Physics 2 Quiz: Lens And Mirror Equations

15 questions · exam conditions

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Lens And Mirror EquationsQuestion 1 of 15

A concave mirror of radius of curvature $R = 30 \text{ cm}$ forms an image of an object located $20 \text{ cm}$ in front of the mirror. A student claims: 'Because the object distance equals the focal length, the image is at infinity.' Which of the following correctly evaluates the student's claim and gives the actual image location?

The claim is incorrect. The focal length is

f = 15 \text{ cm}

, not 20 cm, so using the mirror equation gives a real image at

d_i = 60 \text{ cm}

in front of the mirror.

The claim is correct. The radius of curvature equals the focal length for a concave mirror, so with the object at 30 cm the image would be at infinity, but the object is at 20 cm so the image is virtual and behind the mirror at

d_i = -60 \text{ cm}

The claim is incorrect. The focal length is

f = 15 \text{ cm}

, and the image forms at

d_i = -60 \text{ cm}

, meaning it is virtual and located 60 cm behind the mirror.

The claim is incorrect. The focal length is

f = 15 \text{ cm}

, and with the object beyond the focal point the image is real. Applying the mirror equation yields

d_i = 12 \text{ cm}

in front of the mirror.

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Physics 2 Quiz

Physics 2 Quiz: Lens And Mirror Equations

Practice Lens And Mirror Equations in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Lens And Mirror Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The claim is incorrect. The focal length is $f = 15 \text{ cm}$ , not 20 cm, so using the mirror equation gives a real image at $d_i = 60 \text{ cm}$ in front of the mirror. (correct answer)
The claim is correct. The radius of curvature equals the focal length for a concave mirror, so with the object at 30 cm the image would be at infinity, but the object is at 20 cm so the image is virtual and behind the mirror at $d_i = -60 \text{ cm}$ .
The claim is incorrect. The focal length is $f = 15 \text{ cm}$ , and the image forms at $d_i = -60 \text{ cm}$ , meaning it is virtual and located 60 cm behind the mirror.
The claim is incorrect. The focal length is $f = 15 \text{ cm}$ , and with the object beyond the focal point the image is real. Applying the mirror equation yields $d_i = 12 \text{ cm}$ in front of the mirror.

Explanation: When working with mirror problems, always start by identifying the focal length before evaluating any claims about image location. For a curved mirror, the focal length and radius of curvature are related by

f = \frac{R}{2}

, which is a critical formula students frequently misremember. Here,

R = 30 \text{ cm}

, so

f = \frac{30}{2} = 15 \text{ cm}

. The student's fundamental error is assuming the focal length equals the radius of curvature — it doesn't. With the object at

d_o = 20 \text{ cm}

and

f = 15 \text{ cm}

, you apply the mirror equation:

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

, giving

\frac{1}{d_i} = \frac{1}{15} - \frac{1}{20} = \frac{4-3}{60} = \frac{1}{60}

, so

d_i = 60 \text{ cm}

. The positive value confirms a real image 60 cm in front of the mirror. That's answer A. Answer B incorrectly accepts the student's claim as partially correct and applies confused reasoning — the claim is simply wrong from the start, and no part of it should be validated. Answer C correctly identifies

f = 15 \text{ cm}

but gets the sign of

d_i

wrong; a negative image distance would mean a virtual image behind the mirror, which doesn't result from this calculation. Answer D also correctly finds

f = 15 \text{ cm}

but makes an arithmetic error in the mirror equation, arriving at

12 \text{ cm}

instead of

60 \text{ cm}

. Your go-to study tip: never confuse

R

and

f

. Always write

f = R/2

as your first step in any mirror problem before doing anything else.

