A screen is placed $90\ \text{cm}$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20\ \text{cm}$. Determine the focal length of the lens.

A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at tw

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Concept understanding — Lens Maker's Formula

The Intuition: Why a Lens Bends Light

A lens works because light slows down when it enters glass. When a wavefront hits a curved surface at an angle, different parts of it slow down at different moments, and the wavefront bends. The stronger the curvature, the more it bends.

A lens has two surfaces. Each surface bends light by an amount that depends on its radius of curvature $R$ and the refractive index $n$ of the glass. The net bending — the focal length $f$ — is the combined effect of both surfaces.

If you had a single spherical surface separating air from glass, its contribution to bending power is $\frac{n-1}{R}$ . A lens has two such surfaces: light goes from air into glass at the first surface, then from glass back into air at the second. Because the two surfaces face opposite directions relative to the travelling light, their radii typically carry opposite signs.

Note

This uses the New Cartesian Sign Convention (the one used in NCERT and CBSE): all distances are measured from the optical centre, and the direction the incident light travels in is taken as positive. So $R$ is positive if the centre of curvature lies on the side the light is travelling towards (the outgoing side), and negative if it lies on the side the light is travelling from (the incident side).

The Precise Statement

For a thin lens (thickness negligible compared to the radii), the Lens Maker's Formula is:

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

where:

$f$ is the focal length of the lens (positive for converging, negative for diverging)
$n$ is the refractive index of the lens material relative to the surrounding medium (usually air)
$R_1$ is the radius of curvature of the first surface (the one light reaches first)
$R_2$ is the radius of curvature of the second surface

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

How to Apply It: A Worked Example

Take a biconvex lens made of glass ( $n = 1.5$ ) with both surfaces having the same radius of curvature magnitude, $20$ cm.

Light travels left to right. The first surface bulges toward the incoming light, so its centre of curvature lies to the right of the surface — on the side the light is travelling towards. By the rule above, $R_1 = +20$ cm.

The second surface also bulges outward (away from the lens), so its centre of curvature lies to the left of that surface — on the side the light is travelling from. So $R_2 = -20$ cm.

Plug in:

\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}

So $f = +20$ cm. Positive means converging — correct for a biconvex lens.

Watch out

The most common mistake is getting the sign of $R_2$ wrong. For a biconvex lens, $R_1$ is positive and $R_2$ is negative. For a biconcave lens, it's the reverse: $R_1$ negative, $R_2$ positive. Always sketch the lens and mark where each surface's centre of curvature actually sits.

Why the Formula Works (Brief Derivation) …

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For a fixed object-screen distance $D$ , a convex lens forms a sharp image at two positions separated by $d$ (displacement method). The focal length is $f = \frac{D^2 - d^2}{4D}$ . Here $D = 90\ \text{cm}$ , $d = 20\ \text{cm}$ , so $f = \frac{8100 - 400}{360} = \frac{7700}{360} \approx 21.39\ \text{cm}$ .

The displacement method for finding the focal length of a convex lens is a classic experiment — and a favourite in exams — because it avoids the need to measure object and image distances separately. The key insight: when the object and screen are fixed at a separation $D$ greater than $4f$ , there are two distinct lens positions that produce a sharp image on the screen. One gives a magnified image, the other a diminished one. The distance between these two positions, $d$ , together with $D$ , directly gives $f$ .

Why does this happen? For a given object distance $u$ , the lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ with $v = D - u$ becomes a quadratic in $u$ . Two real roots exist when $D > 4f$ , and the difference between them is exactly $d = \sqrt{D^2 - 4fD}$ . Rearranging gives the neat formula.

Let’s work through it step by step.

Set up the geometry. Object and screen are fixed $90\ \text{cm}$ apart. So $D = 90\ \text{cm}$ . Let the lens be at a distance $u$ from the object. Then the image distance from the lens is $v = D - u$ (since the screen is on the other side). The lens formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{D - u}.

Form the quadratic in $u$ . Combine the fractions:

\frac{1}{f} = \frac{D - u + u}{u(D - u)} = \frac{D}{u(D - u)}.

u(D - u) = Df \quad \Rightarrow \quad -u^2 + Du - Df = 0.

Multiply by $-1$ :

u^2 - Du + Df = 0.

Two solutions — the two lens positions. This quadratic has two roots $u_1$ and $u_2$ (the two object distances for which a sharp image forms). Their sum and product:

u_1 + u_2 = D, \quad u_1 u_2 = Df.

The distance between the two lens positions is $d = |u_1 - u_2|$ .

