no speed.
Bhas only mass.
Chas fixed temperature.
Dhas wave properties.
Answer
D. has wave properties.
Diffraction is a wave phenomenon.
LO17-MCQ05 MCQ medium proportionality • 4 marks

If slit width decreases, diffraction pattern becomes

single slit width wcentral maximum
Single Slit
Aunchanged.
Bcompletely black.
Cmore spread out.
Dless spread out.
Answer
C. more spread out.
Narrower slit gives larger angular spread.
LO17-MCQ06 MCQ medium reasoning • 4 marks

Diffraction of light around a door is not obvious because

Adoors absorb all light.
Blight wavelength is much smaller than door width.
Clight cannot diffract.
Dsound has no wavelength.
Answer
B. light wavelength is much smaller than door width.
Large aperture compared with wavelength gives little spreading.
LO17-MCQ07 MCQ easy identification • 4 marks

A wave spreads into a shadow region after passing an edge. This is

Adiffraction.
Brefraction.
Ctotal internal reflection.
Dmagnification.
Answer
A. diffraction.
Spreading around edge is diffraction.
LO17-MCQ08 MCQ easy experiment • 4 marks

A good light diffraction demonstration uses

single slit width wcentral maximum
Single Slit
Alamp reflecting from mirror.
Blarge lens only.
Cair-water refraction only.
Dlaser through a narrow slit.
Answer
D. laser through a narrow slit.
Narrow slit produces visible pattern.
LO17-MCQ09 MCQ easy Diffraction concepts • 4 marks

Diffraction is best described as:

single slit width wcentral maximum
Single Slit
Areflection from a smooth surface only
Bbending/spreading of a wave as it passes an edge or aperture
Ccomplete absorption of a wave
Dconversion of light into sound
Answer
B. bending/spreading of a wave as it passes an edge or aperture
Diffraction shows the wave nature of light because waves spread after passing openings or edges.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO17-MCQ10 MCQ easy Diffraction concepts • 4 marks

According to Huygens’ principle, each point on a wavefront acts as:

single slit width wcentral maximum
Single Slit
Aa sound detector
Ba perfect absorber
Ca source of secondary wavelets
Da fixed mirror
Answer
C. a source of secondary wavelets
The new wavefront is built from the envelope of secondary wavelets.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO17-SA01 Short Answer • 2 marks

Define diffraction.

Expected Answer

Wave spreading around edges.

Mark Scheme
  • 1 mark: bending/spreading of wave
  • 1 mark: around edge/barrier or through gap
LO17-SA02 Short Answer • 3 marks

Why narrower slit gives wider pattern?

single slit width wcentral maximum
Single Slit
Expected Answer

Narrower slit -> wider spread.

Mark Scheme
  • 1 mark: diffraction increases as slit narrows
  • 1 mark: strong when width comparable to wavelength
  • 1 mark: larger spread angle
LO17-SA03 Short Answer • 3 marks

Describe single-slit diffraction pattern.

Expected Answer

Broad central maximum with weak sides.

Mark Scheme
  • 1 mark: central maximum widest/brightest
  • 1 mark: dark bands on sides
  • 1 mark: side maxima weaker
LO17-SA04 Short Answer • 2 marks

Why does diffraction support wave model?

Expected Answer

Light behaves as a wave.

Mark Scheme
  • 1 mark: diffraction is wave property
  • 1 mark: light spreads/interferes through narrow aperture

Grade 10 Advanced • pages 123-124, 129

LO18: Single-slit diffraction calculations

Apply 2x1=2λ L/w to solve problems on single-slit diffraction.

10 MCQs • 4 Short Answers
LO18-MCQ01 MCQ medium calculation • 4 marks

Light of λ=600 nm passes through a slit of width 0.30 mm. Screen distance is 2.0 m. Find x1.

single slit width wcentral maximum
Single Slit
A0.30 mm
B4.00 mm
C8.00 mm
D4000.00 μm
Answer
B. 4.00 mm
From 2x1=2λ L/w, x1=λL/w.
LO18-MCQ02 MCQ medium calculation • 4 marks

Light of λ=500 nm passes through a slit of width 0.25 mm. Screen distance is 1.5 m. Find x1.

single slit width wcentral maximum
Single Slit
A3.00 mm
B6.00 mm
C3000.00 μm
D0.25 mm
Answer
A. 3.00 mm
From 2x1=2λ L/w, x1=λL/w.
LO18-MCQ03 MCQ medium calculation • 4 marks

Light of λ=650 nm passes through a slit of width 0.40 mm. Screen distance is 1.8 m. Find x1.

single slit width wcentral maximum
Single Slit
A5.85 mm
B2925.00 μm
C0.40 mm
D2.93 mm
Answer
D. 2.93 mm
From 2x1=2λ L/w, x1=λL/w.
LO18-MCQ04 MCQ medium calculation • 4 marks

Light of λ=480 nm passes through a slit of width 0.60 mm. Screen distance is 2.5 m. Find x1.

single slit width wcentral maximum
Single Slit
A2000.00 μm
B0.60 mm
C2.00 mm
D4.00 mm
Answer
C. 2.00 mm
From 2x1=2λ L/w, x1=λL/w.
LO18-MCQ05 MCQ medium calculation • 4 marks

Light of λ=550 nm passes through a slit of width 0.22 mm. Screen distance is 1.2 m. Find x1.

