Senior 4 practice book · Unit 1 of 10

Reflection and Refraction of Light

50 questions that complete the Senior 4 quiz for this unit: 29 core and 21 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

The image in a plane mirror is a real copy of the object standing behind the glass.A plane-mirror image is virtual: reflected rays only seem to come from behind the mirror, so it cannot be caught on a screen.
Light bends towards the normal because it speeds up in glass.Light slows down when it enters an optically denser medium such as glass; that drop in speed is what bends it towards the normal.
Total internal reflection can happen whenever light reaches a surface.It needs two things together: light travelling from the denser medium towards the less dense one, and an angle of incidence greater than the critical angle.
Covering half of a lens cuts off half of the image.Every part of a lens forms the whole image; covering half lets through less light, so the full image is just dimmer.
A prism adds colours to white light.White light already contains all the colours; the glass has a slightly different refractive index for each colour, so it only separates them.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Plane mirrors: image properties, lateral inversion, moving and rotating mirrors, mirror length
Images in inclined and parallel plane mirrors (n = 360°/θ − 1) and the mirror periscope
Curved mirrors: pole, centre of curvature, focus, f = R/2 and ray diagrams
Mirror formula and magnification for concave and convex mirrors
Uses of curved mirrors: reflectors, solar cookers, shaving and road-safety mirrors
Laws of refraction, Snell's law, refractive index and n = c/v
Real and apparent depth
Critical angle, total internal reflection and optical fibres
Other uses of total internal reflection: prism periscopes, binoculars, mirages, diamonds
Refraction through a triangular prism: deviation and minimum deviation
Dispersion of white light: the spectrum and the rainbow
Thin lenses: images, lens formula, magnification, power and lens combinations
Experiments: finding refractive index and focal length from measurements

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 21 harder ones are marked ★ Challenge.

1True or false
If a second identical prism is placed upside down just after the first, it can recombine the spectrum into white light.
Show answer
Answer: True
The second prism bends each colour back by the same amount, so the colours come together again as white light; this showed that the prism does not create the colours.
Common mistakeSome learners think a second prism would spread the colours even more; turned the other way up it undoes the dispersion.
2True or false · ★ Challenge
If the lower half of a converging lens is covered with black card, only the upper half of the image appears on the screen.
Show answer
Answer: False
Rays from every point of the object pass through the uncovered half and still meet at the image, so the whole image forms but it is dimmer.
Common mistakeThe tempting answer "true" pictures each half of the lens making half of the image; in fact each part of the lens forms the complete image.
3True or false
A convex mirror at a sharp bend on a mountain road lets drivers see a wider area of road than a plane mirror of the same size.
Show answer
Answer: True
A convex mirror spreads the reflected rays, so a small mirror shows a wide field of view; the price is a smaller image, so cars look further away than they are.
Common mistakeSome think a plane mirror shows more because its image is full size; the size of the image and the width of the view are different things.
4True or false · ★ Challenge
A cut diamond sparkles because its very small critical angle makes light that enters it undergo total internal reflection many times before it comes out.
Show answer
Answer: True
With n ≈ 2.42 the critical angle is only about 24°, so most rays inside meet the faces above it and are reflected again and again, leaving through the top in bright flashes.
Common mistakeSome learners think a diamond gives out its own light; it only redirects light that enters it, using total internal reflection.
5True or false
Total internal reflection sends back practically all of the light, whereas an ordinary silvered mirror absorbs a small part of it.
Show answer
Answer: True
In total internal reflection no light is refracted out, so almost 100% is reflected; a metal coating always absorbs a few per cent.
Common mistakeSome learners believe silvered mirrors are perfect reflectors; real metal coatings absorb some light and tarnish with time.
6True or false
The word AMBULANCE is printed back to front on the bonnet of an ambulance so that a driver in front reads it the right way round in a rear-view mirror.
Show answer
Answer: True
A plane mirror gives lateral inversion (left and right swapped), so a reversed word looks correct in the mirror.
