Thin Lens and Mirrors

PhysicsRay Optics And Optical InstrumentsFor JEE aspirants

A thin lens refracts light at two curved surfaces so close together that its thickness can be ignored. Its focal length follows from the lens maker's formula $\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ , and images obey the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . This page on thin lenses and mirrors also covers power, lenses in contact, cut and silvered lenses, lens-mirror systems and the displacement method, as asked in JEE Main, JEE Advanced and NEET.

On this page1Lens basics2Lens maker's formula3Lens formula and images4Power and combinations5Cut and silvered lenses6Displacement method7Solved examples

Key Formulas - Quick Reference

★ Must learnLens maker's formula: $\frac{1}{f} = (\frac{n}{n _{s}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
★ Must learnThin lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ; magnification $m = \frac{v}{u}$
★ Must learnPower: $P = \frac{1}{f}$ ( $f$ in metres, $P$ in dioptres); mirror $P = - \frac{1}{f}$
★ Must learnLenses in contact: $P = P_{1} + P_{2} + \dots$ , i.e. $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$
Silvered lens: $P_{eq} = 2 P_{l} + P_{m}$ , i.e. $\frac{1}{F} = \frac{1}{f _{m}} - \frac{2}{f _{l}}$ (acts as a mirror)
Displacement method ( $D > 4 f$ ): $f = \frac{D ^{2} - d ^{2}}{4 D}$ , object height $h_{0} = h_{1} h_{2}$
★ Must learnAchromatic doublet: $\frac{ω _{1}}{f _{1}} + \frac{ω _{2}}{f _{2}} = 0$

1. Lenses: Types and Terms

A lens is a transparent medium bounded by two refracting surfaces, at least one of them curved. It is thin when its thickness is small compared with the radii of curvature and the object and image distances.

Figure 1: Lens shapes. In air a lens thicker at the middle converges light (

f > 0

) and one thinner at the middle diverges it (

f < 0

). The two meniscus shapes (concavo-convex, convexo-concave) look alike: compare the curvatures to decide.

Figure 2: Parts of a lens. The principal axis joins the centres of curvature

C_{1}

and

C_{2}

;

O

is the optical centre, through which a ray passes undeviated.

R_{1}

belongs to the surface the light meets first. (For an equiconvex lens with

n = 1.5

f = R

Principal axis: line through the two centres of curvature. Optical centre $O$ : the point on the axis through which rays pass undeviated (for a thin lens, the centre of the lens).
Foci: rays parallel to the axis converge at (or diverge from) the second focus $F_{2}$ ; rays from the first focus $F_{1}$ emerge parallel. For a thin lens in one medium, both foci are at the same distance $f$ from $O$ .
Sign convention (New Cartesian, as for mirrors): distances from $O$ , positive in the direction of the incident light. Convex (converging) lens $f > 0$ ; concave (diverging) lens $f < 0$ .

Key idea

In air a lens thicker at the centre converges (

f > 0

); one thinner at the centre diverges (

f < 0

2. Lens Maker's Formula

Treat the lens as two refracting surfaces and apply $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ twice, taking the thickness as zero.

First surface (surrounding $n_{s}$ to lens $n$ ): $\frac{n}{v _{1}} - \frac{n _{s}}{u} = \frac{n - n _{s}}{R _{1}}$ .
The image $v_{1}$ is the object for the second surface (lens to surrounding): $\frac{n _{s}}{v} - \frac{n}{v _{1}} = \frac{n _{s} - n}{R _{2}}$ .
Add the equations: $n_{s} (\frac{1}{v} - \frac{1}{u}) = (n - n_{s}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .
For $u = - \infty$ , $v = f$ :
$\frac{1}{f} = (\frac{n}{n _{s}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
and combining the two gives the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

Using the formula:

R_{1}

is the radius of the surface the light meets first;

R > 0

if its centre is on the outgoing side. Equiconvex lens:

R_{1} = R

R_{2} = - R

\frac{1}{f} = \frac{2 ( n - 1 )}{R}

(for

n = 1.5

f = R

). Plano-convex:

f = \frac{R}{n - 1}

. The focal length does not depend on which side faces the light.

