What is the focal length of a convex lens of focal length $30\ \text{cm}$ in contact with a concave lens of focal length $20\ \text{cm}$? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length

Exercises · 9.10

30\ \text{cm}

CBSENCERTSubjective· 2mImportance★★★★★

25% · 18/73 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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) …

Start your 14-day free trial to unlock the full solution →

When two thin lenses are placed in contact, the net power is the algebraic sum of their individual powers. For a convex lens ( $f=+30\ \text{cm}$ ) and a concave lens ( $f=-20\ \text{cm}$ ), the combined focal length is $f_{\text{eq}} = -60\ \text{cm}$ , making the system a diverging lens.

The key to solving this lies in understanding how lenses combine when placed in contact. Instead of working directly with focal lengths, it's far easier to use power — the quantity that adds linearly.

P = \frac{1}{f(\text{in metres})}

A convex lens converges light — by convention, its focal length is positive. A concave lens diverges light — its focal length is negative. This sign convention is crucial.

Let's work through it step by step.

Write the focal lengths with correct signs.

Convex lens: $f_1 = +30\ \text{cm} = +0.30\ \text{m}$

Concave lens: $f_2 = -20\ \text{cm} = -0.20\ \text{m}$
Convert each to power.

$P_1 = \frac{1}{+0.30} = +\frac{10}{3}\ \text{D} \approx +3.33\ \text{D}$

$P_2 = \frac{1}{-0.20} = -5\ \text{D}$
Add the powers.

$P_{\text{eq}} = P_1 + P_2 = \frac{10}{3} - 5 = \frac{10}{3} - \frac{15}{3} = -\frac{5}{3}\ \text{D}$

The negative sign tells us the combination behaves like a diverging lens overall.
Convert back to focal length.

$f_{\text{eq}} = \frac{1}{P_{\text{eq}}} = \frac{1}{-5/3} = -\frac{3}{5}\ \text{m} = -0.60\ \text{m} = -60\ \text{cm}$ …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this exact problem, along with the concept-first reasoning to avoid each.

Mistake 1: Using the wrong sign for the concave lens focal length

The mistake:

Plugging $f_2 = +20\ \text{cm}$ into the lens combination formula.

Why it’s wrong:

By the Cartesian sign convention (used in all Indian board exams):

Convex lens → $f$ is positive.
Concave lens → $f$ is negative.

So for this problem:

$f_1 = +30\ \text{cm}$ (convex)
$f_2 = -20\ \text{cm}$ (concave)

How to avoid:

Always write the sign explicitly before substituting. Memorise:

Convex = positive, Concave = negative (for focal length).

Mistake 2: Adding focal lengths directly

The mistake:

Thinking the combined focal length $F$ is $f_1 + f_2$ .

Why it’s wrong:

Lenses in contact do not add like resistors in series. The correct formula is:

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

This comes from the fact that the total power (in dioptres) adds:

P_{\text{total}} = P_1 + P_2

Since $P = \frac{1}{f\ (\text{in metres})}$ , the focal lengths combine via reciprocals.

How to avoid:

Write the formula every time before substituting numbers. Never shortcut to $F = f_1 + f_2$ .

Mistake 3: Forgetting to convert units

The mistake:

Using cm directly in the power formula without converting to metres.

Why it’s not always wrong:

If you keep everything in cm, the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ still works as long as all $f$ values are in the same unit. The answer $F$ will also be in that unit.

But the trap:

If you later convert $F$ to power in dioptres, you must use metres. Many exam questions ask for power separately.

How to avoid:

For focal length: work entirely in cm (or m), but stay consistent.
For power: always convert to metres: $P = \frac{1}{f\ (\text{m})}$ .

Mistake 4: Misinterpreting the sign of the final answer

The mistake:

Getting $F = +60\ \text{cm}$ and calling the system converging.

Why it’s wrong:

Let’s do the correct calculation:

\frac{1}{F} = \frac{1}{+30} + \frac{1}{-20} = \frac{1}{30} - \frac{1}{20}

\frac{1}{F} = \frac{2 - 3}{60} = -\frac{1}{60}

So:

F = -60\ \text{cm}

Negative focal length → the system behaves as a diverging lens.

How to avoid:

After calculating, check the sign:

$F > 0$ → converging (convex-like)
$F < 0$ → diverging (concave-like)

--- …

Start your 14-day free trial to unlock the full solution →

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.

$f' = -20\ \text{cm}$ …
Start your 14-day free trial to see the full solution →
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: $\mu$ …
Start your 14-day free trial to see the full solution →
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 $({}_{w}n_g-1) \approx 0.13$ . …
Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $P$ . …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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: $\mu_{rel} = 1.5/1 = 1.5$

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

…

Start your 14-day free trial to see the full solution →
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 $n_{rel}=\dfrac{n_{glass}}{n_{liquid}}$ :

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

Start your 14-day free trial to see the full solution →
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:

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →

←Previous question9.9 — An object of size 3.0cm is placed 14cm in front of a concave lens of f…→Next question9.11 — A compound microscope consists of an objective lens of focal length 2.…

Q.What is the focal length of a convex lens of focal length $30\ \text{cm}$ in contact with a concave lens of focal length $20\ \text{cm}$ ? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

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: Power of a Lens System in Contact

Steps

Mistake 1: Using the wrong sign for the concave lens focal length

Mistake 2: Adding focal lengths directly

Mistake 3: Forgetting to convert units

Mistake 4: Misinterpreting the sign of the final answer

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.What is the focal length of a convex lens of focal length 30 cm30\ \text{cm}30 cm in contact with a concave lens of focal length 20 cm20\ \text{cm}20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

🎓Unlock everything free for 14 days

Q.What is the focal length of a convex lens of focal length $30\ \text{cm}$ in contact with a concave lens of focal length $20\ \text{cm}$ ? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Unlock everything free for 14 days