single slit width wcentral maximum
Single Slit
A0.22 mm
B3.00 mm
C6.00 mm
D3000.00 μm
Answer
B. 3.00 mm
From 2x1=2λ L/w, x1=λL/w.
LO18-MCQ06 MCQ easy symbol • 4 marks

In 2x1=2λ L/w, 2x1 represents

single slit width wcentral maximum
Single Slit
Awavelength in glass.
Bwidth of the central maximum.
Cslit separation.
Dfocal length.
Answer
B. width of the central maximum.
x1 is centre to first dark; 2x1 is full central width.
LO18-MCQ07 MCQ medium proportionality • 4 marks

If slit width w is doubled, central maximum width

Ahalves.
Bdoubles.
Cquadruples.
Dstays unchanged.
Answer
A. halves.
Width is inversely proportional to w.
LO18-MCQ08 MCQ medium proportionality • 4 marks

If wavelength increases, the diffraction pattern becomes

Anarrower.
Bunchanged.
Czero width.
Dwider.
Answer
D. wider.
x1 is proportional to λ.
LO18-MCQ09 MCQ hard Single-slit central width • 4 marks

For single-slit diffraction, λ = 500 nm, L = 2.0 m, and slit width w = 100 μm. What is the width of the central bright band?

single slit width wcentral maximum
Single Slit
A40.0 mm
B10.0 mm
C2.0 mm
D20.0 mm
Answer
D. 20.0 mm
Use 2x1 = 2λ L/w = 20.0 mm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO18-MCQ10 MCQ hard Single-slit central width • 4 marks

For single-slit diffraction, λ = 600 nm, L = 1.5 m, and slit width w = 80 μm. What is the width of the central bright band?

single slit width wcentral maximum
Single Slit
A22.5 mm
B11.3 mm
C45.0 mm
D2.3 mm
Answer
A. 22.5 mm
Use 2x1 = 2λ L/w = 22.5 mm.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO18-SA01 Short Answer • 5 marks

λ=600 nm, w=0.30 mm, L=2.0 m. Find x1 and central width.

single slit width wcentral maximum
Single Slit
Expected Answer

x1=4.0 mm, width=8.0 mm

Mark Scheme
  • 1 mark: x1=λL/w
  • 1 mark: convert units
  • 1 mark: x1=4.0 mm
  • 1 mark: central width=2x1
  • 1 mark: 8.0 mm
LO18-SA02 Short Answer • 2 marks

Rearrange 2x1=2λ L/w for w.

Expected Answer

w=λL/x1

Mark Scheme
  • 1 mark: cancel 2 or x1=λL/w
  • 1 mark: w=λL/x1
LO18-SA03 Short Answer • 2 marks

What happens if slit becomes narrower?

Expected Answer

Pattern widens.

Mark Scheme
  • 1 mark: central maximum widens
  • 1 mark: width inversely proportional to w
LO18-SA04 Short Answer • 4 marks

Central width 10 mm, λ=5.0 × 10-7 m, L=1.5 m. Find w.

Expected Answer

1.5 × 10-4 m

Mark Scheme
  • 1 mark: x1=5.0 mm
  • 1 mark: w=λL/x1
  • 1 mark: substitute
  • 1 mark: w=1.5 × 10-4 m

Grade 10 Advanced • pages 121-122

LO19: Double-slit vs single-slit comparison

Compare Young's Double Slit investigation with Single Slit Diffraction regarding spacing, source, width, and intensity.

10 MCQs • 4 Short Answers
LO19-MCQ01 MCQ easy comparison • 4 marks

Compared with double-slit interference, single-slit diffraction has

single slit width wcentral maximum
Single Slit
Aa wider central bright band.
Bonly equal-width bands.
Cno dark bands.
Dno wave behavior.
Answer
A. a wider central bright band.
Single-slit central maximum is broad.
LO19-MCQ02 MCQ easy comparison • 4 marks

In ideal double-slit pattern, fringes are

double slitscreenm=0m=1dL
Double Slit
Awidest only at centre.
Brandomly spaced.
Call the same as side maxima.
Dnearly equally spaced.
Answer
D. nearly equally spaced.
Small-angle double-slit fringes have nearly uniform spacing.
LO19-MCQ03 MCQ easy pattern • 4 marks

In single-slit diffraction, the central maximum is

single slit width wcentral maximum
Single Slit
Asame width as all bands.
Babsent.
Cbrightest and widest.
Ddarkest.
Answer
C. brightest and widest.
Central band dominates.
LO19-MCQ04 MCQ easy source • 4 marks

Two coherent sources are used in

double slitscreenm=0m=1dL
Double Slit
Acritical angle measurement.
BYoung's double-slit experiment.
Csingle-slit diffraction only.
Dthin lens formation.
Answer
B. Young's double-slit experiment.
Double slit uses two coherent slits.
LO19-MCQ05 MCQ medium cause • 4 marks

Single-slit pattern comes from interference from

single slit width wcentral maximum
Single Slit
Amany points across one aperture.
Bone mirror surface only.
Cone boundary refraction.
Drelative velocity.
Answer
A. many points across one aperture.
Wavelets across the slit interfere.
LO19-MCQ06 MCQ medium similarity • 4 marks

Both double-slit and single-slit patterns have dark bands due to

Atotal internal reflection.
Bmagnification.
Cincoherence only.
Ddestructive interference.
Answer
D. destructive interference.
Dark bands arise from cancellation.
LO19-MCQ07 MCQ easy intensity • 4 marks