Common mistakeSome learners think a mirror turns images upside down; a plane mirror swaps left and right only, the image stays upright.
7Multiple choice · ★ Challenge
A student 1.60 m tall, with eyes 1.50 m above the floor, wants to see her whole body in a vertical plane mirror on the dormitory wall. What is the shortest mirror she needs, and where must its top edge be?
1. A0.80 m long, top edge 1.55 m up
2. B1.60 m long, top edge 1.60 m up
3. C0.80 m long, top edge 1.60 m up
4. D0.75 m long, top edge 1.50 m up
Show answer
Answer: A. 0.80 m long, top edge 1.55 m up
The mirror only needs to cover half her height: 1.60 ÷ 2 = 0.80 m. Its top must be halfway between her eyes and the top of her head: (1.50 + 1.60) ÷ 2 = 1.55 m.
Common mistakeChoosing a 1.60 m mirror assumes the mirror must be as tall as the person; the rays from head and feet meet the mirror halfway up, so half the height is enough.
8Fill in the blank
A glass lens that is thicker at the centre than at the edges is a ______ lens.
Show answer
Answer: converging (convex)
A lens thicker in the middle bends parallel rays inwards to meet at a real focus, so it converges light.
Common mistakeLearners sometimes judge by the name "concave" or "convex" of one face only; it is the overall shape (thick or thin middle) that decides.
9Multiple choice · ★ Challenge
A barber sets up two plane mirrors so that a customer sees 7 images of a comb placed between them. What angle is there between the mirrors?
1. A51.4°
2. B40°
3. C60°
4. D45°
Show answer
Answer: D. 45°
n = 360°/θ − 1, so 7 = 360/θ − 1, 360/θ = 8 and θ = 360 ÷ 8 = 45°.
Common mistakeChoosing 51.4° comes from 360 ÷ 7: the number of images is one less than 360°/θ, so you must divide by 7 + 1 = 8.
10Fill in the blank
Whatever the object distance, a convex mirror always forms an image that is virtual, upright and ______.
Show answer
Answer: diminished (smaller than the object)
Rays reflected by a convex mirror always diverge, appearing to come from a point between the pole and F behind the mirror, so the image is always smaller.
Common mistakeSome learners expect a convex mirror to magnify like a convex lens; a convex mirror diverges light, so its image is always smaller.
11True or false
Two plane mirrors facing each other and exactly parallel would, in theory, produce an endless row of images of an object placed between them.
Show answer
Answer: True
For parallel mirrors θ = 0°, so 360°/θ is infinite: each image is reflected again in the other mirror, though in practice the images grow dimmer.
Common mistakeLearners often think only two images form, one in each mirror; each image acts as an object for the other mirror, again and again.
12Multiple choice · ★ Challenge
The refractive index of glass is 1.50 and that of water is 1.33. What is the refractive index for light travelling from water into glass (the refractive index of glass relative to water)?
1. A0.89
2. B2.00
3. C1.13
4. D2.83
Show answer
Answer: C. 1.13
n(water to glass) = n_glass ÷ n_water = 1.50 ÷ 1.33 = 1.13.
Common mistakeChoosing 0.89 divides the wrong way round (1.33 ÷ 1.50); light going into the denser glass slows down, so the relative index must be greater than 1.
13True or false
A ray of light entering a glass block along the normal does not change direction, but it still slows down inside the glass.
Show answer
Answer: True
At normal incidence i = 0°, so r = 0° and the ray goes straight on; the speed still drops from c to c/n inside the glass.
Common mistakeLearners often think no bending means nothing happens; the change of speed occurs whatever the angle, and bending is only seen for oblique rays.
14Multiple choice · ★ Challenge
An equilateral glass prism (angle A = 60°) gives a minimum deviation of 40° for yellow light. What is the refractive index of the glass for this light?
