Magnification

Q: What is the magnification formula for a mirror and a lens?

Lateral magnification is image height divided by object height. For a spherical mirror it equals minus v over u, and for a thin lens it equals v over u, with the New Cartesian sign convention. A negative value means an inverted real image and a positive value an erect virtual image.

Q: Why is there a minus sign in mirror magnification but not in lens magnification?

Both come from similar triangles. For a mirror, the object and a real image lie on the same side, so u and v have the same sign while the image is inverted, which needs a minus sign. For a lens a real image lies on the opposite side, so v over u is already negative.

Q: How do you find magnification without finding the image distance?

Combine the magnification with the mirror or lens formula. For a mirror, m equals f divided by (f minus u); for a lens, m equals f divided by (f plus u). Both can also be written as (f minus v) divided by f when the image distance is known instead.

Q: What is longitudinal magnification?

Longitudinal magnification describes a short object lying along the principal axis. Differentiating the formulas gives minus m squared for a mirror and plus m squared for a lens. For a long rod the magnification changes along its length, so each end must be imaged separately and the positions subtracted.

Q: How fast does an image move when the object moves along the axis?

Relative to the mirror or lens, the image moves at m squared times the object's speed along the axis. For a mirror the image moves in the opposite direction, for a lens in the same direction. At a distance 2f from a concave mirror or convex lens the image moves at exactly the object's speed.

Q: What is the total magnification of a combination of lenses or mirrors?

Each element forms an image that becomes the object for the next one, and the total magnification is the product of the individual magnifications. Its sign gives the final orientation; for example two inverting lenses give an erect final image.

Q: Is magnification important for JEE Main and JEE Advanced?

Yes. JEE Main asks image size, the two answers to image n times the object, and magnification of combinations. JEE Advanced adds image velocity along and across the axis, longitudinal magnification of rods, moving mirrors and lenses, and images of points off the axis.

Q: What magnification questions come in NEET?

NEET usually asks the magnification formula for mirrors and lenses, finding object position for a given magnification, the nature of images from the sign of m, and the power or focal length from a measured image size. Keep the sign convention consistent throughout.

PhysicsRay Optics And Optical InstrumentsFor JEE aspirants

Magnification tells how large an image is compared with its object and whether it is erect or inverted: $m = \frac{h _{i}}{h _{o}}$ , which equals $- \frac{v}{u}$ for a mirror and $\frac{v}{u}$ for a thin lens. Magnification also decides how long the image of a rod along the axis is and how fast an image moves when the object moves. These ideas of magnification carry marks in JEE Main, JEE Advanced and NEET every year.

On this page1Lateral magnification2Magnification in terms of f3Along the axis4Image velocity5Several elements6Flowchart and map7Solved examples

Key Formulas - Quick Reference

★ Must learnMirror: $m = \frac{h _{i}}{h _{o}} = - \frac{v}{u} = \frac{f}{f - u} = \frac{f - v}{f}$
★ Must learnThin lens: $m = \frac{v}{u} = \frac{f}{f + u} = \frac{f - v}{f}$
Refracting surface: $m = \frac{n _{1} v}{n _{2} u}$
★ Must learnShort object along the axis: mirror $m_{L} = - m^{2}$ , lens $m_{L} = + m^{2}$ ; area magnification $m^{2}$
★ Must learnImage velocity along the axis: mirror $v_{I} = - m^{2} v_{O}$ , lens $v_{I} = + m^{2} v_{O}$ (relative to the element)
Across the axis: $v_{I, ⊥} = m v_{O, ⊥}$ (relative to the element)
Several elements: $m = m_{1} m_{2} m_{3} \dots$

1. Lateral (Transverse) Magnification

For an object perpendicular to the axis, the lateral magnification is $m = \frac{h _{i}}{h _{o}}$ , heights measured with sign (up positive). For a single mirror or lens with a real object, $m < 0$ means a real, inverted image and $m > 0$ a virtual, erect one; $∣ m ∣ > 1$ means enlarged.