Question 2

A convex mirror (radius 24 cm) has an object 6 cm away. Image is:

real, 4 cm, inverted
virtual, 4 cm, upright (correct answer)
real, 12 cm, inverted
virtual, 12 cm, upright

Explanation: For a convex mirror, focal length is half the radius with a negative sign, so f = -12 cm. Using 1/f = 1/do + 1/di with do = 6 cm gives 1/di = -1/12 - 1/6 = -1/4, so di = -4 cm. Negative distance means virtual, and magnification is positive, so the image is upright. The tempting error is using 24 cm as the focal length, which would wrongly give 12 cm.

Question 3

A diverging lens (focal length 10 cm) forms an upright image one-third the size of the object. Object distance?

40 cm
10 cm
6.7 cm
20 cm (correct answer)

Explanation: Since the lens is diverging, its focal length is -10 cm. An upright virtual image has m = +1/3, and the lens convention gives m = -di/do, so di = -do/3. Plugging into 1/f = 1/do + 1/di gives -1/10 = 1/do - 3/do = -2/do, so do = 20 cm. The tempting 40 cm comes from dropping the negative sign in m = -di/do.

Question 4

An object is 6 cm from a +10 cm lens. Image is:

virtual, 15 cm, upright (correct answer)
real, 15 cm, inverted
virtual, 3.75 cm, upright
real, 3.75 cm, inverted

Explanation: Using 1/f = 1/do + 1/di with f = 10 cm and do = 6 cm gives 1/di = 1/10 - 1/6 = -1/15, so di = -15 cm. The negative sign means the image is virtual, and magnification is -di/do = +2.5, so it's upright. Since the object sits inside the focal length, that's exactly what you expect. The tempting error is treating di as positive; that would incorrectly make the image real and inverted.

Question 5

A concave mirror with f=20 cm makes a real image twice as large. Object distance?

10 cm; inside f, virtual
40 cm; at the center C
30 cm; between f and C (correct answer)
60 cm; beyond the center

Explanation: A real image is inverted, so magnification is -2 and image distance is twice the object distance. Using 1/f = 1/do + 1/di with f=20 and di=2do gives 1/20 = 3/(2do), so do=30 cm, placing the object between f and C. The trap is 10 cm inside f, which magnifies but gives only a virtual image, not a real one.

Question 6

A 4-cm object is 20 cm from a +15 cm lens. Image height?

12 cm, virtual, upright
1.3 cm, real, inverted
12 cm, real, inverted (correct answer)
1.3 cm, virtual, upright

Explanation: Use 1/f = 1/do + 1/di with f = +15 cm and do = 20 cm, so 1/di = 1/15 - 1/20 = 1/60, giving di = 60 cm. Since di is positive, the image is real. Magnification is di/do = 60/20 = 3, so image height is 3 x 4 cm = 12 cm, and the image is inverted. The tempting wrong answer is 12 cm virtual, upright, which confuses the sign of di and forgets that an object beyond the focal point forms a real, inverted image.

Question 7

A diverging lens with $|f| = 15 \text{ cm}$ is used to examine an object. An observer notes that the image appears upright and is located $10 \text{ cm}$ from the lens on the same side as the object. Which of the following gives the correct object distance and lateral magnification?

$d_o = 30 \text{ cm}$ , $m = +1/3$ ; the image is virtual, upright, and reduced to one-third the object's size. (correct answer)
$d_o = 6 \text{ cm}$ , $m = +5/3$ ; the image is virtual, upright, and enlarged to five-thirds the object's size.
$d_o = 30 \text{ cm}$ , $m = -1/3$ ; the image is real, inverted, and reduced to one-third the object's size.
$d_o = 6 \text{ cm}$ , $m = -5/3$ ; the image is real, inverted, and enlarged to five-thirds the object's size.