Using the identity $(u_1 - u_2)^2 = (u_1 + u_2)^2 - 4u_1 u_2$ , we get

d^2 = D^2 - 4Df.

Solve for $f$ . Rearranging: …

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Here are the common mistakes students make on this classic displacement method problem, along with how to avoid each.

1. Confusing the two lens positions

Mistake:

Students think the two positions are symmetric about the midpoint, or they try to guess which distance is $u$ and which is $v$ without using the formula.

Why it happens:

The problem says the lens forms a sharp image at two positions separated by $20\ \text{cm}$ . Many assume one position is obvious — it is not.

How to avoid:

Use the displacement method formula directly:

If $D$ = distance between object and screen, and $d$ = separation between the two lens positions, then

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

Here:

$D = 90\ \text{cm}$
$d = 20\ \text{cm}$

So:

f = \frac{90^2 - 20^2}{4 \times 90} = \frac{8100 - 400}{360} = \frac{7700}{360} \approx 21.39\ \text{cm}

Key takeaway: Memorise the formula — it saves time and avoids confusion.

2. Forgetting that $u$ and $v$ swap between positions

Mistake:

Students set up $u$ and $v$ for one position, then use the same values for the second position.

Why it happens:

They don’t realise that in the displacement method, the two positions correspond to interchanging object and image distances.

How to avoid:

Remember:

At position 1: $u_1$ , $v_1$
At position 2: $u_2 = v_1$ , $v_2 = u_1$

This is the core symmetry. The lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ gives the same $f$ for both.

3. Using the lens formula incorrectly with the given numbers

Mistake:

Plugging $u = 90$ or $v = 90$ directly into $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .

Why it happens:

They think the 90 cm is either $u$ or $v$ , but it’s actually $u + v = D$ .

How to avoid:

Always write:

u + v = D

Then use the fact that the two positions give $u_1 - u_2 = d$ (or $v_1 - v_2 = d$ ).

Solve the system:

u + v = 90

v - u = 20

This gives $v = 55$ , $u = 35$ (or vice versa). Then:

\frac{1}{f} = \frac{1}{35} + \frac{1}{55} \implies f \approx 21.39\ \text{cm}

4. Sign convention errors

Mistake: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$10\ \text{cm}$
›Reveal solutionSolution

Cutting a symmetric concave lens through its middle (perpendicular to the principal axis) leaves each piece with only one curved surface, so each plano-concave piece has half the power — and therefore double the focal length: 20 cm, option (A).

Concept and Intuition

The lens maker's formula relates a thin lens's focal length to its two radii of curvature and the refractive index of the material:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
A symmetric biconcave lens has two curved surfaces, and each contributes equally to the total power. When the lens is cut through its middle by a plane perpendicular to the principal axis, each piece keeps one original curved surface and gains a flat face. A flat surface has an infinite radius of curvature, so it contributes nothing to the power — each piece is left with only half the original bending power.

Step-by-Step Solution

1. Apply the formula to the original biconcave lens.

Use the New Cartesian sign convention with light travelling left to right. For a biconcave lens of equal radii of magnitude $R$ : the first surface's centre of curvature lies on the incident (left) side, so $R_1 = -R$ ; the second surface's centre of curvature lies on the outgoing (right) side, so $R_2 = +R$ . Then
$\frac{1}{f} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{+R}\right) = -\frac{2(\mu-1)}{R}$
The negative sign confirms a diverging lens. With $f = -10\ \text{cm}$ :
$\frac{\mu-1}{R} = \frac{1}{20}\ \text{cm}^{-1}$
2. Apply it to one plano-concave piece.

Each piece keeps one curved surface ( $R_1 = -R$ ) and gains a flat cut face ( $R_2 = \infty$ ):
$\frac{1}{f'} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{\mu-1}{R} = -\frac{1}{20}\ \text{cm}^{-1}$
3. Read off the result.

$f' = -20\ \text{cm}$ …
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CBSE 2026Set 55/3/11 markMCQ
$\mu$
›Reveal solutionSolution
We calculate the focal lengths of the plano-convex and equi-concave lenses separately using the lens maker's formula, applying the correct sign conventions for radii of curvature. Then, we combine these focal lengths to find the equivalent focal length of the system. The focal length of the combination is $\boxed{-\dfrac{R}{\mu-1}}$ .

Figure — plano-convex and equi-concave lens combination

Concept and Intuition

To find the focal length of a combination of thin lenses kept in contact, we first need to determine the focal length of each individual lens. The fundamental tool for this is the Lens Maker's Formula.