Figure 1: The ray to the pole reflects at equal angles, so triangles

A B P

and

A^{'} B^{'} P

are similar:

\frac{h _{i}}{h _{o}} = \frac{- v}{u}

. Here

u = - 80

v = - 48 cm

m = - 0.60

(real, inverted, diminished).

The ray to the pole of a mirror reflects at equal angles, so the triangles on the two sides are similar: $\frac{h _{i}}{h _{o}} = \frac{- v}{u}$ . For a lens the undeviated ray through the optical centre gives $\frac{h _{i}}{h _{o}} = \frac{v}{u}$ .

Figure 2: The undeviated ray through the optical centre gives similar triangles, so

m = \frac{h _{i}}{h _{o}} = \frac{v}{u}

for a lens (no minus sign). Here

m = - 2

: real, inverted, twice the size.

Mirror

$m = - \frac{v}{u}$ . Real image: $u$ and $v$ both negative, so $m < 0$ (inverted). Virtual image: $v > 0$ , $m > 0$ (erect).

Lens

$m = + \frac{v}{u}$ . Real image: $u < 0$ , $v > 0$ , so $m < 0$ (inverted). Virtual image: $v < 0$ , $m > 0$ (erect).

A spherical refracting surface gives, from $n_{1} sin i = n_{2} sin r$ with small angles,

m = \frac{n _{1} v}{n _{2} u}

The ray through the centre of curvature is undeviated, which gives the equivalent form $m = \frac{v - R}{u - R}$ .

Figure 3: Single surface:

m = \frac{n _{1} v}{n _{2} u} = \frac{1 \times 90}{1.5 \times ( - 30 )} = - 2

. The ray through

C

is undeviated, giving the equivalent form

m = \frac{v - R}{u - R}

. Heights are drawn twice the horizontal scale.

Key idea

The sign of

m

gives orientation, its size gives enlargement: mirror

- \frac{v}{u}

, lens

+ \frac{v}{u}

, surface

\frac{n _{1} v}{n _{2} u}

2. Magnification in Terms of Focal Length

Eliminating $v$ with the mirror or lens formula gives $m$ from the object distance alone:

Mirror: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ , so $m = - \frac{v}{u} = \frac{f}{f - u}$ ; eliminating $u$ instead, $m = \frac{f - v}{f}$ .
Lens: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ , so $m = \frac{v}{u} = \frac{f}{f + u}$ , and $m = \frac{f - v}{f}$ .

Figure 4:

m

against object distance, exact. Concave mirror and convex lens (solid) share

m = \frac{f}{f - ∣ u ∣}

: enlarged and erect inside

f

, infinite at

f

- 1

2 f

. Convex mirror and concave lens (dashed):

m = \frac{f}{f + ∣ u ∣}

, always erect and diminished.

Exam Trick

"Image $n$ times the object" has two answers for a converging element. Put $m = - n$ (real) and $m = + n$ (virtual) in $m = \frac{f}{f - u}$ (mirror) or $\frac{f}{f + u}$ (lens). A diverging element or a convex mirror can only give $0 < m < 1$ for a real object.

Figure 5: "Image 3 times the object" with a concave mirror (

f = - 20 cm

) has two answers. From

u = f (1 - \frac{1}{m})

: (a)

m = - 3

gives

u = - 26.7 cm

(between

F

and

C

v = - 80 cm

; (b)

m = + 3

gives

u = - 13.3 cm

(inside

F

v = + 40 cm

. Both objects are

\frac{∣ f ∣}{3}

from

F

, one on each side.

Quick Recall: tap to check

A concave mirror (

f = - 20 cm

) forms a real image 3 times the object. Where is the object?

m = - 3 = \frac{- 20}{- 20 - u}

gives

u = - 26.7 cm

Can a convex lens give

m = + \frac{1}{2}

for a real object?

No. For a real object its erect images are virtual and enlarged (

m > 1

Key idea

m = \frac{f}{f - u}

(mirror) and

\frac{f}{f + u}

(lens) avoid finding

v

first.

3. Objects Along the Axis: Longitudinal Magnification

For a short object of length $d u$ lying along the axis, differentiate the formula. Mirror: $- \frac{d v}{v ^{2}} - \frac{d u}{u ^{2}} = 0$ , so

m_{L} = \frac{d v}{d u} = - \frac{v ^{2}}{u ^{2}} = - m^{2}

For a lens, $- \frac{d v}{v ^{2}} + \frac{d u}{u ^{2}} = 0$ , so $m_{L} = + m^{2}$ .

