JEE Mains Physics · Ray Optics

Lenses in a Medium, Cut Lenses and Silvered Lenses

In a medium a lens's power scales with μ(lens)/μ(medium) − 1; a lens cut through its axis keeps its focal length while one cut across it doubles it; a silvered lens is a mirror whose power is twice the lens's power plus that of the silvered face.

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Why this matters

Eighteen PYQs, thirteen of them multiple choice, and two from 2026. Ten put a lens in water or another liquid, including a liquid denser than the glass and a liquid trapped between two lenses. Four cut a lens into pieces, along the axis or across it. Four silver one face of a lens, so that it acts as a mirror.

Concept 1 of 3: A lens in a liquid

A lens bends light because its glass is optically denser than its surroundings. In water the difference is smaller, so the lens is weaker and its focal length longer. In a liquid with the same index as the glass the lens vanishes optically. In a denser liquid a convex lens spreads light out and acts as a diverging lens.

Definition

$\dfrac{1}{f_m} = \left(\dfrac{\mu_l}{\mu_m} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ , with $\mu_l$ the lens and $\mu_m$ the medium.
The radii do not change, so $\dfrac{f_m}{f_{\text{air}}} = \dfrac{\mu_l - 1}{\mu_l/\mu_m - 1}$ .
$\mu_m = \mu_l$ : no refraction, f is infinite. $\mu_m > \mu_l$ : f changes sign, and a convex lens diverges.
For a lens described in air, first find the curvature factor $\dfrac{1}{R_1} - \dfrac{1}{R_2}$ from the air data, then apply the new factor.
A liquid filling the gap between two lenses is a third lens (concave, between two convex faces); add all three powers.

Lens in a medium

\frac{1}{f_m} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right), \qquad \frac{f_m}{f_{\text{air}}} = \frac{\mu_l - 1}{\mu_l/\mu_m - 1}

Worked example

A convex glass lens (

\mu = 1.5

) has a focal length of 12 cm in air. Find its focal length (a) in water,

\mu = 4/3

, and (b) in a liquid of

\mu = 1.75

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q15Moderate

Example 1 · Ray Optics · Lenses in a Medium, Cut Lenses and Silvered Lenses

A biconvex lens of refractive index 1.5 has a focal length of

20\text{ }cm

in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

The radii stay; only the factor changes

Immersing a lens does not change its shape. Find the curvature factor from the air data, then multiply it by the new μ(lens)/μ(medium) − 1.

A convex lens can diverge

If the liquid is optically denser than the glass, μ(lens)/μ(medium) − 1 is negative and the converging lens becomes a diverging one.

Use the ratio of the factors

The focal length in water is neither f(air) × μ(water) nor f(air)/μ(water). It follows from the ratio of the two (relative index − 1) factors.

Drill 9 more on a lens in a liquid

Concept 2 of 3: Focal length of the pieces of a cut lens

A cut that contains the principal axis leaves both curved faces as they were, only smaller, so each half has the full focal length. A cut across the axis, through the centre, splits a biconvex lens into two plano-convex lenses, each with one curved face, so each has half the power and twice the focal length. Put the pieces back together and the original lens returns.

Definition

Focal length depends on the curvatures and μ, not on the size of the lens. A smaller aperture only makes the image dimmer.
A plane containing the axis: each piece keeps f and P.
A plane perpendicular to the axis, through the centre of an equiconvex lens: each piece is plano-convex with $f' = 2f$ and $P' = P/2$ ; the two in contact give back P.
Cut both ways: apply the two rules in turn.

How the lens is cut	Each piece is	Focal length of a piece	Power of a piece
Along a plane containing the principal axis	Half of the same lens, both curved faces kept	$f$	$P$
Across, perpendicular to the axis, through the centre	A plano-convex lens	$2f$	$P/2$
Along the axis, then one half across it	A plano-convex quarter	$2f$	$P/2$
Half the lens covered, not cut	The whole lens, with less light	$f$	$P$ The image is complete, only dimmer.
Two plano-convex halves put back together	The original lens	$f$	$P$

For an equiconvex lens of focal length f and power P. Only a cut across the axis changes the focal length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q111Moderate

Example 2 · Ray Optics · Lenses in a Medium, Cut Lenses and Silvered Lenses

A bi convex lens of focal length

10\text{ }cm

is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is

\_\_\_\_

A smaller lens is not a weaker lens

Cutting along the axis halves the size, not the power. The focal length depends only on the curvatures and on μ.