Explanation: When working with lens problems, always start with the thin lens equation and the sign conventions: for a diverging lens, the focal length is negative, so

f = -15 \text{ cm}

. When an image forms on the same side as the object, it's virtual, meaning

d_i

is negative — here,

d_i = -10 \text{ cm}

. Plugging into the thin lens equation:

\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}

\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10} = \frac{1}{30}

d_o = 30 \text{ cm}

. The lateral magnification is:

m = -\frac{d_i}{d_o} = -\frac{-10}{30} = +\frac{1}{3}

A positive magnification confirms the image is upright, and a magnitude less than 1 means it's reduced — perfectly consistent with a diverging lens. Answer A is correct. Answer B uses

d_o = 6 \text{ cm}

, which comes from incorrectly treating

d_i

as positive (+10 cm) instead of negative. This violates the sign convention for virtual images. Answer C gets the correct object distance but assigns a negative magnification, implying an inverted image — but diverging lenses always produce upright virtual images when the object is real. Answer D compounds both errors: wrong

d_o

from ignoring the sign of

d_i

, and a negative magnification. Your key strategy: Always assign signs before calculating. For any image on the same side as the incoming light (virtual image),

d_i

must be negative — this single step prevents most errors in lens problems.

Question 8

A flat mirror is replaced by a convex mirror of focal length magnitude $|f| = 20 \text{ cm}$ at the same location. An object originally at $d_o = 40 \text{ cm}$ remains stationary. By how much does the image distance change, and in which direction does the image shift?

The image moves $20 \text{ cm}$ closer to the mirror surface, shifting from $40 \text{ cm}$ behind the flat mirror to $20 \text{ cm}$ behind the convex mirror.
The image moves $26.7 \text{ cm}$ closer to the mirror surface, shifting from $40 \text{ cm}$ behind the flat mirror to $13.3 \text{ cm}$ behind the convex mirror. (correct answer)
The image moves $13.3 \text{ cm}$ closer to the mirror surface, shifting from $40 \text{ cm}$ behind the flat mirror to $26.7 \text{ cm}$ behind the convex mirror.
The image moves $30 \text{ cm}$ farther from the mirror surface, shifting from $40 \text{ cm}$ behind the flat mirror to $70 \text{ cm}$ behind the convex mirror.

Explanation: Whenever you swap mirror types in a problem, your first move should be to apply the mirror equation

\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}

carefully, paying close attention to sign conventions. Flat mirror baseline: A flat mirror always produces an image directly behind the surface at the same distance as the object. With

d_o = 40 \text{ cm}

, the image sits

40 \text{ cm}

behind the mirror, so

d_i = -40 \text{ cm}

(virtual, behind the mirror). Convex mirror calculation: For a convex mirror, the focal length is negative:

f = -20 \text{ cm}

. Plugging into the mirror equation:

\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-20} - \frac{1}{40} = -\frac{2}{40} - \frac{1}{40} = -\frac{3}{40}

d_i = -\frac{40}{3} \approx -13.3 \text{ cm}

The image is now

13.3 \text{ cm}

behind the convex mirror. The shift is

40 - 13.3 = 26.7 \text{ cm}

closer to the surface — confirming answer B. Why the others fail: Choice A assumes the image lands at

-20 \text{ cm}

, as if the image simply moves to the focal point, ignoring the full mirror equation. Choice C gets the subtraction backwards — it reports the new image distance (

13.3 \text{ cm}

) as the shift and vice versa, swapping the two numbers. Choice D treats the convex mirror like a concave mirror with a positive focal length, producing a nonsensical result where the virtual image moves farther away. Study tip: Always assign the correct sign to

f

before calculating — convex mirrors have

f < 0

, and forgetting this single step is the most common trap on mirror problems.

Question 9

A thin converging lens forms a real image with a lateral magnification of $m = -2$ . If the object is then moved so that it is $4 \text{ cm}$ closer to the lens (without changing the lens), the new magnification becomes $m = -4$ . What is the focal length of the lens?