Figure 6: Longitudinal magnification (

f = - 30 cm

, both ends imaged exactly). Beyond

C

the image is shorter; between

F

and

C

it is stretched. For a short rod, image length

\approx m^{2} \times

object length, with the order of the ends reversed (

A^{'}

and

B^{'}

swap sides).

For a long rod, $m^{2}$ changes along the rod: image each end separately and subtract.
For a mirror the image of a rod along the axis is reversed end to end (the end nearer the mirror is imaged farther away); for a lens the order is kept.
A small flat object perpendicular to the axis has its area magnified by $m^{2}$ .

Exam Trick

Small rod along the axis: multiply its length by $m^{2}$ . A $1 mm$ object $60 cm$ from a lens of $f = 20 cm$ has $m = - \frac{1}{2}$ , so its image is $\frac{1}{4} mm$ long. Use exact end-by-end imaging only when the rod is long.

Key idea

Along the axis the image length scales as

m^{2}

(short rods); long rods need both ends imaged.

4. Velocity of the Image

Dividing $d v = m_{L} d u$ by $d t$ gives the image velocity along the axis, measured relative to the mirror or lens:

Mirror: $v_{I} - v_{M} = - m^{2} (v_{O} - v_{M})$ : image and object move in opposite directions along the axis.
Lens: $v_{I} - v_{L} = + m^{2} (v_{O} - v_{L})$ : they move in the same direction.
Across the axis (height changing, $u$ fixed): $\frac{d h _{i}}{d t} = m \frac{d h _{o}}{d t}$ for both.

Figure 7: For motion along the axis,

\frac{d v}{d u} = \frac{v ^{2}}{u ^{2}} = m^{2}

. The image moves at the object's speed only at

2 f

, much slower far away, and very fast near

f

Figure 8: Differentiating the two formulas: mirror

\frac{d v}{d t} = - \frac{v ^{2}}{u ^{2}} \frac{d u}{d t}

(opposite directions along the axis); lens

\frac{d v}{d t} = + \frac{v ^{2}}{u ^{2}} \frac{d u}{d t}

(same direction). Velocities are relative to the mirror or lens.

JEE Advanced

Object moving in both directions. Split the object velocity into components along and across the axis. The axial part of the image velocity is $\mp m^{2} v_{O, x}$ (mirror $-$ , lens $+$ ). Across the axis, $h_{i} = m h_{o}$ with $m$ itself changing as $u$ changes, so $\frac{d h _{i}}{d t} = m \frac{d h _{o}}{d t} + h_{o} \frac{d m}{d t}$ ; the second term matters when the object has height and also moves along the axis. For a plane mirror, $m = 1$ and the image moves with the normal component reversed.

Quick Recall: tap to check

An object moves towards a concave mirror. Which way does its real image move?

Away from the mirror (opposite direction), at

m^{2}

times the object's speed relative to the mirror.

An object at

2 f

from a convex lens moves along the axis at

1 cm s^{- 1}

. How fast does the image move?

1 cm s^{- 1}

, in the same direction (

m^{2} = 1

2 f

Key idea

Image velocity along the axis is

m^{2}

times the object velocity (relative to the element): reversed for mirrors, same direction for lenses.

5. Points off the Axis and Several Elements

For a point object at $(u, h)$ the image lies at $(v, mh)$ : the mirror or lens formula gives the $x$ coordinate, the magnification gives the $y$ coordinate.

Figure 9: Point object off the axis (

R = 10 cm

, pole at the origin).

x

from the mirror formula,

y

from

m

v = - \frac{40}{7} cm

m = - \frac{v}{u} = - \frac{1}{7}

, so the image is at

(- \frac{40}{7}, - \frac{1}{7}) cm

In a system of mirrors, lenses and surfaces, each image acts as the object for the next element. The total magnification is the product $m = m_{1} m_{2} m_{3} \dots$ ; its sign gives the final orientation relative to the original object.