1. A0.77
2. B1.67
3. C1.53
4. D1.14
Show answer
Answer: C. 1.53
n = sin[(A + D)/2] ÷ sin(A/2) = sin 50° ÷ sin 30° = 0.766 ÷ 0.500 = 1.53.
Common mistakeChoosing 0.77 forgets to divide by sin(A/2) = 0.5; a refractive index below 1 should always tell you something was missed.
15Fill in the blank
When a beam of light passes from air into glass, its speed and wavelength decrease but its ______ stays the same.
Show answer
Answer: frequency
The frequency is fixed by the source; since v = fλ, a lower speed means a proportionally shorter wavelength.
Common mistakeThinking the colour changes in glass assumes the frequency changes; the frequency (and so the colour) stays the same in every medium.
16Multiple choice
An object stands between the centre of curvature C and the focus F of a concave mirror. Which description of the image is correct?
1. AReal, inverted, diminished, between C and F
2. BVirtual, upright, magnified, behind the mirror
3. CReal, inverted, magnified, beyond C
4. DReal, inverted, same size, at C
Show answer
Answer: C. Real, inverted, magnified, beyond C
For an object between C and F the reflected rays meet beyond C, giving a real, inverted, magnified image (the reverse of an object beyond C).
Common mistakeChoosing "diminished, between C and F" describes an object beyond C: the object and image positions swap, so between C and F the image is beyond C and larger.
17Short answer · ★ Challenge
On a hot afternoon a driver on the Kigali–Rwamagana road sees what looks like a pool of water on the road ahead, which disappears as she gets closer. Explain this mirage.
Show answer
Model answer: The air just above the hot tarmac is hotter and less dense, so its refractive index is lower than the air above it. Light from the sky heading down towards the road passes into layers of lower and lower refractive index and bends further from the normal, until its angle of incidence exceeds the critical angle and it is totally internally reflected upwards into the eye. The brain traces the ray back in a straight line and "sees" the sky on the road, which looks like water.
Common mistakeThe mistake is to say the road is wet or that the heat reflects light; the mirage is refraction plus total internal reflection in layers of air of different density.
18Multiple choice · ★ Challenge
A cheap prismatic periscope is made with plastic right-angled prisms of refractive index 1.35. Why does the image look very dim?
1. AIts critical angle is about 48°, more than the 45° angle of incidence
2. BIts critical angle is about 42°, less than the 45° angle of incidence
3. CLight travels too slowly in the plastic to reach the eye
4. DIts critical angle is about 48°, so the plastic absorbs the light
Show answer
Answer: A. Its critical angle is about 48°, more than the 45° angle of incidence
sin C = 1/1.35 = 0.741, so C = 47.8°. The ray meets the hypotenuse at 45° < 47.8°, so most light refracts out instead of being totally reflected.
Common mistakeChoosing 42° uses the critical angle of ordinary glass (n = 1.5); a lower refractive index gives a LARGER critical angle.
19Fill in the blank
Inside a prism of refracting angle A, the angle of refraction at the first face r₁ and the angle of incidence at the second face r₂ always add up to give r₁ + r₂ = ______.
Show answer
Answer: A (the refracting angle of the prism)
The two normals and the two faces form a quadrilateral, and geometry gives r₁ + r₂ = A for any ray through the prism.
Common mistakeA frequent slip is writing r₁ + r₂ = 90° or 180°; the angles inside always add up to the prism angle A.
20Multiple choice
A family in Bugesera cooks with a solar cooker made of a large concave mirror pointed at the Sun. Where should the cooking pot be placed?
1. AAt the centre of curvature, as the image is the same size
2. BAt the principal focus, where the Sun’s rays meet
3. CAt the pole, because the mirror itself gets hottest
4. DBetween F and the pole, where the image is magnified
Show answer
Answer: B. At the principal focus, where the Sun’s rays meet
Rays from the distant Sun arrive parallel to the axis and are reflected through F, so the energy is concentrated there.
Common mistakeChoosing the centre of curvature confuses the image rule for an object at C with what happens to parallel sunlight, which meets at F = R/2.