$f = 8 \text{ cm}$
$f = 12 \text{ cm}$
$f = 6 \text{ cm}$
$f = 16 \text{ cm}$ (correct answer)

Explanation: Whenever you see a lens problem combining magnification with a positional change, your tools are the thin lens equation and the magnification formula working together. Recall that lateral magnification is

m = -\frac{d_i}{d_o}

, and the thin lens equation is

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

. Setting up the system: For the first position,

m = -2

means

d_i = 2d_o

. Substituting into the thin lens equation:

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{2d_o} = \frac{3}{2d_o}

, giving

d_o = \frac{3f}{2}

. For the second position, the object moves 4 cm closer, so

d_o' = d_o - 4

. Now

m = -4

means

d_i' = 4d_o'

, so:

\frac{1}{f} = \frac{1}{d_o'} + \frac{1}{4d_o'} = \frac{5}{4d_o'}

, giving

d_o' = \frac{5f}{4}

. Solving: Since

d_o - d_o' = 4

\frac{3f}{2} - \frac{5f}{4} = 4 \implies \frac{6f - 5f}{4} = 4 \implies \frac{f}{4} = 4 \implies f = 16 \text{ cm}

That confirms answer D. Why the other answers fail: Choice A (

f = 8

) likely comes from forgetting the factor of 3/2 and using simplified ratios. Choice B (

f = 12

) results from an arithmetic error when combining the fractions. Choice C (

f = 6

) comes from misapplying the magnification formula, perhaps using

m = +d_i/d_o

without the negative sign, which scrambles the algebra entirely. Study tip: Always express

d_i

in terms of

d_o

using magnification first, then substitute into the thin lens equation — this reduces the problem to one unknown immediately.

Question 10

A lens system consists of a single thin converging lens of focal length $f$ . An object is placed at distance $d_o = 3f/2$ from the lens.

The object is now moved to a new position such that the image distance doubles (i.e., $d_i^{\text{new}} = 2 d_i^{\text{old}}$ ). What is the new object distance in terms of $f$ ?

$d_o^{\text{new}} = \frac{4f}{3}$ ; the object has moved slightly closer to the lens, and the magnification magnitude decreases from the original value.
$d_o^{\text{new}} = 2f$ ; the object has moved farther from the lens to $2f$ , placing it at the center of curvature analog for a thin lens.
$d_o^{\text{new}} = \frac{6f}{5}$ ; the object has moved closer to the lens, and the new magnification magnitude is larger than the original. (correct answer)
$d_o^{\text{new}} = \frac{3f}{4}$ ; the object has moved inside the focal length, converting the image from real to virtual.

Explanation: Whenever you see a thin-lens problem, your anchor is the thin-lens equation:

\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}

. The key move is finding the original image distance first, then applying the new condition. Step 1 — Find the original image distance. With

d_o = \frac{3f}{2}

\frac{1}{d_i^{\text{old}}} = \frac{1}{f} - \frac{2}{3f} = \frac{1}{3f} \implies d_i^{\text{old}} = 3f

Step 2 — Apply the new condition. Doubling gives

d_i^{\text{new}} = 6f

. Plug into the lens equation:

\frac{1}{d_o^{\text{new}}} = \frac{1}{f} - \frac{1}{6f} = \frac{5}{6f} \implies d_o^{\text{new}} = \frac{6f}{5}

This confirms C. Since

\frac{6f}{5} < \frac{3f}{2}

, the object moved closer to the lens. The new magnification magnitude is

\frac{6f}{6f/5} = 5

, larger than the original

\frac{3f}{3f/2} = 2

, which matches C's claim perfectly. Choice A gives

\frac{4f}{3}

, which produces

d_i = 4f

, not

6f

— the doubling condition simply isn't satisfied. Choice B places the object at

2f

, yielding

d_i = 2f

, far short of

6f

. Choice D puts the object inside the focal length (

\frac{3f}{4} < f

), which would produce a virtual image with a negative

d_i

— nowhere near doubling a real positive image distance. Study tip: Always solve for the original image distance explicitly before applying any "change" condition. Skipping that step is the fastest route to choosing a tempting but wrong distractor.

Question 11

A concave mirror produces an image that has a lateral magnification of $m = +2$ . Which of the following must be true about the object's location relative to the mirror's focal point $F$ and center of curvature $C$ ?