Single-slit side bright bands are

single slit width wcentral maximum
Single Slit
Aall equal.
Bnot present.
Cweaker than the central maximum.
Dbrighter than central maximum.
Answer
C. weaker than the central maximum.
Intensity decreases outward.
LO19-MCQ08 MCQ easy identification • 4 marks

A pattern with broad central bright region and weak side regions is likely

Alens magnification.
Bsingle-slit diffraction.
Cideal double-slit only.
DSnell refraction.
Answer
B. single-slit diffraction.
Broad central maximum is single-slit feature.
LO19-MCQ09 MCQ easy Diffraction concepts • 4 marks

A narrower single slit produces a central bright band that is:

double slitscreenm=0m=1dL
Double Slit
Anarrower
Bunchanged
Cwider
Dalways absent
Answer
C. wider
For single-slit diffraction, central width is proportional to 1/w.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO19-MCQ10 MCQ medium Mixed exam review • 4 marks

Why do dark fringes appear in a light interference pattern?

double slitscreenm=0m=1dL
Double Slit
ALight waves arrive out of phase and cancel.
BThe screen absorbs only blue light.
CLight stops moving.
DThe slits become lenses.
Answer
A. Light waves arrive out of phase and cancel.
Dark fringes are positions of destructive interference.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO19-SA01 Short Answer • 4 marks

Compare double-slit and single-slit spacing/source.

Expected Answer

Double: two slits equal spacing; single: one slit broad centre.

Mark Scheme
  • 1 mark: double slit uses two coherent slits
  • 1 mark: double slit nearly equal spacing
  • 1 mark: single slit one aperture
  • 1 mark: single slit broad central maximum
LO19-SA02 Short Answer • 3 marks

Compare intensities.

Expected Answer

Single-slit intensity falls outward.

Mark Scheme
  • 1 mark: double slit ideal fringes similar/modulated
  • 1 mark: single slit central brightest
  • 1 mark: side maxima weaker
LO19-SA03 Short Answer • 2 marks

Why broad central band suggests single slit?

single slit width wcentral maximum
Single Slit
Expected Answer

Broad centre is single-slit feature.

Mark Scheme
  • 1 mark: single slit has broad central maximum
  • 1 mark: double slit has evenly spaced fringes
LO19-SA04 Short Answer • 2 marks

State one similarity.

Expected Answer

Both show interference.

Mark Scheme
  • 1 mark: both are wave phenomena
  • 1 mark: both have constructive/destructive interference

Grade 10 Advanced • pages 124-126

LO20: Diffraction grating definitions

Define a diffraction grating, reflection grating, and grating spectroscope.

10 MCQs • 4 Short Answers
LO20-MCQ01 MCQ easy definition • 4 marks

A diffraction grating is

mλ = d sinθm=0m=1m=-1
Grating
Aa surface or plate with many equally spaced slits/lines.
Ba single convex lens.
Ca plane mirror only.
Da thick glass block.
Answer
A. a surface or plate with many equally spaced slits/lines.
A grating has many regular lines.
LO20-MCQ02 MCQ easy definition • 4 marks

A reflection grating forms spectra using

mλ = d sinθm=0m=1m=-1
Grating
Arefraction through one lens only.
BTIR in a fibre.
Ca single slit only.
Dreflection from many equally spaced grooves.
Answer
D. reflection from many equally spaced grooves.
Reflection gratings use grooved reflective surfaces.
LO20-MCQ03 MCQ easy application • 4 marks

A grating spectroscope is used to

mλ = d sinθm=0m=1m=-1
Grating
Adraw lens images.
Bstop diffraction.
Cseparate light into wavelengths and observe spectra.
Dmeasure mass only.
Answer
C. separate light into wavelengths and observe spectra.
Spectroscopes analyze spectra.
LO20-MCQ04 MCQ easy symbol • 4 marks

The spacing d of a grating is distance between

mλ = d sinθm=0m=1m=-1
Grating
Acentral bright and first dark in single slit.
Badjacent slits/lines.
Cscreen and grating.
Dobject and lens.
Answer
B. adjacent slits/lines.
d is line separation.
LO20-MCQ05 MCQ medium reasoning • 4 marks

A grating with more lines per mm has

Asmaller spacing d.
Blarger spacing d.
Cno diffraction.
Dzero wavelength.
Answer
A. smaller spacing d.
More lines in same length means smaller spacing.
LO20-MCQ06 MCQ hard explanation • 4 marks

Gratings produce sharp bright lines because

mλ = d sinθm=0m=1m=-1
Grating
Athey absorb all colors.
Bthey change light into sound.
Cthey use no interference.
Dmany waves interfere constructively at specific angles.
Answer
D. many waves interfere constructively at specific angles.
Many slits create narrow maxima.
LO20-MCQ07 MCQ easy application • 4 marks

Which device commonly uses a diffraction grating?

Abarometer
Bpendulum
Cspectroscope
Dspring balance
Answer
C. spectroscope
Spectroscopes use gratings.
LO20-MCQ08 MCQ easy application • 4 marks

Which everyday object can act as a reflection grating?