21Multiple choice · ★ Challenge
Light of frequency 5.0 × 10¹⁴ Hz passes from air into water (n = 4/3, c = 3.0 × 10⁸ m/s). Which pair of values is correct for the light in the water?
1. ASpeed 2.25 × 10⁸ m/s, frequency 3.75 × 10¹⁴ Hz
2. BSpeed 4.0 × 10⁸ m/s, frequency 5.0 × 10¹⁴ Hz
3. CSpeed 4.0 × 10⁸ m/s, frequency 6.7 × 10¹⁴ Hz
4. DSpeed 2.25 × 10⁸ m/s, frequency 5.0 × 10¹⁴ Hz
Show answer
Answer: D. Speed 2.25 × 10⁸ m/s, frequency 5.0 × 10¹⁴ Hz
v = c/n = 3.0 × 10⁸ ÷ (4/3) = 2.25 × 10⁸ m/s. The frequency is set by the source and does not change; only the speed and wavelength fall.
Common mistakeChoosing 3.75 × 10¹⁴ Hz divides the frequency by n as well; it is the wavelength (450 nm here), not the frequency, that shrinks by the factor n.
22Fill in the blank
When a ray passes through a prism at minimum deviation, it travels through the prism ______ to the base.
Show answer
Answer: parallel
At minimum deviation the ray passes symmetrically, so i = e, r₁ = r₂ = A/2, and inside the prism the ray is parallel to the base of an isosceles prism.
Common mistakeSome learners think minimum deviation means the ray goes straight through undeviated; the ray is still bent, just by the smallest possible angle.
23Multiple choice
Why is the bulb of a car headlamp placed at the principal focus of its concave reflector?
1. AThe light converges to one bright spot on the road
2. BThe image of the bulb is upright and magnified
3. CThe reflected light leaves as a parallel beam
4. DNo light is reflected back towards the bulb
Show answer
Answer: C. The reflected light leaves as a parallel beam
Rays from a source at F are reflected parallel to the principal axis, giving a strong beam that does not spread out and stays bright far down the road.
Common mistakeChoosing "converges to a spot" reverses the rule: parallel rays converge to F, but rays starting at F come out parallel.
24Multiple choice · ★ Challenge
A crown glass has n = 1.51 for red light and n = 1.53 for violet light. A narrow beam of white light enters a block of it at an angle of incidence of 50°. What angle separates the red and violet refracted rays inside the glass?
1. AAbout 0.02°
2. BAbout 0.44°
3. CExactly 0°
4. DAbout 4.5°
Show answer
Answer: B. About 0.44°
r(red) = sin⁻¹(sin 50° ÷ 1.51) = 30.48°; r(violet) = sin⁻¹(sin 50° ÷ 1.53) = 30.04°. The difference is about 0.44°.
Common mistakeChoosing 0.02° subtracts the refractive indices instead of the angles; dispersion is measured as the angle between the coloured rays.
25Multiple choice
When a narrow beam of white light passes through a glass prism, which colour is deviated most, and why?
1. ARed, because red light has the longest wavelength of all
2. BViolet, because violet light travels fastest inside glass
3. CRed, because glass has its largest refractive index for red
4. DViolet, because glass has its largest refractive index for violet
Show answer
Answer: D. Violet, because glass has its largest refractive index for violet
The refractive index of glass is slightly larger for violet than for red, so violet slows down most and is bent most.
Common mistakeChoosing "violet travels fastest" has the reason backwards: violet travels SLOWEST in glass, which is why it is bent the most.
26Multiple choice · ★ Challenge
A fisherman on Lake Kivu sees a fish that seems to be 0.90 m below the surface (n of water = 4/3). He thrusts a spear straight at where the fish appears to be. What happens, and what should he do?