Lens Maker's Formula

The focal length $f$ of a thin lens made of a material with refractive index $\mu$ (relative to the surrounding medium, usually air, for which $\mu_{air} = 1$ ) is given by:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Sign Convention for Radii of Curvature

We will use the following convention for $R_1$ and $R_2$ in the lens maker's formula, assuming light travels from left to right:

$R_1$ (First Surface):

If the first surface is convex (bulges towards the right), $R_1$ is positive ( $+R$ ).

If the first surface is concave (bulges towards the left), $R_1$ is negative ( $-R$ ).

If the first surface is flat (plano), $R_1$ is infinite ( $\infty$ ).

$R_2$ (Second Surface):

If the second surface is convex (bulges towards the left), $R_2$ is negative ( $-R$ ).

If the second surface is concave (bulges towards the right), $R_2$ is positive ( $+R$ ).

If the second surface is flat (plano), $R_2$ is infinite ( $\infty$ ).

Watch out

The sign convention for $R_1$ and $R_2$ is a common source of error. Always be consistent with the convention you choose. The one outlined above ensures that converging lenses have positive focal lengths and diverging lenses have negative focal lengths when $\mu > 1$ .

Combination of Thin Lenses in Contact

When two thin lenses with focal lengths $f_1$ and $f_2$ are placed coaxially in contact, the focal length $F$ of the combination is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Step-by-step Solution

Identify the properties of the plano-convex lens (Lens 1).

Refractive index: $\mu$

First surface: Flat. According to our sign convention, $R_1 = \infty$ .

Second surface: Convex. It bulges towards the left (as seen from the second surface, or its center of curvature is to the left). According to our sign convention, $R_2 = -R$ .

Calculate the focal length of the plano-convex lens ( $f_1$ ).

Using the lens maker's formula:

$\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substitute the values for $R_1$ and $R_2$:
$\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right)$ $\frac{1}{f_1} = (\mu - 1) \left( 0 + \frac{1}{R} \right)$ $\frac{1}{f_1} = \frac{\mu - 1}{R}$
Therefore, the focal length of the plano-convex lens is:
$f_1 = \frac{R}{\mu - 1}$
This is a positive focal length, as expected for a converging lens.

3. Identify the properties of the equi-concave lens (Lens 2).

* Refractive index: $\mu$ …
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CBSE 2026Set DS1 mark
$Q. In which the power of a lens will be large — in air or water?$
›Reveal solutionSolution
Power is larger in air, because the glass–water relative refractive index is smaller than the glass–air one.

Concept. By the lens-maker's formula the power of a lens depends on the refractive index of the lens relative to its surroundings:

$P = \frac1f = \left({}_{m}n_g - 1\right)\left(\frac1{R_1}-\frac1{R_2}\right),$

where ${}_{m}n_g = n_g/n_m$ is the index of glass with respect to the medium.

In air ( $n_m \approx 1$ ): ${}_{a}n_g \approx 1.5$ , so $({}_{a}n_g-1) \approx 0.5$ .

In water ( $n_m \approx 1.33$ ): ${}_{w}n_g = 1.5/1.33 \approx 1.13$ , so $({}_{w}n_g-1) \approx 0.13$ . …
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CBSE 2025Set D1 markMCQ
$Q.A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the material of the lens. Then its focal length will (A) become zero (B) become infinity (C) reduce (D) increase$
›Reveal solutionSolution

Lensmaker's formula has a factor (n_lens/n_medium − 1); if the two indices are equal this factor is zero, so 1/f = 0 and f → ∞.

By the lensmaker's formula in a medium,

1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂)

If the liquid's refractive index equals the lens material's index, then n_lens/n_medium = 1, so the factor (1 − 1) = 0.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When monochromatic red light is used instead of blue light in a convex lens, its focal length$
(a) does not change
(b) increases
(c) decreases
(d) remain same
›Reveal solutionSolution

Red light has a lower refractive index than blue (dispersion), and $f \propto 1/(n-1)$ , so lower $n$ gives a larger $f$ .

By the lens maker's formula, $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ , so $f \propto \dfrac{1}{(n-1)}$ for fixed geometry. Due to dispersion, the refractive index of a material is slightly higher for blue light than for red light ( $n_{blue} > n_{red}$ , since blue light bends more). Using red …

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CBSE 2024Set FS1 mark
$Q. Find the ratio of focal length of lens in air and that of lens when it is immersed in liquid.$
›Reveal solutionSolution

$\dfrac{f_{air}}{f_{liquid}}=\dfrac{n_g-n_l}{n_l\,(n_g-1)}$ , where $n_g$ , $n_l$ are the refractive indices of glass and the liquid.