Exam Trick

Put the lens in a liquid and its focal length scales with $\frac{n - 1}{n / n _{s} - 1}$ . A glass lens ( $n = 1.5$ ) in water ( $n_{s} = \frac{4}{3}$ ) becomes $\frac{0.5}{0.125} = 4$ times weaker. If $n_{s} = n$ the lens vanishes optically ( $f = \infty$ ); if $n_{s} > n$ a convex lens diverges.

Figure 3: One equiconvex glass lens (

n = 1.5

R = 20 cm

) in four surroundings, rays to scale.

\frac{1}{f} = (\frac{n}{n _{s}} - 1) \frac{2}{R}

gives

f = + 20

+ 80

\infty

and

- 70 cm

. A convex lens converges only while it is optically denser than its surroundings.

3. Lens Formula and Image Formation

A ray through the optical centre goes straight on.
A ray parallel to the axis passes (convex) or appears to pass (concave) through $F_{2}$ .
A ray through (or towards) $F_{1}$ emerges parallel to the axis.

Figure 4:

u = - 30 cm

f = + 10 cm

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

gives

v = + 15 cm

m = \frac{v}{u} = - 0.5

: real, inverted, diminished, between

F_{2}

and

2 F_{2}

Figure 5: Object inside

F

(

u = - 6

f = 10

v = - 15 cm

(same side as the object),

m = + 2.5

. Virtual, erect, magnified: the simple microscope.

Figure 6: Concave lens,

u = - 20

f = - 10

v = - 6.67 cm

m = + 0.33

. A concave lens always gives a virtual, erect, diminished image of a real object, between

O

and

F

Convex lens: object at	Image at	Nature
Infinity	$F_{2}$	Real, inverted, point-sized
Beyond $2 F_{1}$	Between $F_{2}$ and $2 F_{2}$	Real, inverted, diminished
$2 F_{1}$	$2 F_{2}$	Real, inverted, same size
Between $F_{1}$ and $2 F_{1}$	Beyond $2 F_{2}$	Real, inverted, enlarged
$F_{1}$	Infinity	Real, inverted, highly enlarged
Between $F_{1}$ and $O$	Same side as object	Virtual, erect, enlarged
Virtual object ( $u > 0$ )	Between $O$ and $F_{2}$	Real, erect, diminished

A concave lens gives a virtual, erect, diminished image of any real object, between $O$ and $F_{1}$ ; with a virtual object inside its focal length it can give a real image.

Figure 7:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

plotted exactly for

f = \pm 10 cm

. Convex (solid): real image (

v > 0

) whenever

u < - f

. Concave (dashed): a real object (

u < 0

) always gives

- f < v < 0

. Objects with

u > 0

are virtual (converging light).

Exam Trick

Minimum object-image distance for a real image is $4 f$ , reached when $u = - 2 f$ , $v = 2 f$ . If the screen is closer than $4 f$ to the object, no position of the lens gives a sharp image; farther than $4 f$ , there are exactly two positions (the displacement method).

Quick Recall: tap to check

A convex lens (

f = 10 cm

) forms an image twice the size of the object. Where is the object?

15 cm

away (real image,

m = - 2

) or

5 cm

away (virtual image,

m = + 2

Does the focal length of a lens change if it is turned round?

No, for a thin lens in one medium; swapping the faces swaps and changes the sign of

R_{1}

and

R_{2}

, leaving

f

unchanged.

Key idea

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

with signs handles every lens case; real images need

u < - f

for a convex lens.

4. Power and Combinations of Lenses

The power of a lens measures how strongly it bends light: $P = \frac{1}{f}$ with $f$ in metres, in dioptres ( $1 D = 1 m^{- 1}$ ). Converging lenses have positive power. For mirrors, $P = - \frac{1}{f}$ , so a concave (converging) mirror also has positive power.

Thin lenses in contact: the first lens forms an image that acts as the object for the second: $\frac{1}{v _{1}} - \frac{1}{u} = \frac{1}{f _{1}}$ and $\frac{1}{v} - \frac{1}{v _{1}} = \frac{1}{f _{2}}$ . Adding, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ , so the powers simply add: $P = P_{1} + P_{2} + \dots$ The magnification of a combination is the product $m = m_{1} m_{2} \dots$

Figure 8: Lenses in contact add powers:

\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{20} + \frac{1}{30}

, so

F = 12 cm

(

P = 5 D + 3.33 D = 8.33 D

). Dashed: where the first lens alone would focus.