Arubber band
BCD or DVD surface
Cblack cloth
Dplain paper
Answer
B. CD or DVD surface
Close tracks diffract reflected light.
LO20-MCQ09 MCQ easy Diffraction concepts • 4 marks

A diffraction grating is useful because it:

mλ = d sinθm=0m=1m=-1
Grating
Aprevents refraction completely
Bconverts light to heat only
Cdestroys all interference
Dseparates wavelengths into sharp spectral lines
Answer
D. separates wavelengths into sharp spectral lines
Many equally spaced slits produce strong constructive interference at specific angles.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO20-MCQ10 MCQ easy Diffraction concepts • 4 marks

Optical discs can show rainbow colors because their tracks act like:

mλ = d sinθm=0m=1m=-1
Grating
Areflection diffraction gratings
Bconcave lenses
Cmagnetic poles
Dsound absorbers
Answer
A. reflection diffraction gratings
Closely spaced tracks reflect different wavelengths in different directions.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO20-SA01 Short Answer • 2 marks

Define diffraction grating.

mλ = d sinθm=0m=1m=-1
Grating
Expected Answer

Many equally spaced lines/slits.

Mark Scheme
  • 1 mark: many equally spaced slits/lines/grooves
  • 1 mark: diffracts light to form spectra/maxima
LO20-SA02 Short Answer • 2 marks

Define reflection grating and give example.

Expected Answer

A reflective grooved grating.

Mark Scheme
  • 1 mark: works by reflecting from grooves
  • 1 mark: CD/DVD or reflective spectrometer grating
LO20-SA03 Short Answer • 3 marks

Define grating spectroscope and purpose.

Expected Answer

Instrument for spectra.

Mark Scheme
  • 1 mark: instrument with grating
  • 1 mark: separates wavelengths/colors
  • 1 mark: observe/measure spectra
LO20-SA04 Short Answer • 3 marks

500 lines/mm. Calculate grating spacing.

Expected Answer

2.0 × 10-6 m

Mark Scheme
  • 1 mark: 500,000 lines/m
  • 1 mark: d=1/N
  • 1 mark: d=2.0 × 10-6 m

Grade 10 Advanced • pages 126

LO21: Diffraction grating equation

Explain constructive interference from a diffraction grating using mλ =d sinθ, where m=1,2,3...

10 MCQs • 4 Short Answers
LO21-MCQ01 MCQ medium calculation • 4 marks

A grating has d=1.0 × 10-6 m. Light wavelength is 500 nm at order m=1. Find θ.

mλ = d sinθm=0m=1m=-1
Grating
A60.0°
B28.6°
C5.0 × 10-13°
D30.0°
Answer
D. 30.0°
Use mλ =d sinθ.
LO21-MCQ02 MCQ medium calculation • 4 marks

A grating has d=2.0 × 10-6 m. Light wavelength is 600 nm at order m=1. Find θ.

mλ = d sinθm=0m=1m=-1
Grating
A17.2°
B1.2 × 10-12°
C17.5°
D72.5°
Answer
C. 17.5°
Use mλ =d sinθ.
LO21-MCQ03 MCQ medium calculation • 4 marks

A grating has d=1.5 × 10-6 m. Light wavelength is 450 nm at order m=2. Find θ.

mλ = d sinθm=0m=1m=-1
Grating
A1.3 × 10-12°
B36.9°
C53.1°
D34.4°
Answer
B. 36.9°
Use mλ =d sinθ.
LO21-MCQ04 MCQ medium calculation • 4 marks

A grating has d=2.5 × 10-6 m. Light wavelength is 650 nm at order m=1. Find θ.

mλ = d sinθm=0m=1m=-1
Grating
A15.1°
B74.9°
C14.9°
D1.6 × 10-12°
Answer
A. 15.1°
Use mλ =d sinθ.
LO21-MCQ05 MCQ medium calculation • 4 marks

A grating has d=2.0 × 10-6 m. Light wavelength is 520 nm at order m=2. Find θ.

mλ = d sinθm=0m=1m=-1
Grating
A58.7°
B29.8°
C2.1 × 10-12°
D31.3°
Answer
D. 31.3°
Use mλ =d sinθ.
LO21-MCQ06 MCQ easy formula • 4 marks

The grating equation for maxima is

mλ = d sinθm=0m=1m=-1
Grating
Amλ = d/sinθ
Bλ = md sinθ
Cm = λd sinθ
Dmλ = d sin θ
Answer
D. mλ = d sin θ
Constructive maxima satisfy mλ =d sinθ.
LO21-MCQ07 MCQ medium reasoning • 4 marks

Red light has larger diffraction angle than blue for same grating because red has

Ano frequency.
Bsmaller amplitude only.
Clonger wavelength.
Dshorter wavelength.
Answer
C. longer wavelength.
sinθ is proportional to λ.
LO21-MCQ08 MCQ hard order • 4 marks

Highest possible order when d=2.0 μm and λ=600 nm is

A1
B3
C2
D4
Answer
B. 3
m≤d/λ=3.33, so maximum integer is 3.
LO21-MCQ09 MCQ hard Diffraction grating equation • 4 marks

A diffraction grating has 500 lines/mm. For λ = 600 nm and order m = 1, what is the diffraction angle?

mλ = d sinθm=0m=1m=-1
Grating
ANo maximum
B9.5°
C17.5°
D27.5°
Answer
C. 17.5°
d = 1/(500 lines/mm) = 2.00 × 10-6 m. mλ = d sinθ, so θ = 17.5°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO21-MCQ10 MCQ hard Diffraction grating equation • 4 marks

A diffraction grating has 600 lines/mm. For λ = 500 nm and order m = 1, what is the diffraction angle?

mλ = d sinθm=0m=1m=-1
Grating
A9.5°
BNo maximum
C17.5°
D27.5°
Answer
C. 17.5°
d = 1/(600 lines/mm) = 1.67 × 10-6 m. mλ = d sinθ, so θ = 17.5°.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO21-SA01 Short Answer • 4 marks

State grating equation and symbols.

mλ = d sinθm=0m=1m=-1
Grating
Expected Answer

mλ =d sinθ

Mark Scheme
  • 1 mark: mλ =d sinθ
  • 1 mark: m order
  • 1 mark: λ wavelength
  • 1 mark: d spacing and θ angle
LO21-SA02 Short Answer • 4 marks

d=1.5 × 10-6 m, λ=500 nm. Find first-order angle.