Figure 10: Successive imaging.

L_{1}

u = - 15

v = + 30

m_{1} = - 2

I_{1}

20 cm

before

L_{2}

v = + 20

m_{2} = - 1

. Total

m = m_{1} m_{2} = + 2

: real and erect.

6. Flowchart and Mind Map

Decide whether the question is about the size of the image or its motion, then pick the matching relation.

Figure 11: Problem-solving flowchart. Size questions use

m

(or

m^{2}

along the axis); motion questions use

\frac{v ^{2}}{u ^{2}}

along the axis and

m

across it, always relative to the mirror or lens.

Figure 12: Magnification on one page. Revise from the map, then test yourself on the solved examples.

7. Solved Examples

Solved Example 1

An object

4 cm

tall stands

10 cm

in front of a convex mirror of focal length

10 cm

. Find the position and size of the image.

Solution:

$f = + 10$ , $u = - 10$ : $\frac{1}{v} = \frac{1}{10} + \frac{1}{10}$ , $v = + 5 cm$ . $m = - \frac{v}{u} = \frac{1}{2}$ , so $h_{i} = 2 cm$ .

Answer: $5 cm$ behind the mirror, $2 cm$ tall, virtual and erect.

Solved Example 2

A concave mirror of radius

10 cm

has its pole at the origin and axis along

x

. Find the image of a point object at

(- 40 cm, 1 cm)

Solution:

$f = - 5$ , $u = - 40$ : $\frac{1}{v} = - \frac{1}{5} + \frac{1}{40} = - \frac{7}{40}$ , $v = - \frac{40}{7} cm$ . $m = - \frac{v}{u} = - \frac{1}{7}$ , so $y = - \frac{1}{7} cm$ .

Answer: $(- \frac{40}{7}, - \frac{1}{7}) cm$ (Figure 9).

Solved Example 3

A concave mirror of focal length

20 cm

forms a real image three times the size of the object. The object distance is
(A)

13.3 cm

(B)

20 cm

(C)

26.7 cm

(D)

60 cm

Solution:

Real image: $m = - 3 = \frac{f}{f - u} = \frac{- 20}{- 20 - u}$ , so $60 + 3 u = - 20$ and $u = - 26.7 cm$ . (A virtual image three times as large would need $u = - 13.3 cm$ : option (A) is that trap.)

Answer: (C).

Solved Example 4

A rod lies along the axis of a convex lens (

f = 20 cm

) with its ends

40 cm

and

35 cm

from the lens. Find the length of its image.

Solution:

End at $u = - 40$ : $v = 40 cm$ . End at $u = - 35$ : $\frac{1}{v} = \frac{1}{20} - \frac{1}{35} = \frac{3}{140}$ , $v = 46.7 cm$ . The rod is long (5 cm), so image both ends: length $= 46.7 - 40$ .

Answer: $\frac{20}{3} \approx 6.7 cm$ . (The short-rod rule would give $5 \times 1 = 5 cm$ , visibly wrong here.)

Solved Example 5

A small object

1 mm

long lies along the axis

60 cm

from a convex lens of focal length

20 cm

. Find the length of its image.

Solution:

$v = \frac{u f}{u + f} = \frac{( - 60 ) ( 20 )}{- 40} = 30 cm$ , $m = \frac{v}{u} = - \frac{1}{2}$ . Short object: $d v = m^{2} d u = \frac{1}{4} \times 1 mm$ .

Answer: $0.25 mm$ .

Solved Example 6

A point object moves along the axis towards a concave mirror (

f = - 10 cm

) at

2 cm s^{- 1}

, while the mirror moves towards the object at

1 cm s^{- 1}

. Find the image velocity when the object is

30 cm

from the mirror.

Solution:

Take $+ x$ from object to mirror: $v_{O} = + 2$ , $v_{M} = - 1$ . $v = \frac{u f}{u - f} = \frac{( - 30 ) ( - 10 )}{- 20} = - 15 cm$ , $m^{2} = \frac{225}{900} = \frac{1}{4}$ . $v_{I} - v_{M} = - \frac{1}{4} (v_{O} - v_{M})$ : $v_{I} = - 1 - \frac{3}{4}$ .