Elements in contact

Add powers: $P = P_{1} + P_{2}$ . One formula, one image. Valid for thin lenses touching each other.

Elements separated

Image by image: the image of the first becomes the object of the next, with $u$ re-measured from the new element. Powers do not simply add.

Key idea

In contact, powers add; separated, trace image by image.

5. Cut, Silvered and Mirror-Lens Systems

5.1 Cutting a lens

Cut along the axis (into upper and lower halves) and kept together: each half has the same $f$ ; there is still one image, only dimmer. If the halves are separated, each has its own axis and forms its own image.
Cut perpendicular to the axis (an equiconvex lens split into two plano-convex lenses): each half has $\frac{1}{f ^{'}} = \frac{n - 1}{R} = \frac{1}{2 f}$ , so $f^{'} = 2 f$ and $P^{'} = \frac{P}{2}$ .

Figure 9: Cutting a lens. Perpendicular to the axis: each half has

f^{'} = 2 f

. Along the axis: each half keeps

f

but has its own axis; with

u = - 15 cm

f = 10 cm

(

m = - 2

) and halves

1 cm

off the axis, the two images are

2 \times (1 + 2) = 6 cm

apart.

5.2 Silvered lens

If the back surface of a lens is silvered (or the lens rests on a mirror), light passes through the lens, reflects, and passes through the lens again. The system is a single mirror with

P_{eq} = 2 P_{l} + P_{m}, \frac{1}{F} = \frac{1}{f _{m}} - \frac{2}{f _{l}}

( $F < 0$ : concave mirror, $F > 0$ : convex mirror, $F = \infty$ : plane mirror).

Figure 10: Silvered lens. Light crosses the lens twice and reflects once, so

P_{eq} = 2 P_{lens} + P_{mirror}

, i.e.

\frac{1}{F} = \frac{1}{f _{m}} - \frac{2}{f _{l}}

. An equiconvex lens (

f_{l} = 10 cm

) on a plane mirror (

f_{m} = \infty

) acts as a concave mirror with

F = - 5 cm

. An object

10 cm

away sits at the lens focus, which is the mirror's

C

: the rays leave the lens parallel, hit the mirror normally and retrace, so the image forms on the object.

JEE Advanced

Lens plus mirror systems. For a lens in front of a mirror, the image is formed on the object itself when the lens sends the light either to the mirror's pole (a point on the mirror is its own image) or to its centre of curvature (the light then strikes the mirror normally and retraces). Write the lens formula for each case; both give valid object positions (Solved Example 7). For a silvered plano-convex lens the curved-side silvering and plane-side silvering give different mirrors: work out $P = 2 P_{l} + P_{m}$ for each.

6. Displacement Method and Chromatic Aberration

With the object and screen a fixed distance $D > 4 f$ apart, a convex lens gives a sharp image at two positions, a distance $d$ apart, with $u$ and $v$ interchanged.

Figure 11: Displacement method (

D = 90 cm

f = 20 cm

). The two lens positions are

d = 30 cm

apart, with

u

and

v

interchanged. Then

f = \frac{D ^{2} - d ^{2}}{4 D} = \frac{8100 - 900}{360} = 20 cm

and the object height is

h_{1} h_{2}

(

m = - 2

and

- 0.5

$u + v = D$ and $v - u = d$ (magnitudes), so $u = \frac{D - d}{2}$ and $v = \frac{D + d}{2}$ .
Lens formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{D}{uv}$ , giving
$f = \frac{D ^{2} - d ^{2}}{4 D}$
The magnifications are $\frac{v}{u}$ and $\frac{u}{v}$ ; their product is $1$ , so $h_{1} h_{2} = h_{0}^{2}$ and $h_{0} = h_{1} h_{2}$ .

Chromatic aberration. Because $n$ depends on colour, the focal length does too: violet focuses closer than red. A white object gives images with coloured fringes. Cementing a convex crown lens to a concave flint lens with $\frac{ω _{1}}{f _{1}} + \frac{ω _{2}}{f _{2}} = 0$ brings two colours to the same focus: an achromatic doublet.

Figure 12: Chromatic aberration (dispersion exaggerated). Since

\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

and

n_{v} > n_{r}

, violet focuses nearer:

f_{v} < f_{r}

. A convex crown lens cemented to a concave flint lens with

\frac{ω _{1}}{f _{1}} + \frac{ω _{2}}{f _{2}} = 0

(achromatic doublet) brings red and violet to one focus.