Expected Answer

19.5°

Mark Scheme
  • 1 mark: mλ =d sinθ
  • 1 mark: sinθ=λ/d
  • 1 mark: sinθ=0.333
  • 1 mark: θ=19.5°
LO21-SA03 Short Answer • 3 marks

Why red and blue diffract at different angles?

Expected Answer

Red diffracts more.

Mark Scheme
  • 1 mark: different wavelengths
  • 1 mark: equation contains λ
  • 1 mark: larger λ gives larger θ
LO21-SA04 Short Answer • 3 marks

Find maximum order for λ=650 nm, d=2.0 × 10-6 m.

Expected Answer

m=3

Mark Scheme
  • 1 mark: mλ≤d
  • 1 mark: m≤3.08
  • 1 mark: highest integer m=3

Grade 10 Advanced • pages 126

LO22: Grating spectroscope applications

Explain how a grating spectroscope works and give applications of diffraction gratings such as gemstone analysis.

10 MCQs • 4 Short Answers
LO22-MCQ01 MCQ easy principle • 4 marks

A grating spectroscope works by

mλ = d sinθm=0m=1m=-1
Grating
Astopping all wavelengths except red.
Busing TIR only.
Cdiffracting different wavelengths to different angles.
Dfocusing all colors to one point.
Answer
C. diffracting different wavelengths to different angles.
Grating equation makes angle depend on wavelength.
LO22-MCQ02 MCQ easy purpose • 4 marks

In a spectroscope, the grating is used to

mλ = d sinθm=0m=1m=-1
Grating
Amake all wavelengths same direction.
Bseparate light into its spectrum.
Cincrease mass of light.
Dremove frequency.
Answer
B. separate light into its spectrum.
The grating disperses light.
LO22-MCQ03 MCQ medium application • 4 marks

Gemstones can be analyzed with a spectroscope because materials

Aabsorb or transmit characteristic wavelengths.
Bdo not interact with light.
Call have the same spectrum.
Dturn light into sound.
Answer
A. absorb or transmit characteristic wavelengths.
Spectra act as fingerprints.
LO22-MCQ04 MCQ easy spectra • 4 marks

An emission spectrum shows

Aone shadow only.
Ba virtual lens image.
Cno bright lines.
Dbright lines at specific wavelengths on dark background.
Answer
D. bright lines at specific wavelengths on dark background.
Emission spectra have bright lines.
LO22-MCQ05 MCQ medium spectra • 4 marks

An absorption spectrum shows

Aa virtual image.
Breflected angle equals incident angle.
Cdark lines missing from a continuous spectrum.
Donly one central maximum.
Answer
C. dark lines missing from a continuous spectrum.
Absorbed wavelengths appear as dark lines.
LO22-MCQ06 MCQ medium instrument • 4 marks

A spectroscope slit should be narrow to

Astop all light.
Bproduce sharp spectral lines.
Cremove diffraction.
Dmake light incoherent.
Answer
B. produce sharp spectral lines.
Narrow slit improves resolution.
LO22-MCQ07 MCQ easy application • 4 marks

A valid use of diffraction gratings is

Aidentifying elements from emission lines.
Bmeasuring spring constant only.
Cfinding car velocity only.
Dweighing gemstones directly.
Answer
A. identifying elements from emission lines.
Spectra identify elements/materials.
LO22-MCQ08 MCQ medium application • 4 marks

A gem spectroscope helps identify

Amass directly.
Bgravity only.
Cobject distance.
Dwavelengths absorbed or transmitted by the gemstone.
Answer
D. wavelengths absorbed or transmitted by the gemstone.
Gem absorption spectra reveal composition.
LO22-MCQ09 MCQ easy Diffraction concepts • 4 marks

A grating spectroscope is designed to:

mλ = d sinθm=0m=1m=-1
Grating
Ameasure mass directly
Bremove all colors from light
Cseparate light into component wavelengths for analysis
Dincrease the speed of sound
Answer
C. separate light into component wavelengths for analysis
Spectroscopes use dispersion/diffraction to analyze the wavelengths in light.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO22-MCQ10 MCQ easy Diffraction concepts • 4 marks

Optical discs can show rainbow colors because their tracks act like:

mλ = d sinθm=0m=1m=-1
Grating
Areflection diffraction gratings
Bconcave lenses
Cmagnetic poles
Dsound absorbers
Answer
A. reflection diffraction gratings
Closely spaced tracks reflect different wavelengths in different directions.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO22-SA01 Short Answer • 4 marks

Explain how grating spectroscope produces spectrum.

mλ = d sinθm=0m=1m=-1
Grating
Expected Answer

Grating separates wavelengths.

Mark Scheme
  • 1 mark: light through narrow slit/collimator
  • 1 mark: hits grating
  • 1 mark: different wavelengths construct at different angles
  • 1 mark: separate spectral lines observed
LO22-SA02 Short Answer • 3 marks

Describe use in gemstone analysis.