The object is located between $F$ and $C$ , because a magnification of $+2$ indicates a real, upright, enlarged image — which for a concave mirror requires the object to be between the focal point and center of curvature.
The object is located beyond $C$ , because an upright image with $|m| > 1$ can only be formed by a concave mirror when the object is farther than twice the focal length from the mirror.
The object is located between the mirror's surface and $F$ , because a magnification of $+2$ indicates a virtual, upright, enlarged image — which for a concave mirror occurs only when the object is within the focal length. (correct answer)
The object could be located anywhere beyond $F$ , because the sign of magnification is determined solely by the sign convention chosen and does not constrain the object location for a concave mirror.

Explanation: When analyzing mirror problems, your first move should be to decode what the magnification value tells you about the image — specifically its orientation and whether it's real or virtual. A lateral magnification of

m = +2

carries two pieces of information: the positive sign means the image is upright (same orientation as the object), and

|m| = 2

means the image is enlarged. For a concave mirror, an upright image is always virtual — real images from concave mirrors are always inverted (negative magnification). A virtual, upright, enlarged image from a concave mirror occurs only when the object is placed between the mirror's surface and the focal point $F$ . In this region, reflected rays diverge and appear to come from a magnified image behind the mirror. This confirms C is correct. A is wrong on two counts: an object between

F

and

C

produces a real, inverted image with

m < -1

— not upright. A positive magnification cannot come from that region. B is wrong because objects beyond

C

produce real, inverted, diminished images (

-1 < m < 0

). An upright image with

|m| > 1

is impossible in this region for a concave mirror. D is a dangerous distractor. The sign of magnification is not arbitrary — it directly encodes image orientation within the standard sign convention, which absolutely constrains object placement. The sign convention is fixed, not flexible. Study tip: Memorize this concave mirror rule: positive

m

always means virtual + upright, which always means the object is inside

F

. If you see

m > 0

for a concave mirror, the object is between the surface and the focal point — no exceptions.

Question 12

A student constructs an astronomical telescope using two converging lenses. The objective lens has focal length $f_{\text{obj}} = 80 \text{ cm}$ and the eyepiece has focal length $f_{\text{eye}} = 4 \text{ cm}$ . The telescope is adjusted for a relaxed eye (final image at infinity), meaning the intermediate image formed by the objective falls exactly at the front focal point of the eyepiece.

An object of height $h = 2 \text{ mm}$ is located $400 \text{ cm}$ from the objective lens. What is the height of the intermediate image formed by the objective lens alone?

$h_i \approx -0.4 \text{ mm}$ ; the intermediate image is real, inverted, and reduced, corresponding to a magnification of $-0.20$ .
$h_i \approx +0.5 \text{ mm}$ ; the intermediate image is virtual, upright, and reduced relative to the object.
$h_i \approx -1.0 \text{ mm}$ ; the intermediate image is real, inverted, and reduced, corresponding to a magnification of $-0.50$ .
$h_i \approx -0.5 \text{ mm}$ ; the intermediate image is real, inverted, and reduced relative to the object. (correct answer)

Explanation: When solving for an intermediate image in a telescope, treat the objective lens as a standalone thin lens problem using the lens equation

\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}

. Here,

f_{\text{obj}} = 80 \text{ cm}

and

d_o = 400 \text{ cm}

. Plugging in:

\frac{1}{d_i} = \frac{1}{80} - \frac{1}{400} = \frac{5}{400} - \frac{1}{400} = \frac{4}{400} = \frac{1}{100}

d_i = 100 \text{ cm}

. The lateral magnification is:

m = -\frac{d_i}{d_o} = -\frac{100}{400} = -0.25

The intermediate image height is

h_i = m \cdot h = -0.25 \times 2 \text{ mm} = -0.5 \text{ mm}

. The negative sign confirms the image is real and inverted — making D correct. Choice A claims a magnification of