Exam Trick

In an achromatic doublet the convex lens is made of the less dispersive glass. From $\frac{ω _{1}}{f _{1}} = - \frac{ω _{2}}{f _{2}}$ the lens with smaller $ω$ must have the smaller $∣ f ∣$ (larger power), and it must be the convex one for the pair to converge.

Quick Recall: tap to check

Object and screen are

80 cm

apart. Can a lens of focal length

25 cm

form a sharp image?

No: that needs

D \geq 4 f = 100 cm

An equiconvex lens is placed on a plane mirror. What does the system behave as?

A concave mirror of focal length

\frac{f _{l}}{2}

Key idea

f = \frac{D ^{2} - d ^{2}}{4 D}

turns two screen positions into a focal length;

h_{1} h_{2}

gives the object size.

7. Flowchart and Mind Map

Decide first whether the elements touch: in contact they combine into one; separated they must be taken one at a time.

Figure 13: Problem-solving flowchart. One lens: lens formula. Elements in contact: add powers. Separated elements: image by image, re-measuring distances from each new element.

Figure 14: Thin lenses on one page. Revise from the map, then test yourself on the solved examples.

8. Solved Examples

Solved Example 1

A biconvex glass lens (

n = 1.5

) has radii

60 cm

and

15 cm

. Find its focal length in air.

Solution:

$R_{1} = + 60$ , $R_{2} = - 15$ : $\frac{1}{f} = 0.5 (\frac{1}{60} + \frac{1}{15}) = 0.5 \times \frac{5}{60} = \frac{1}{24}$ .

Answer: $f = + 24 cm$ (converging); the same whichever face meets the light.

Solved Example 2

A meniscus lens (

n = 1.5

) has surfaces of radii

30 cm

and

60 cm

, both with centres on the same side. Find its focal length, and show it is unchanged if the lens is reversed.

Solution:

Light meets the $30 cm$ surface first: $R_{1} = + 30$ , $R_{2} = + 60$ : $\frac{1}{f} = 0.5 (\frac{1}{30} - \frac{1}{60}) = \frac{1}{120}$ . Reversed: $R_{1} = - 60$ , $R_{2} = - 30$ : $\frac{1}{f} = 0.5 (- \frac{1}{60} + \frac{1}{30}) = \frac{1}{120}$ .

Answer: $f = + 120 cm$ both ways (concavo-convex, converging).

Solved Example 3

Light converging towards a point

30 cm

behind a convex lens (

f = 10 cm

) is intercepted by the lens. Where does it focus?

Solution:

Virtual object: $u = + 30$ . $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}$ .

Answer: $v = + 7.5 cm$ , a real image closer than $F$ .

Solved Example 4

An object

2 cm

tall is

30 cm

from a convex lens of focal length

10 cm

. Find the position, size and nature of the image.

Solution:

$\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$ , $v = + 15 cm$ . $m = \frac{v}{u} = \frac{15}{- 30} = - 0.5$ , so $h_{i} = - 1 cm$ .

Answer: $15 cm$ behind the lens, $1 cm$ tall, real and inverted (Figure 4).

Solved Example 5

A convex lens of focal length

10 cm

forms an image twice the size of the object. Find the object distance.

Solution:

Real image ( $m = - 2$ , $v = - 2 u$ ): $\frac{1}{- 2 u} - \frac{1}{u} = \frac{1}{10}$ , $- \frac{3}{2 u} = \frac{1}{10}$ , $u = - 15 cm$ .

Virtual image ( $m = + 2$ , $v = 2 u$ ): $\frac{1}{2 u} - \frac{1}{u} = \frac{1}{10}$ , $u = - 5 cm$ .

Answer: $15 cm$ (real image) or $5 cm$ (virtual image).

Solved Example 6

A combination is made of an equiconcave glass lens (

n = 1.5

R = 10 cm

), an equiconvex water lens (

n = \frac{4}{3}

R = 10 cm

) and another equiconcave glass lens identical to the first, all in contact in air. Find the equivalent focal length.