Expected Answer

Spectra identify material.

Mark Scheme
  • 1 mark: pass light through/from gemstone
  • 1 mark: observe absorption/transmission lines
  • 1 mark: compare with known spectra
LO22-SA03 Short Answer • 2 marks

Why different elements have different spectral lines?

Expected Answer

Characteristic energy levels.

Mark Scheme
  • 1 mark: different energy levels
  • 1 mark: emit/absorb specific wavelengths
LO22-SA04 Short Answer • 2 marks

State two advantages of grating for spectrum measurement.

Expected Answer

Sharp measurable lines.

Mark Scheme
  • 1 mark: sharp separated lines/high resolution
  • 1 mark: wavelength can be calculated using mλ =d sinθ

Grade 10 Advanced • pages 148

LO23: Frame of reference

Define a frame of reference.

10 MCQs • 4 Short Answers
LO23-MCQ01 MCQ easy definition • 4 marks

A frame of reference is

Aa grating spacing.
Ba coordinate system or viewpoint used to measure motion.
Ca type of lens.
Da material boundary.
Answer
B. a coordinate system or viewpoint used to measure motion.
Motion is measured relative to a frame.
LO23-MCQ02 MCQ easy reference • 4 marks

A passenger sitting in a moving bus is at rest relative to

Athe bus.
Ba tree beside road.
Cthe Sun only.
Da car moving opposite.
Answer
A. the bus.
In the bus frame, velocity is zero.
LO23-MCQ03 MCQ medium concept • 4 marks

The same object can have different velocities because

Aspeed has no units.
Bmass changes in Grade 10 problems.
Cdistance is always zero.
Dvelocity is relative to the observer/frame.
Answer
D. velocity is relative to the observer/frame.
Velocity depends on reference frame.
LO23-MCQ04 MCQ easy application • 4 marks

A car moves east at 20 m/s. A seated passenger's velocity relative to the car is

A20 m/s west.
B40 m/s east.
C0 m/s.
D20 m/s east.
Answer
C. 0 m/s.
Passenger is at rest in car frame.
LO23-MCQ05 MCQ easy definition • 4 marks

Relative velocity means

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
Afocal length divided by image distance.
Bvelocity of one object measured in another object's frame.
Cmass divided by time.
Dspeed of light in glass.
Answer
B. velocity of one object measured in another object's frame.
Relative velocity compares frames.
LO23-MCQ06 MCQ medium reference • 4 marks

A boat at rest relative to flowing water moves relative to the bank

Awith the water.
Bopposite the water always.
Cnot at all.
Dnorth only.
Answer
A. with the water.
It shares the water's ground velocity.
LO23-MCQ07 MCQ easy reasoning • 4 marks

Why state a reference frame?

AForces do not exist without it.
BWavelength becomes zero.
CThe calculator requires it.
DMeasured velocity depends on the frame chosen.
Answer
D. Measured velocity depends on the frame chosen.
Velocity statements are incomplete without a frame.
LO23-MCQ08 MCQ easy concept • 4 marks

Which can be a reference frame?

Aonly a stationary object.
Bonly the Sun.
Cground, moving car, or boat
Donly ground.
Answer
C. ground, moving car, or boat
Any specified observer/system can be a frame.
LO23-MCQ09 MCQ easy Relative velocity concepts • 4 marks

A frame of reference is:

Aa sound wave pattern
Bthe mass of an object
Ca type of lens
Da coordinate system or viewpoint used to describe motion
Answer
D. a coordinate system or viewpoint used to describe motion
Velocity is always described relative to a chosen reference frame.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO23-MCQ10 MCQ easy Relative velocity concepts • 4 marks

Relative velocity is useful because:

Ait applies only to light
Bthe observed velocity can change when the observer changes
Cit makes acceleration zero
Dit removes the need for vectors
Answer
B. the observed velocity can change when the observer changes
Different observers can measure different velocities for the same object.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO23-SA01 Short Answer • 2 marks

Define frame of reference.

Expected Answer

Viewpoint for measuring motion.

Mark Scheme
  • 1 mark: viewpoint/coordinate system/observer
  • 1 mark: used to measure position/velocity
LO23-SA02 Short Answer • 3 marks

Explain how passenger can be at rest and moving.

Expected Answer

Different frames give different velocities.

Mark Scheme
  • 1 mark: at rest relative bus
  • 1 mark: moving relative ground
  • 1 mark: motion depends on frame
LO23-SA03 Short Answer • 2 marks

Give two reference frames for boat on river.

Expected Answer

Bank and water frames.

Mark Scheme
  • 1 mark: bank/ground
  • 1 mark: water/current/boat
LO23-SA04 Short Answer • 2 marks

Statement 'ball has velocity 5 m/s'. What is missing?

Expected Answer

Velocity needs direction and frame.

Mark Scheme
  • 1 mark: direction
  • 1 mark: reference frame

Grade 10 Advanced • pages 148-149, 152

LO24: Relative velocity in one dimension

Calculate relative velocity using vector addition and subtraction in one dimension: va/b + vb/c = va/c.

10 MCQs • 4 Short Answers
LO24-MCQ01 MCQ medium calculation • 4 marks

Object A moves 20 m/s east; object B moves 12 m/s east. Taking east positive, find velocity of A relative to B.

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A8 m/s east
B32 m/s east
C8 m/s west
D-8 m/s east
Answer
A. 8 m/s east
Use v_A/B=v_A/G-v_B/G with signs.
LO24-MCQ02 MCQ medium calculation • 4 marks

Object A moves 20 m/s east; object B moves 12 m/s west. Taking east positive, find velocity of A relative to B.