1. AHe misses; the fish is really 0.68 m deep, so he should aim above it
2. BHe misses; the fish is really 1.2 m deep, so he should aim above it
3. CHe hits it, because refraction does not change the depth seen
4. DHe misses; the fish is really 1.2 m deep, so he should aim below it
Show answer
Answer: D. He misses; the fish is really 1.2 m deep, so he should aim below it
Real depth = n × apparent depth = 4/3 × 0.90 = 1.2 m. The fish is deeper than it looks, so he must aim below where he sees it.
Common mistakeChoosing 0.68 m divides by n instead of multiplying: the real depth is always the larger one, since the water makes things look shallower.
27Short answer
State two advantages of using right-angled glass prisms instead of plane mirrors in periscopes and binoculars.
Show answer
Model answer: (1) Total internal reflection reflects almost all of the light, so the image is brighter. (2) There is no silver coating that can tarnish or peel off. (3) A back-silvered mirror gives faint extra images from its front glass surface; a prism gives only one image.
Common mistakeSaying "prisms magnify the image" is wrong; reflecting prisms only turn the light, they do not change the size of the image.
28Short answer · ★ Challenge
Describe how a pin and the no-parallax method can be used to find the radius of curvature, and hence the focal length, of a concave mirror.
Show answer
Model answer: Clamp the mirror and hold a pin above the principal axis in front of it. Move the pin until its inverted image and the pin itself coincide with no parallax (they stay together when you move your head from side to side). The pin is then at the centre of curvature C, so the pin–mirror distance is R; then f = R/2.
Common mistakeA common slip is to give the measured distance as the focal length; the pin and its image coincide at C, which is 2f from the mirror.
29Short answer
Give one everyday use of (a) a plane mirror, (b) a concave mirror and (c) a convex mirror, and for each state the property of the image that makes it suitable.
Show answer
Model answer: (a) Plane mirror: a dressing mirror, because the image is upright and the same size. (b) Concave mirror: a shaving or make-up mirror, because a face inside F gives an upright, magnified image. (c) Convex mirror: a driving or road-bend mirror, because it gives an upright image and a wide field of view.
Common mistakeMany learners say a convex mirror is used because it magnifies; it is chosen for its wide field of view, and its image is in fact diminished.
30Multiple choice
Which of these is NOT a correct rule for drawing rays in a concave mirror ray diagram?
1. AA ray parallel to the axis is reflected through F
2. BA ray hitting the pole goes straight back along itself
3. CA ray passing through F is reflected parallel to the axis
4. DA ray passing through C is reflected back along itself
Show answer
Answer: B. A ray hitting the pole goes straight back along itself
At the pole the normal is the principal axis, so a ray hitting the pole reflects at an equal angle on the other side of the axis; only a ray through C returns along itself.
Common mistakeLearners mix up the pole and the centre of curvature: the ray through C meets the mirror along a radius (a normal), which is why it comes straight back.
31Short answer · ★ Challenge
In a school laboratory, a narrow beam of light is reflected from a small mirror fixed to a twisting wire onto a scale 2.0 m away. When the wire twists, the light spot moves 14 cm along the scale. Through what angle did the mirror turn?
Show answer
Model answer: The reflected ray turned through angle 2θ where tan 2θ = 0.14 ÷ 2.0 = 0.070, so 2θ = 4.0°. A mirror turned through θ turns the reflected ray through 2θ, so the mirror turned θ = 2.0°.
Common mistakeAnswering 4.0° forgets that the reflected ray turns through twice the angle of the mirror, so the mirror angle is half of the ray angle.
32Multiple choice
A right-angled isosceles glass prism (n = 1.5) turns a ray through 90° in a prismatic periscope. Why does the ray reflect at the long face (hypotenuse)?
1. AThe long face is coated with silver like a plane mirror
2. BIt meets that face at 90°, so it is reflected straight back
3. CIt meets that face at 45°, which is less than the critical angle
4. DIt meets that face at 45°, more than the critical angle of 41.8°
Show answer
Answer: D. It meets that face at 45°, more than the critical angle of 41.8°
The ray enters a short face normally, hits the hypotenuse at 45°, and since 45° > 41.8° it is totally internally reflected through 90°.