-0.20

, which would require

d_i = 80 \text{ cm}

— but that's only true when the object is at infinity, not at 400 cm. Choice B describes a virtual, upright image, which is impossible here since the object is beyond the focal point; virtual images from converging lenses only form when

d_o < f

. Choice C gives

h_i = -1.0 \text{ mm}

, corresponding to

m = -0.50

, which implies

d_i = 200 \text{ cm}

— an arithmetic error likely from misapplying the lens equation. As a strategy, always compute

d_i

first before finding magnification — skipping straight to guessing the magnification from the focal length ratio (

f_{\text{eye}}/f_{\text{obj}}

) is a trap that ignores the actual object distance.

Question 13

A student uses a converging lens of focal length $f = 10 \text{ cm}$ as a simple magnifier. The near point of the student's eye is $N = 25 \text{ cm}$ .

The student first places the object at the front focal point of the lens (image at infinity, relaxed-eye viewing). She then repositions the object so that the virtual image falls exactly at her near point ( $25 \text{ cm}$ from the lens on the same side as the object). By what factor does the angular magnification increase when she shifts from the relaxed-eye configuration to the near-point configuration?

The magnification increases by a factor of $\dfrac{6}{5}$ , from $M_{\infty} = 2.5$ to $M_{N} = 3.0$ .
The magnification increases by a factor of $\dfrac{7}{5}$ , from $M_{\infty} = 2.5$ to $M_{N} = 3.5$ . (correct answer)
The magnification increases by a factor of $\dfrac{4}{3}$ , from $M_{\infty} = 3.0$ to $M_{N} = 4.0$ .
The magnification increases by a factor of $\dfrac{8}{5}$ , from $M_{\infty} = 2.5$ to $M_{N} = 4.0$ .

Explanation: When working with simple magnifiers, you need two formulas: the relaxed-eye magnification

M_{\infty} = \dfrac{N}{f}

, and the near-point magnification

M_N = 1 + \dfrac{N}{f}

, where

N = 25\text{ cm}

is the near-point distance and

f

is the focal length. For the relaxed configuration (object at the focal point, image at infinity):

M_{\infty} = \dfrac{N}{f} = \dfrac{25}{10} = 2.5

. For the near-point configuration (virtual image at 25 cm):

M_N = 1 + \dfrac{N}{f} = 1 + \dfrac{25}{10} = 3.5

. The ratio of these magnifications is

\dfrac{M_N}{M_{\infty}} = \dfrac{3.5}{2.5} = \dfrac{7}{5}

, confirming that answer B is correct. Choice A gets the ratio wrong — it uses

M_N = 3.0

instead of 3.5, which would only be correct if

N = 20\text{ cm}

. This is a calculation error from misapplying the near-point formula. Choice C inflates both values by using

f

incorrectly, arriving at

M_{\infty} = 3.0

— someone may have mistakenly added 1 to the relaxed formula as well. Choice D correctly identifies

M_{\infty} = 2.5

but uses

M_N = 4.0

, which would require

N = 30\text{ cm}

, not 25 cm — a near-point substitution error. Study tip: Memorize the "+1" distinction — the near-point formula adds 1 because the object is closer than the focal point, so the lens does extra converging work. On the exam, always check whether the image is at infinity or at the near point before selecting your magnification formula.

Question 14

A thin converging lens of focal length $f = 12 \text{ cm}$ is used to project an image onto a screen. A student places an object at $d_o = 16 \text{ cm}$ from the lens and adjusts the screen position until a sharp image is formed.

What is the lateral magnification of the image, and which of the following correctly describes the nature of the image?

$m = -3$ ; the image is real, inverted, and enlarged by a factor of 3 relative to the object. (correct answer)
$m = +3$ ; the image is virtual, upright, and enlarged by a factor of 3 relative to the object.
$m = -3/4$ ; the image is real, inverted, and reduced to three-quarters of the object's size.
$m = +4/3$ ; the image is virtual, upright, and enlarged by a factor of four-thirds relative to the object.