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A32 m/s east
B32 m/s west
C8 m/s east
D8 m/s west
Answer
A. 32 m/s east
Use v_A/B=v_A/G-v_B/G with signs.
LO24-MCQ03 MCQ medium calculation • 4 marks

Object A moves 15 m/s west; object B moves 5 m/s west. Taking east positive, find velocity of A relative to B.

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A10 m/s west
B10 m/s east
C20 m/s west
D20 m/s east
Answer
A. 10 m/s west
Use v_A/B=v_A/G-v_B/G with signs.
LO24-MCQ04 MCQ medium calculation • 4 marks

Object A moves 15 m/s west; object B moves 5 m/s east. Taking east positive, find velocity of A relative to B.

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A20 m/s east
B20 m/s west
C10 m/s west
D10 m/s east
Answer
B. 20 m/s west
Use v_A/B=v_A/G-v_B/G with signs.
LO24-MCQ05 MCQ medium calculation • 4 marks

Object A moves 30 m/s east; object B moves 30 m/s east. Taking east positive, find velocity of A relative to B.

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A0 m/s
B60 m/s east
C0 m/s west
D0 m/s east
Answer
A. 0 m/s
Use v_A/B=v_A/G-v_B/G with signs.
LO24-MCQ06 MCQ easy formula • 4 marks

The one-dimensional relative velocity relation is

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
Ava/b + vb/c = va/c
Bv_a/b = v_a/c + v_b/c always
Cv_a/b = speed_a + speed_b only
Dv_a/c = v_a/b - v_b/c always
Answer
A. va/b + vb/c = va/c
Relative velocities add vectorially.
LO24-MCQ07 MCQ medium calculation • 4 marks

A train moves east at 25 m/s. Passenger walks west at 2 m/s relative to train. Passenger velocity relative ground is

A27 m/s east
B23 m/s west
C2 m/s west
D23 m/s east
Answer
D. 23 m/s east
v=25-2=23 m/s east.
LO24-MCQ08 MCQ easy calculation • 4 marks

Two cars move toward each other at 18 m/s and 22 m/s. Relative speed is

A22 m/s
B18 m/s
C40 m/s
D4 m/s
Answer
C. 40 m/s
Opposite directions add for relative speed.
LO24-MCQ09 MCQ easy 1D relative velocity • 4 marks

A student walks forward inside a bus at 1.5 m/s relative to the bus. The bus moves forward at 10.0 m/s relative to the ground. What is the student’s velocity relative to the ground?

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A11.5 m/s forward
B1.5 m/s forward
C11.5 m/s backward
D8.5 m/s forward
Answer
A. 11.5 m/s forward
Use va/g = va/b + vb/g = 1.5 + 10.0 = 11.5 m/s forward.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO24-MCQ10 MCQ easy 1D relative velocity • 4 marks

A student walks forward inside a bus at 2.0 m/s relative to the bus. The bus moves forward at 12.0 m/s relative to the ground. What is the student’s velocity relative to the ground?

Car ACar BvA/B = vA/G - vB/G
Relative Velocity 1D
A10.0 m/s forward
B2.0 m/s forward
C14.0 m/s forward
D14.0 m/s backward
Answer
C. 14.0 m/s forward
Use va/g = va/b + vb/g = 2.0 + 12.0 = 14.0 m/s forward.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO24-SA01 Short Answer • 3 marks

Car A east 28 m/s, B east 16 m/s. Find A relative B.

Expected Answer

12 m/s east

Mark Scheme
  • 1 mark: east positive
  • 1 mark: v=28-16
  • 1 mark: 12 m/s east
LO24-SA02 Short Answer • 3 marks

A east 20 m/s, B west 15 m/s. Find A relative B.

Expected Answer

35 m/s east

Mark Scheme
  • 1 mark: vA=+20, vB=-15
  • 1 mark: 20-(-15)
  • 1 mark: 35 m/s east
LO24-SA03 Short Answer • 3 marks

Passenger walks forward 1.5 m/s in train east 18 m/s. Find ground velocity.

Expected Answer

19.5 m/s east

Mark Scheme
  • 1 mark: add relative velocities
  • 1 mark: 1.5+18
  • 1 mark: 19.5 m/s east
LO24-SA04 Short Answer • 2 marks

Why are signs important in 1D relative velocity?

Expected Answer

Signs encode direction.

Mark Scheme
  • 1 mark: represent direction
  • 1 mark: opposite directions otherwise give wrong result

Grade 10 Advanced • pages 150-152

LO25: Relative velocity in two dimensions

Calculate relative velocity in two dimensions using vector addition/subtraction graphically and arithmetically.