Common mistakeChoosing "silvered" misses the point of using a prism: no coating is needed because total internal reflection does the job.
33Multiple choice · ★ Challenge
A gemstone tester finds that light inside a transparent stone is just totally internally reflected when it meets the stone–air surface at 39°. What is the refractive index of the stone?
1. A1.29
2. B1.59
3. C0.63
4. D2.31
Show answer
Answer: B. 1.59
The critical angle is 39°, so n = 1/sin C = 1 ÷ sin 39° = 1 ÷ 0.629 = 1.59.
Common mistakeChoosing 1.29 uses cos 39° instead of sin 39°; the critical angle is measured from the normal, and sin C = 1/n.
34Multiple choice
The glass wall of an aquarium has n = 1.50 and the water inside has n = 1.33. What is the critical angle for light inside the glass meeting the glass–water boundary?
1. A41.8°
2. B62.5°
3. C48.8°
4. D27.5°
Show answer
Answer: B. 62.5°
sin C = n_water ÷ n_glass = 1.33 ÷ 1.50 = 0.887, so C = 62.5°.
Common mistakeChoosing 41.8° uses sin C = 1/1.50, which is the glass–AIR critical angle; with water on the other side, sin C = n₂/n₁.
35Multiple choice
In a test of a cooking oil, a ray in air meets the oil surface at an angle of incidence of 50° and is refracted at 35° to the normal. What is the refractive index of the oil?
1. A1.34
2. B1.43
3. C0.75
4. D1.27
Show answer
Answer: A. 1.34
n = sin i ÷ sin r = sin 50° ÷ sin 35° = 0.766 ÷ 0.574 = 1.34.
Common mistakeChoosing 1.43 divides the angles themselves (50 ÷ 35); Snell’s law uses the sines of the angles, not the angles.
36Short answer · ★ Challenge
A learner studies a car's convex side mirror of focal length 24 cm. For an object 72 cm away she writes 1/v = 1/24 − 1/72, gets v = 36 cm and concludes 'a real image 36 cm in front of the mirror'. Identify her error and give the correct image position.
Show answer
Model answer: She used f = +24 cm, but a convex mirror has a virtual focus, so f = −24 cm (real-is-positive). 1/v = −1/24 − 1/72 = −4/72, so v = −18 cm: a virtual, upright image 18 cm behind the mirror (diminished, m = 18/72 = 0.25).
Common mistakeThe error is the sign of f: a convex mirror can never form a real image of a real object, so a positive v for it is a warning sign.
37Short answer
After an afternoon shower in Musanze, a learner sees a rainbow in the east with the Sun low in the west behind her. Explain how the raindrops produce the colours.
Show answer
Model answer: Sunlight enters each raindrop and is refracted and dispersed, because water has a slightly different refractive index for each colour. The light is reflected at the back of the drop and refracted again as it leaves, which spreads the colours further. Each colour leaves at its own angle (about 42° for red, 40° for violet) to the incoming sunlight, so the observer sees a band of colours in the part of the sky opposite the Sun.
Common mistakeSaying the drops "paint" or "colour" the light is wrong: the colours were already in the white sunlight; the drops only separate them by dispersion.
38Multiple choice · ★ Challenge
A concave make-up mirror of focal length 20 cm shows a girl an upright image of her face twice the real size. How far is her face from the mirror? (Use 1/f = 1/u + 1/v, real-is-positive.)
1. A10 cm
2. B30 cm
3. C40 cm
4. D15 cm
Show answer
Answer: A. 10 cm
Upright means virtual, so v = −2u: 1/20 = 1/u − 1/(2u) = 1/(2u), giving 2u = 20 and u = 10 cm.
Common mistakeChoosing 30 cm comes from taking v = +2u, which is the REAL (inverted) image twice the size; an upright image is virtual, so v must be negative.
39Short answer
Explain why light stays inside an optical fibre when the fibre is bent gently around a corner of a building, but some light escapes if the fibre is bent very sharply.