Concept. The lens maker's formula uses the index of the lens relative to its surroundings:

$\frac{1}{f}=\left(\,{}_{med}n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

In air ( $n_{air}=1$ ): $\dfrac{1}{f_{air}}=(n_g-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).$

In liquid: the glass index relative to the liquid is $n_g/n_l$ , so …

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CBSE 2024Set ANNUAL1 mark
$Q. The power of a lens is greater in water or air?$
›Reveal solutionSolution
A glass lens is more powerful in air than in water, because water's refractive index is closer to that of glass.

Power of a lens, $P = 1/f$ , and from the lens maker's formula:

$\frac{1}{f} = \left(\frac{{}_a\mu_g}{{}_a\mu_m} - 1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where ${}_a\mu_m$ is the refractive index of the surrounding medium (relative to air).

In air, ${}_a\mu_m = 1$ , so the relative refractive index of glass w.r.t. the medium is large ( $\approx 1.5$ ), giving a small $f$ and large $P$ . …
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CBSE 2024Set ANNUAL1 markMCQ
$Q.The refractive index of the material of a double equiconvex lens is 2.5. If R be its radius of curvature, then its focal length is$
(a) 0
(b) R/3
(c) 2R
(d) 3R.
›Reveal solutionSolution

Applying the lens maker's formula to an equiconvex lens (equal radii of curvature, opposite sign) with refractive index 2.5 gives $f = R/3$ .

The lens maker's formula is

$\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

For a double equiconvex lens, both surfaces bulge outward with the same radius of curvature magnitude $R$ . Using the standard sign convention (distances measured from the optical centre, in the direction of incident light being positive): the first surface is convex towards the incoming light, so $R_1 = +R$ ; the second surface's centre of …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The focal length of a glass (μ=1.5) lens in air is 20cm. If it is dipped in water (μ = 4/3), its focal length in water will be -$
(a) 80 cm
(b) 40 cm
(c) 60 cm
(d) 20 cm
›Reveal solutionSolution

A lens's focal length depends on its refractive index relative to the surrounding medium; a converging lens weakens (longer $f$ ) when moved from air into a denser medium like water.

Lensmaker's equation: $\dfrac{1}{f} = \left(\dfrac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

In air: $\mu_{rel} = 1.5/1 = 1.5$

$\dfrac{1}{20} = (1.5-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \Rightarrow \left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) = \dfrac{1}{10}$

…

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CBSE 2023Set BS1 mark
$n = \dfrac{3}{2}$
›Reveal solutionSolution

With glass and surrounding liquid of the same index, $f\to\infty$ ; the lens stops acting as a lens.

The lens-maker's formula in a medium uses the relative refractive index $n_{rel}=\dfrac{n_{glass}}{n_{liquid}}$ :

$\frac{1}{f}=(n_{rel}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$ …

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CBSE 2023Set F1 markMCQ
$Q.The radius of curvature of each surface of a biconvex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length of the lens is (A) 20 m (B) 1/20 m (C) 20 cm (D) 1/20 cm$
›Reveal solutionSolution

Lens maker's formula gives f = 20 cm for a biconvex lens with R = 20 cm and n = 1.5.

For a thin lens, the lens maker's formula is

$\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

For a biconvex lens with each surface radius 20 cm, using sign convention $R_1 = +20$ cm and $R_2 = -20$ cm:

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q.A convex lens is immersed in a liquid of refractive index greater than that of glass. It will behave as a -$
(a) convergent lens
(b) divergent lens
(c) plane glass
(d) homogeneous liquid
›Reveal solutionSolution

When the surrounding medium is optically denser than the lens material, a convex lens's effect reverses.

…

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Q.A screen is placed $90\ \text{cm}$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20\ \text{cm}$ . Determine the focal length of the lens.

Concept understanding — Lens Maker's Formula

The Intuition: Why a Lens Bends Light

The Precise Statement

How to Apply It: A Worked Example

Why the Formula Works (Brief Derivation) …

Method: Displacement Method (for Convex Lens)

Step 1: Understand the given data

Step 2: Recall the formula

Step 3: Substitute the values

Step 4: Simplify

1. Confusing the two lens positions

2. Forgetting that $u$ and $v$ swap between positions

3. Using the lens formula incorrectly with the given numbers

4. Sign convention errors

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Lens Maker's Formula

Sign Convention for Radii of Curvature

Combination of Thin Lenses in Contact

Step-by-step Solution

Q.A screen is placed 90 cm90\ \text{cm}90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm20\ \text{cm}20 cm. Determine the focal length of the lens.

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Q.A screen is placed $90\ \text{cm}$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20\ \text{cm}$ . Determine the focal length of the lens.

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