Solution:

Glass lens: $\frac{1}{f _{1}} = 0.5 (- \frac{2}{10})$ , $f_{1} = - 10 cm$ . Water lens: $\frac{1}{f _{2}} = \frac{1}{3} \times \frac{2}{10}$ , $f_{2} = + 15 cm$ . $\frac{1}{F} = - \frac{1}{10} + \frac{1}{15} - \frac{1}{10} = - \frac{4}{30}$ .

Answer: $F = - 7.5 cm$ (diverging).

Solved Example 7

A convex lens (

f = 10 cm

) and a concave mirror (

R = 20 cm

) face each other

40 cm

apart on a common axis. Where must a point object be placed, on the far side of the lens, for its final image to fall on itself?

Solution:

Case 1: lens images the object on the mirror's pole ( $v = 40 cm$ ): $\frac{1}{x} = \frac{1}{10} - \frac{1}{40} = \frac{3}{40}$ , $x = 13.3 cm$ .

Case 2: lens images it at the mirror's centre of curvature, $20 cm$ in front of the mirror ( $v = 20 cm$ ): the light meets the mirror normally and retraces. $\frac{1}{x} = \frac{1}{10} - \frac{1}{20}$ , $x = 20 cm$ .

Answer: $x = 13.3 cm$ or $x = 20 cm$ from the lens.

Solved Example 8

An equiconvex lens (

n = 1.5

R = 10 cm

) lies on a horizontal plane mirror. How high above it must a point object be for the image to coincide with the object?

Solution:

$f_{l} = \frac{R}{2 ( n - 1 )} = 10 cm$ ; $f_{m} = \infty$ . $\frac{1}{F} = \frac{1}{\infty} - \frac{2}{10}$ , so $F = - 5 cm$ : a concave mirror of focal length $5 cm$ . The image coincides with the object at its centre of curvature, $2∣ F ∣$ away.

Answer: $10 cm$ above the lens.

Solved Example 9

An object and a screen are

90 cm

apart. A convex lens gives sharp images at two positions

30 cm

apart, the images being

2 cm

and

0.5 cm

high. Find

f

and the object height.

Solution:

$f = \frac{D ^{2} - d ^{2}}{4 D} = \frac{8100 - 900}{360} = 20 cm$ . Lens positions: $u = 30 cm$ and $60 cm$ from the object. $h_{0} = 2 \times 0.5 = 1 cm$ .

Answer: $f = 20 cm$ , object $1 cm$ tall (Figure 11).

Solved Example 10

Lenses of power

+ 5 D

and

- 3 D

are placed in contact. The combination is
(A) converging,

f = 50 cm

(B) diverging,

f = 50 cm

f = 12.5 cm

(D) diverging,

f = 33 cm

Solution:

$P = + 5 - 3 = + 2 D$ , positive so converging; $f = \frac{1}{2} m = 50 cm$ .

Answer: (A).

Solved Example 11

A glass lens (

n = 1.5

) has focal length

20 cm

in air. Its focal length in water (

n = \frac{4}{3}

) is
(A)

20 cm

(B)

40 cm

(C)

80 cm

(D)

160 cm

Solution:

$\frac{f _{w}}{f _{a}} = \frac{n - 1}{n / n _{w} - 1} = \frac{0.5}{1.125 - 1} = 4$ , so $f_{w} = 80 cm$ .

Answer: (C).

Solved Example 12

An achromatic doublet of focal length

+ 40 cm

is made of crown glass (

ω = 0.02

) and flint glass (

ω^{'} = 0.04

). Find the focal lengths of the two lenses.

Solution:

$\frac{0.02}{f _{1}} + \frac{0.04}{f _{2}} = 0$ gives $f_{2} = - 2 f_{1}$ . Then $\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{2 f _{1}} = \frac{1}{40}$ , so $f_{1} = 20 cm$ .

Answer: crown convex $f_{1} = + 20 cm$ , flint concave $f_{2} = - 40 cm$ .

Solved Example 13

A convex lens (

f = 10 cm

) is cut along its axis and the halves are moved

1 cm

above and below the axis. A point object lies on the original axis

15 cm

away. Find the separation of the two images.

Solution:

Each half has $f = 10 cm$ : $v = 30 cm$ , $m = - 2$ . For the upper half the object is $1 cm$ below its axis, so the image is $2 cm$ above that axis, i.e. $3 cm$ above the original axis; symmetrically the other image is $3 cm$ below.

Answer: $6 cm$ (Figure 9).