10 MCQs • 4 Short Answers
LO25-MCQ01 MCQ medium calculation • 4 marks

Velocity components are 3 m/s east and 4 m/s north. Find resultant velocity.

east componentnorth componentresultant
Relative Velocity 2D
A7.0 m/s north of east
B1.0 m/s east
C5.0 m/s at 36.9° north of east
D5.0 m/s at 53.1° north of east
Answer
D. 5.0 m/s at 53.1° north of east
Use Pythagoras for magnitude and tanθ =vy/vx.
LO25-MCQ02 MCQ medium calculation • 4 marks

Velocity components are 6 m/s east and 8 m/s north. Find resultant velocity.

east componentnorth componentresultant
Relative Velocity 2D
A2.0 m/s east
B10.0 m/s at 36.9° north of east
C10.0 m/s at 53.1° north of east
D14.0 m/s north of east
Answer
C. 10.0 m/s at 53.1° north of east
Use Pythagoras for magnitude and tanθ =vy/vx.
LO25-MCQ03 MCQ medium calculation • 4 marks

Velocity components are 5 m/s east and 12 m/s north. Find resultant velocity.

east componentnorth componentresultant
Relative Velocity 2D
A13.0 m/s at 22.6° north of east
B13.0 m/s at 67.4° north of east
C17.0 m/s north of east
D7.0 m/s east
Answer
B. 13.0 m/s at 67.4° north of east
Use Pythagoras for magnitude and tanθ =vy/vx.
LO25-MCQ04 MCQ medium calculation • 4 marks

Velocity components are 9 m/s east and 12 m/s north. Find resultant velocity.

east componentnorth componentresultant
Relative Velocity 2D
A15.0 m/s at 53.1° north of east
B21.0 m/s north of east
C3.0 m/s east
D15.0 m/s at 36.9° north of east
Answer
A. 15.0 m/s at 53.1° north of east
Use Pythagoras for magnitude and tanθ =vy/vx.
LO25-MCQ05 MCQ medium calculation • 4 marks

Velocity components are 8 m/s east and 15 m/s north. Find resultant velocity.

east componentnorth componentresultant
Relative Velocity 2D
A23.0 m/s north of east
B7.0 m/s east
C17.0 m/s at 28.1° north of east
D17.0 m/s at 61.9° north of east
Answer
D. 17.0 m/s at 61.9° north of east
Use Pythagoras for magnitude and tanθ =vy/vx.
LO25-MCQ06 MCQ medium application • 4 marks

A plane flies east at 80 m/s in air; wind is 60 m/s north. Ground velocity is

east componentnorth componentresultant
Relative Velocity 2D
A140 m/s east
B20 m/s north
C100 m/s at 53.1° north of east
D100 m/s at 36.9° north of east
Answer
D. 100 m/s at 36.9° north of east
Vector sum: sqrt(80²+60²), tanθ=60/80.
LO25-MCQ07 MCQ medium application • 4 marks

A swimmer heads north at 1.2 m/s; river flows east at 0.9 m/s. Speed relative bank is

A0.3 m/s
B1.08 m/s
C1.5 m/s
D2.1 m/s
Answer
C. 1.5 m/s
sqrt(1.2²+0.9²)=1.5.
LO25-MCQ08 MCQ easy method • 4 marks

In 2D relative velocity, perpendicular components are combined using

east componentnorth componentresultant
Relative Velocity 2D
Athin lens equation.
BPythagoras and trigonometry.
Cordinary addition only.
DSnell's law.
Answer
B. Pythagoras and trigonometry.
Components form a right triangle.
LO25-MCQ09 MCQ hard 2D relative velocity • 4 marks

A boat moves north at 4.0 m/s relative to water. The river flows east at 3.0 m/s. What is the boat’s speed relative to the ground?

east componentnorth componentresultant
Relative Velocity 2D
A1.0 m/s
B4.0 m/s
C5.0 m/s
D7.0 m/s
Answer
C. 5.0 m/s
North and east components are perpendicular: v = √(4.0² + 3.0²) = 5.0 m/s.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO25-MCQ10 MCQ hard 2D relative velocity direction • 4 marks

For the same boat (4.0 m/s north, river 3.0 m/s east), what is the direction of the resultant velocity measured east of north?

east componentnorth componentresultant
Relative Velocity 2D
A51.9° east of north
B36.9° east of north
C53.1° east of north
D36.9° west of north
Answer
B. 36.9° east of north
tanθ = east/north = 3.0/4.0; θ = 36.9° east of north.
Source: Adapted from Abdul Elah Sheik Omar’s G 10 A Q-Bank
LO25-SA01 Short Answer • 5 marks

Plane 120 m/s east and wind 50 m/s north. Find ground velocity.

east componentnorth componentresultant
Relative Velocity 2D
Expected Answer

130 m/s at 22.6° N of E

Mark Scheme
  • 1 mark: components identified
  • 1 mark: sqrt(120²+50²)
  • 1 mark: 130 m/s
  • 1 mark: tanθ=50/120
  • 1 mark: 22.6° north of east
LO25-SA02 Short Answer • 5 marks

Swimmer north 1.6 m/s, current east 1.2 m/s. Find speed/direction.

Expected Answer

2.0 m/s at 36.9° E of N

Mark Scheme
  • 1 mark: components east/north
  • 1 mark: magnitude sqrt
  • 1 mark: 2.0 m/s
  • 1 mark: angle using tan
  • 1 mark: 36.9° east of north
LO25-SA03 Short Answer • 4 marks

Describe graphical 2D relative velocity method.

Expected Answer

Scale tip-to-tail vectors.

Mark Scheme
  • 1 mark: choose scale
  • 1 mark: draw first vector
  • 1 mark: draw second tip-to-tail
  • 1 mark: measure resultant
LO25-SA04 Short Answer • 4 marks

Drone 6 m/s east, wind 8 m/s south. Find ground velocity.

Expected Answer

10 m/s at 53.1° south of east

Mark Scheme
  • 1 mark: components 6 east 8 south
  • 1 mark: magnitude 10 m/s
  • 1 mark: angle 53.1°
  • 1 mark: south of east