Across the axis, the power halves

Each plano-convex piece has one curved face instead of two, so its power is half and its focal length twice the original.

Drill 3 more on focal length of the pieces of a cut lens

Concept 3 of 3: A silvered lens as a mirror

Silver one face of a lens and light enters, crosses the lens, reflects off the silver and crosses the lens again. So the system is a mirror. Its power is the lens's power twice, for the two passes, plus the power of the silvered face as a mirror. An object at the centre of curvature of this equivalent mirror is imaged onto itself.

Definition

Take converging powers as positive. $P = 2P_L + P_M$ , where $P_L$ is the lens's power and $P_M = 2/R$ is the silvered face's power as a concave mirror seen from inside. The system is a concave mirror of focal length $F = 1/P$ .
Plane face silvered: $P_M = 0$ , so $F = f_L/2$ .
Equiconvex lens, both radii R, one face silvered: $P = \dfrac{4(\mu - 1)}{R} + \dfrac{2}{R} = \dfrac{2(2\mu - 1)}{R}$ .
Image on the object itself: place the object at the equivalent mirror's centre of curvature, a distance 2F from the lens.
In a liquid, find the lens's power with $\mu_l/\mu_m - 1$ . A silvered plane face still adds nothing.

Silvered lens

P = 2P_L + P_M, \qquad F = \frac{1}{P}, \qquad F = \frac{f_L}{2}\ (\text{plane face silvered})

Worked example

A plano-convex lens (

\mu = 1.5

) has a curved face of radius 30 cm. Find the focal length of the equivalent mirror when (a) its plane face is silvered, and (b) its curved face is silvered instead.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q20Moderate

Example 3 · Ray Optics · Lenses in a Medium, Cut Lenses and Silvered Lenses

Given is a thin convex lens of glass (refractive index

\mu

) and each side having radius of curvature

R

. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?

The lens counts twice

Light crosses the lens on the way in and again on the way out, so the lens's power enters as 2P(lens), not P(lens).

A silvered plane face adds no power

A plane mirror has zero power, so with the flat face silvered the system's focal length is just half the lens's.

The silvered curved face is concave from inside

Light inside the glass sees a silvered convex surface as a concave mirror of radius R, with power 2/R.

Drill 3 more on a silvered lens as a mirror

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

A lens in a liquid
Lens in a medium
$\frac{1}{f_m} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right), \qquad \frac{f_m}{f_{\text{air}}} = \frac{\mu_l - 1}{\mu_l/\mu_m - 1}$
A silvered lens as a mirror
Silvered lens
$P = 2P_L + P_M, \qquad F = \frac{1}{P}, \qquad F = \frac{f_L}{2}\ (\text{plane face silvered})$

Reference tables (1)

Focal length of the pieces of a cut lens5 rows

How the lens is cut	Each piece is	Focal length of a piece	Power of a piece
Along a plane containing the principal axis	Half of the same lens, both curved faces kept	$f$	$P$
Across, perpendicular to the axis, through the centre	A plano-convex lens	$2f$	$P/2$
Along the axis, then one half across it	A plano-convex quarter	$2f$	$P/2$
Half the lens covered, not cut	The whole lens, with less light	$f$	$P$ The image is complete, only dimmer.
Two plano-convex halves put back together	The original lens	$f$	$P$

For an equiconvex lens of focal length f and power P. Only a cut across the axis changes the focal length.

Watch out for (8)

The radii stay; only the factor changes
→ A lens in a liquid
A convex lens can diverge
→ A lens in a liquid
Use the ratio of the factors
→ A lens in a liquid
A smaller lens is not a weaker lens
→ Focal length of the pieces of a cut lens
Across the axis, the power halves
→ Focal length of the pieces of a cut lens
The lens counts twice
→ A silvered lens as a mirror
A silvered plane face adds no power
→ A silvered lens as a mirror
The silvered curved face is concave from inside
→ A silvered lens as a mirror

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