Practice Questions

Find the focal length of a plano-convex lens ( $n = 1.5$ , $R = 20 cm$ ).Answer: $40 cm$
Lenses of focal lengths $+ 20 cm$ and $- 30 cm$ are in contact. Find the focal length and power of the combination.Answer: $+ 60 cm$ ; $+ 1.67 D$
An equiconvex lens of focal length $20 cm$ is cut perpendicular to its axis into two halves. Find the focal length and power of each.Answer: $40 cm$ ; $+ 2.5 D$
Find the power of a concave mirror of focal length $20 cm$ (using $P = - \frac{1}{f}$ ).Answer: $+ 5 D$
In a displacement experiment $D = 100 cm$ and $d = 20 cm$ . Find $f$ .Answer: $24 cm$
The plane face of a plano-convex lens ( $n = 1.5$ , $R = 20 cm$ ) is silvered. What does it act as?Answer: A concave mirror of focal length $20 cm$ ( $P = 2 P_{l}$ )
Where must an object be placed in front of a convex lens of $f = 12 cm$ to get an image three times its size on a screen?Answer: $16 cm$ from the lens ( $v = 48 cm$ )

Common Mistakes to Avoid

Watch out

Using $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (the mirror formula) for a lens. For a lens it is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .
Getting the signs of $R_{1}$ and $R_{2}$ wrong: an equiconvex lens has $R_{1} = + R$ , $R_{2} = - R$ , not both positive.
Forgetting the surrounding medium: in a liquid use $\frac{n}{n _{s}} - 1$ , not $n - 1$ .
Adding focal lengths for lenses in contact. Add powers (reciprocals), not focal lengths.
Adding powers of separated lenses. For separated elements, trace image by image.
For a silvered lens, counting the lens power once. The light crosses the lens twice: $P = 2 P_{l} + P_{m}$ .
Using $f$ in centimetres in $P = \frac{1}{f}$ . Convert to metres to get dioptres.
Thinking half a lens gives half an image. Each part forms the complete image, only fainter.

Frequently Asked Questions

What is the lens maker's formula?

The lens maker's formula gives the focal length of a thin lens from its refractive index and radii: 1 over f equals (n minus 1) times (1 over R1 minus 1 over R2), where R1 is the surface the light meets first. In a liquid, replace n by the ratio of the lens index to the liquid index.

What is the difference between the lens formula and the mirror formula?

The thin lens formula is 1 over v minus 1 over u equals 1 over f, while the mirror formula is 1 over v plus 1 over u equals 1 over f, both with the New Cartesian sign convention. The difference comes from light passing through a lens but returning from a mirror.

What is the power of a lens?

Power is the reciprocal of the focal length in metres, measured in dioptres. A converging lens has positive power and a diverging lens negative power. For thin lenses in contact the powers add, so a +5 D and a -3 D lens together act like a single +2 D lens of focal length 50 cm.

What happens to the focal length when a lens is cut in half?

It depends on the cut. Cutting along the principal axis leaves each half with the same focal length; each half still forms a full image, only dimmer. Cutting perpendicular to the axis, as when an equiconvex lens is split into two plano-convex lenses, doubles the focal length of each half.

How does a silvered lens behave?

When one surface of a lens is silvered, light passes through the lens, reflects and passes through the lens again. The system behaves as a single mirror whose power is twice the lens power plus the mirror power. An equiconvex lens on a plane mirror acts as a concave mirror of half the lens focal length.

What is the displacement method for finding the focal length?

With object and screen fixed a distance D apart, greater than four times the focal length, a convex lens gives a sharp image at two positions a distance d apart. Then f equals (D squared minus d squared) divided by 4D, and the object height is the square root of the product of the two image heights.

Are thin lenses important for JEE Main and JEE Advanced?

Yes. JEE Main regularly asks lens maker and lens formula numericals, power of combinations and focal length in a liquid. JEE Advanced adds silvered lenses, lens-mirror systems where the image falls on the object, cut lenses, the displacement method and achromatic doublets.

Which lens questions are common in NEET?

NEET mostly asks the lens maker formula, image position and magnification with the lens formula, power in dioptres, power of lenses in contact, the change of focal length when a lens is immersed in water, and the nature of images formed by convex and concave lenses.

Previous year questions on Thin Lens and Mirrors

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

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