Show answer
Model answer: In a gentle bend the light still meets the core–cladding boundary at angles of incidence greater than the critical angle, so it is totally internally reflected every time. In a very sharp bend the angle of incidence becomes smaller than the critical angle, so part of the light is refracted out into the cladding and lost.
Common mistakeSome learners think the cladding is a mirror that traps all light; nothing traps the light if the angle of incidence falls below the critical angle.
40Short answer · ★ Challenge
A learner measures the angle of deviation D of a ray through a glass prism for angles of incidence from 30° to 70°. Describe how D changes as i increases and sketch or describe the shape of the graph of D against i.
Show answer
Model answer: D first decreases as i increases, reaches a minimum value (the minimum deviation, when the ray passes symmetrically through the prism with i = e) and then increases again. The graph of D against i is a U-shaped curve with its lowest point at minimum deviation.
Common mistakeMany expect D to rise steadily with i; in fact it falls first, and only the minimum value is used to find n with the minimum-deviation formula.
41Short answer
Explain how a simple periscope made from a cardboard tube and two plane mirrors lets a learner see over a wall. State how the mirrors are set and describe the final image.
Show answer
Model answer: The two mirrors are parallel to each other, each at 45° to the tube. Light from the object strikes the top mirror at 45° and is turned through 90° down the tube; the lower mirror turns it through 90° again into the eye. The final image is virtual and upright (the same way up as the object), and the same size.
Common mistakeA common error is to set the mirrors at 90° to each other; they must be parallel, both at 45° to the tube, or the light leaves the tube the wrong way.
42Short answer · ★ Challenge
An illuminated slide and a screen are fixed 90 cm apart. A converging lens of focal length 20 cm is moved between them. Show that there are two lens positions that give a sharp image, and find them. (real-is-positive)
Show answer
Model answer: u + v = 90 and 1/u + 1/v = 1/20. Then (u + v)/(uv) = 1/20, so uv = 20 × 90 = 1800. u and v are the roots of x² − 90x + 1800 = 0, giving 30 cm and 60 cm. So the lens can be 30 cm from the slide (image 60 cm away, magnified ×2) or 60 cm from the slide (image 30 cm away, diminished ×0.5).
Common mistakeMany learners stop after one answer; because u and v can be swapped in the lens formula, every arrangement with the object–screen distance above 4f has two sharp positions.
43Multiple choice
A convex security mirror in a supermarket has a radius of curvature of 1.2 m. A shopper stands 3.0 m in front of it. Where is the image of the shopper? (real-is-positive)
1. A0.50 m behind the mirror
2. B0.75 m in front of the mirror
3. C0.86 m behind the mirror
4. D0.60 m behind the mirror
Show answer
Answer: A. 0.50 m behind the mirror
f = −R/2 = −0.60 m. 1/v = 1/f − 1/u = −1/0.6 − 1/3.0 = −2.0, so v = −0.50 m: a virtual image 0.50 m behind the mirror.
Common mistakeChoosing 0.75 m in front comes from taking f = +0.6 m; in the real-is-positive convention a convex mirror has a negative (virtual) focal length.
44Multiple choice
A diverging lens of focal length 15 cm is used to view a matchbox 30 cm away. Which description of the image is correct? (real-is-positive, f = −15 cm)
1. AReal, inverted, 30 cm from the lens, the same size
2. BVirtual, upright, 30 cm from the lens, the same size
3. CVirtual, upright, 10 cm from the lens, one-third size
4. DReal, inverted, 10 cm from the lens, one-third size
Show answer
Answer: C. Virtual, upright, 10 cm from the lens, one-third size
1/v = 1/f − 1/u = −1/15 − 1/30 = −3/30, so v = −10 cm (virtual, same side as the object); m = 10/30 = 1/3, upright.
Common mistakeChoosing "30 cm, same size" treats the lens as converging with the object at 2f; a diverging lens always gives a virtual, upright, diminished image.