Explanation: When you see a thin lens problem asking about image nature and magnification, your two key tools are the thin lens equation and the magnification formula. The approach is always the same: find the image distance first, then calculate magnification. Start with the thin lens equation:

\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}

. Plugging in

d_o = 16 \text{ cm}

and

f = 12 \text{ cm}

\frac{1}{d_i} = \frac{1}{12} - \frac{1}{16} = \frac{4-3}{48} = \frac{1}{48}

d_i = 48 \text{ cm}

. Since

d_i

is positive, the image forms on the opposite side of the lens from the object — confirming it's real. Now apply the magnification formula:

m = -\frac{d_i}{d_o} = -\frac{48}{16} = -3

. The negative sign means the image is inverted, and the magnitude of 3 means it's enlarged by a factor of 3. This confirms A. Choice B is tempting if you drop the negative sign from the magnification formula — a very common arithmetic error. A positive magnification would indicate a virtual, upright image, which contradicts the physics here. Choice C comes from incorrectly computing

m = -d_o/d_i

(flipping the ratio), giving

-16/48 = -1/3

... actually producing

-3/4

if someone misreads the lens equation entirely — this reflects a setup error. Choice D results from confusing the lens equation and solving for a virtual image scenario, which doesn't apply when the object is beyond the focal length. A reliable study tip: whenever

d_o > f

for a converging lens, the image is always real, inverted, and on the opposite side — the only question is whether it's enlarged or reduced relative to

2f

Question 15

A convex mirror is used as a security mirror in a store. The mirror has a focal length of magnitude $|f| = 0.80 \text{ m}$ . A person stands $3.2 \text{ m}$ in front of the mirror.

A second person stands directly behind the first and is twice as far from the mirror (i.e., $6.4 \text{ m}$ away). By what factor does the image size of the second person differ from the image size of the first person, assuming both people are the same actual height $h$ ?

The second person's image is smaller by a factor of approximately $0.56$ , meaning the second person's image height is about $56\%$ of the first person's image height. (correct answer)
The second person's image is smaller by a factor of $0.50$ , meaning the second person's image height is exactly half the first person's image height.
The second person's image is smaller by a factor of approximately $0.64$ , meaning the second person's image height is about $64\%$ of the first person's image height.
The second person's image is smaller by a factor of approximately $0.44$ , meaning the second person's image height is about $44\%$ of the first person's image height.

Explanation: Whenever you see a mirror or lens problem asking about image size, your key tool is the magnification equation. For mirrors, the lateral magnification is

m = -\frac{d_i}{d_o}

, and the image height is

h_i = m \cdot h

. To find

d_i

, you first apply the mirror equation:

\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}

. For a convex mirror, the focal length is negative:

f = -0.80 \text{ m}

. For Person 1 (

d_o = 3.2 \text{ m}

\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-0.80} - \frac{1}{3.2} = -1.25 - 0.3125 = -1.5625

d_i = -0.64 \text{ m}

, giving magnification

|m_1| = \frac{0.64}{3.2} = 0.20

. For Person 2 (

d_o = 6.4 \text{ m}

\frac{1}{d_i} = -1.25 - 0.15625 = -1.40625

d_i = -0.711 \text{ m}

, giving magnification

|m_2| = \frac{0.711}{6.4} \approx 0.111

. The ratio of image sizes is

\frac{|m_2|}{|m_1|} = \frac{0.111}{0.20} \approx 0.556

, which is approximately 0.56 — confirming answer A. Answer B assumes the magnification simply scales in direct proportion to object distance (

\frac{3.2}{6.4} = 0.50

), ignoring how

d_i

also shifts nonlinearly. Answer C likely arises from an arithmetic error in computing

d_i

. Answer D overcorrects, perhaps misapplying signs or inverting a ratio. Your takeaway: never assume image size scales linearly with object distance in mirror/lens problems — you must solve for

d_i

first, then compute magnification. Skipping that step is the most common trap on optics questions.