Answer: $v_{I} = - \frac{7}{4} cm s^{- 1}$ , i.e. $1.75 cm s^{- 1}$ in the direction the mirror moves (towards the object). Relative to the mirror the image moves away from it at $0.75 cm s^{- 1}$ .

Solved Example 7

A convex lens (

f = 10 cm

) moves at

2 cm s^{- 1}

and a point object at

1 cm s^{- 1}

, both in the same direction along the axis. Find the image velocity when the object is

15 cm

from the lens.

Solution:

$v = \frac{u f}{u + f} = \frac{( - 15 ) ( 10 )}{- 5} = 30 cm$ , $m^{2} = 4$ . $v_{I} - v_{L} = + m^{2} (v_{O} - v_{L}) = 4 (1 - 2) = - 4$ , so $v_{I} = 2 - 4$ .

Answer: $v_{I} = - 2 cm s^{- 1}$ (against the direction of motion of both).

Solved Example 8

A point object

15 cm

from a convex lens (

f = 10 cm

) moves perpendicular to the axis at

2 cm s^{- 1}

. Find the velocity of its image.

Solution:

$v = 30 cm$ , $m = \frac{v}{u} = - 2$ . Across the axis $v_{I, ⊥} = m v_{O, ⊥} = - 2 \times 2$ .

Answer: $4 cm s^{- 1}$ in the opposite direction.

Solved Example 9

A gun of mass

m_{1}

fires a bullet of mass

m_{2}

with horizontal speed

v_{0}

. The gun is fitted with a concave mirror of focal length

f

facing the receding bullet. Find the speed of separation of the bullet and its image just after firing.

Solution:

Momentum: the gun (and mirror) recoils at $v_{1} = \frac{m _{2} v _{0}}{m _{1}}$ . Relative to the mirror the bullet moves away at $v_{0} + v_{1}$ . Just after firing the bullet is at the mirror ( $u \to 0$ , so $m^{2} = \frac{v ^{2}}{u ^{2}} \to 1$ ), and the image moves away from the mirror on the other side at the same relative speed $v_{0} + v_{1}$ .

Answer: $2 (v_{0} + v_{1}) = 2 (1 + \frac{m _{2}}{m _{1}}) v_{0}$ .

Solved Example 10

Two convex lenses of focal length

10 cm

are

50 cm

apart. An object is

15 cm

in front of the first. Find the final image and the total magnification.

Solution:

$L_{1}$ : $v_{1} = 30 cm$ , $m_{1} = - 2$ . For $L_{2}$ the object is $50 - 30 = 20 cm$ in front: $v_{2} = 20 cm$ , $m_{2} = - 1$ .

Answer: $20 cm$ beyond $L_{2}$ , $m = m_{1} m_{2} = + 2$ : real, erect, double size (Figure 10).

Solved Example 11

A small square of area

1 cm^{2}

is placed perpendicular to the axis

30 cm

from a concave mirror of focal length

10 cm

. The area of its image is
(A)

0.25 cm^{2}

(B)

0.5 cm^{2}

(C)

1 cm^{2}

(D)

4 cm^{2}

Solution:

$v = - 15 cm$ , $m = - \frac{v}{u} = - \frac{1}{2}$ . Both sides scale by $∣ m ∣$ , so the area scales by $m^{2} = \frac{1}{4}$ .

Answer: (A).

Solved Example 12

An object

2 cm

tall is

30 cm

in front of a convex spherical glass surface (

n = 1.5

R = 10 cm

). Find the size of the image.

Solution:

$\frac{1.5}{v} - \frac{1}{- 30} = \frac{0.5}{10}$ gives $v = 90 cm$ . $m = \frac{n _{1} v}{n _{2} u} = \frac{90}{1.5 \times ( - 30 )} = - 2$ .

Answer: $4 cm$ , real and inverted, inside the glass (Figure 3).