45Short answer · ★ Challenge
A concave mirror throws a sharp real image of a candle flame onto a screen 60 cm from the mirror. The image is 4 times as tall as the flame. Find the distance of the candle from the mirror and the focal length of the mirror.
Show answer
Model answer: m = v/u, so u = v ÷ m = 60 ÷ 4 = 15 cm. 1/f = 1/u + 1/v = 1/15 + 1/60 = 5/60, so f = 12 cm.
Common mistakeWriting u = 60 × 4 = 240 cm multiplies instead of dividing; a magnified real image is FURTHER from the mirror than the object, so u is smaller than v.
46Multiple choice
Musa stands 1.5 m in front of a wall mirror in a Kigali hair salon. He then walks 0.5 m towards the mirror. How far is he now from his image?
1. A1.0 m
2. B3.0 m
3. C2.5 m
4. D2.0 m
Show answer
Answer: D. 2.0 m
He is now 1.5 − 0.5 = 1.0 m from the mirror, and the image is 1.0 m behind it, so the separation is 2 × 1.0 = 2.0 m.
Common mistakeChoosing 1.0 m gives his distance to the mirror, not to the image, which is as far behind the mirror as he is in front.
47Multiple choice
Two plane mirrors are hinged together at 60° to each other and a bottle top is placed between them. How many images of the bottle top are seen?
1. A6
2. B4
3. C5
4. D3
Show answer
Answer: C. 5
n = 360° ÷ θ − 1 = 360 ÷ 60 − 1 = 6 − 1 = 5 images.
Common mistakeChoosing 6 forgets to subtract 1: the formula 360°/θ counts the object itself as one of the "images".
48Multiple choice · ★ Challenge
A learner measures a pin under different depths of a sugar solution: real depth 6.0, 9.0, 12.0 cm; apparent depth 4.6, 6.9, 9.2 cm. Which result is the best value of the refractive index?
1. A1.30, the gradient of real depth against apparent depth
2. B0.77, the gradient of real depth against apparent depth
3. C2.80, the difference between the real and apparent depths
4. D1.50, the value of n for glass, as the solution is clear
Show answer
Answer: A. 1.30, the gradient of real depth against apparent depth
Gradient = (12.0 − 6.0) ÷ (9.2 − 4.6) = 6.0 ÷ 4.6 = 1.30, and n = real depth ÷ apparent depth is this gradient.
Common mistakeChoosing 0.77 takes apparent ÷ real; a refractive index is always greater than 1, so the real depth must be divided by the apparent depth.
49Short answer
Using a glass block and pins, a learner records these angles: i = 20°, 40°, 60°; r = 13°, 25°, 35°. Show that the results obey Snell’s law and give the refractive index of the glass.
Show answer
Model answer: sin i ÷ sin r: sin 20° ÷ sin 13° = 0.342 ÷ 0.225 = 1.52; sin 40° ÷ sin 25° = 0.643 ÷ 0.423 = 1.52; sin 60° ÷ sin 35° = 0.866 ÷ 0.574 = 1.51. The ratio is constant, as Snell’s law says, so n ≈ 1.52.
Common mistakeDividing i by r gives 1.54, 1.60, 1.71 (not constant), which wrongly suggests Snell’s law fails; the law is about the ratio of the sines.
50Short answer · ★ Challenge
A pin lies at the bottom of a tank filled to a depth of 20.0 cm with a clear liquid. Viewed from directly above, the pin seems to be only 15.0 cm below the surface. Find the refractive index of the liquid and the speed of light in it (c = 3.0 × 10⁸ m/s).
Show answer
Model answer: n = real depth ÷ apparent depth = 20.0 ÷ 15.0 = 1.33. v = c/n = 3.0 × 10⁸ ÷ 1.33 = 2.25 × 10⁸ m/s.
Common mistakeUsing n = 15 ÷ 20 = 0.75 gives a value below 1, which is impossible for a material; the real depth always goes on top.