Practice Questions

Find the magnification by a convex mirror of focal length $15 cm$ for an object $10 cm$ away.Answer: $m = \frac{15}{15 + 10} = 0.6$
A concave lens ( $f = - 20 cm$ ) has an object $20 cm$ away. Find $m$ .Answer: $m = \frac{- 20}{- 20 - 20} = 0.5$
A $2 mm$ object lies along the axis $30 cm$ from a concave mirror of focal length $10 cm$ . Find the image length.Answer: $m^{2} = \frac{1}{4}$ : $0.5 mm$
An object $30 cm$ from a concave mirror ( $f = 20 cm$ ) approaches it at $3 cm s^{- 1}$ . Find the image speed and direction.Answer: $v = - 60 cm$ , $m^{2} = 4$ : $12 cm s^{- 1}$ away from the mirror
An object $20 cm$ from a convex lens ( $f = 10 cm$ ) moves towards the lens at $1 cm s^{- 1}$ . Find the image velocity.Answer: $1 cm s^{- 1}$ away from the lens (same direction as the object)
A rod lies along the axis of a concave mirror between $F$ and $C$ . Describe its image.Answer: Real, beyond $C$ , longer than the rod, ends reversed
A $3 cm$ object is $15 cm$ from a convex lens of $f = 10 cm$ . Find the image height.Answer: $v = 30 cm$ , $m = - 2$ : $6 cm$ , inverted

Common Mistakes to Avoid

Watch out

Using $m = - \frac{v}{u}$ for a lens. For a lens $m = + \frac{v}{u}$ ; the minus sign belongs to mirrors.
Dropping signs: $m$ must carry its sign to tell erect from inverted.
Using $m$ (not $m^{2}$ ) for an object along the axis.
Applying $m_{L} = m^{2}$ to a long rod; image the two ends separately.
Using ground-frame velocities in $v_{I} = \mp m^{2} v_{O}$ when the mirror or lens moves. Use velocities relative to the element.
Assuming the image of a mirror moves the same way as the object along the axis; for mirrors it is opposite.
Adding magnifications of successive elements instead of multiplying them.
For a refracting surface forgetting the indices: $m = \frac{n _{1} v}{n _{2} u}$ , not $\frac{v}{u}$ .

Frequently Asked Questions

What is the magnification formula for a mirror and a lens?

Lateral magnification is image height divided by object height. For a spherical mirror it equals minus v over u, and for a thin lens it equals v over u, with the New Cartesian sign convention. A negative value means an inverted real image and a positive value an erect virtual image.

Why is there a minus sign in mirror magnification but not in lens magnification?

Both come from similar triangles. For a mirror, the object and a real image lie on the same side, so u and v have the same sign while the image is inverted, which needs a minus sign. For a lens a real image lies on the opposite side, so v over u is already negative.

How do you find magnification without finding the image distance?

Combine the magnification with the mirror or lens formula. For a mirror, m equals f divided by (f minus u); for a lens, m equals f divided by (f plus u). Both can also be written as (f minus v) divided by f when the image distance is known instead.

What is longitudinal magnification?

Longitudinal magnification describes a short object lying along the principal axis. Differentiating the formulas gives minus m squared for a mirror and plus m squared for a lens. For a long rod the magnification changes along its length, so each end must be imaged separately and the positions subtracted.

How fast does an image move when the object moves along the axis?

Relative to the mirror or lens, the image moves at m squared times the object's speed along the axis. For a mirror the image moves in the opposite direction, for a lens in the same direction. At a distance 2f from a concave mirror or convex lens the image moves at exactly the object's speed.

What is the total magnification of a combination of lenses or mirrors?

Each element forms an image that becomes the object for the next one, and the total magnification is the product of the individual magnifications. Its sign gives the final orientation; for example two inverting lenses give an erect final image.

Is magnification important for JEE Main and JEE Advanced?

Yes. JEE Main asks image size, the two answers to image n times the object, and magnification of combinations. JEE Advanced adds image velocity along and across the axis, longitudinal magnification of rods, moving mirrors and lenses, and images of points off the axis.

What magnification questions come in NEET?

NEET usually asks the magnification formula for mirrors and lenses, finding object position for a given magnification, the nature of images from the sign of m, and the power or focal length from a measured image size. Keep the sign convention consistent throughout.

Previous year questions on Magnification

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

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