(a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed $8.0\ \text{cm}$ apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object $1.5\ \text{cm}$ in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is $40\ \text{cm}$. Determine the magnification produced by the two-lens system, and the size of the image.

(a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are

Exercises · 9.20

8.0\ \text{cm}

(b) An object

1.5\ \text{cm}

in size is placed on the side of the convex lens in the arrangement

(a) above. The distance between the object and the convex lens is

40\ \text{cm}

. Determine the magnification produced by the two-lens system, and the size of the image.

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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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Tracing parallel light through the separated pair gives an emergent beam that appears to come from $220\ \text{cm}$ (light entering the convex side) or $420\ \text{cm}$ (entering the concave side); the two differ, so a single 'effective focal length' is not useful. In (b) the system gives $m=\tfrac{15}{23}\approx0.65$ and an image $\approx0.98\ \text{cm}$ tall.

(a) Effective focal length

The lenses of Exercise 9.10 are $f_1=+30\ \text{cm}$ (convex) and $f_2=-20\ \text{cm}$ (concave), now $d=8.0\ \text{cm}$ apart. A single equivalent focal length only describes a pair faithfully when the lenses are in contact; with a gap we trace the beam lens by lens.

Light on the convex lens first. Parallel rays head for the convex focus, $30\ \text{cm}$ to its right. That point is $30-8=22\ \text{cm}$ beyond the concave lens and acts as a virtual object for it ( $u=+22\ \text{cm}$ ):

\frac{1}{v} = \frac{1}{f_2} + \frac{1}{u} = -\frac{1}{20} + \frac{1}{22} = -\frac{1}{220} \Rightarrow v = -220\ \text{cm}.

The emergent beam diverges as if from a point $220\ \text{cm}$ to the left of the concave lens.

Light on the concave lens first. Parallel rays diverge as if from the concave focus, $20\ \text{cm}$ to its left — a real object for the convex lens $8\ \text{cm}$ away, $u=-(20+8)=-28\ \text{cm}$ :

\frac{1}{v} = \frac{1}{f_1} + \frac{1}{u} = \frac{1}{30} - \frac{1}{28} = -\frac{1}{420} \Rightarrow v = -420\ \text{cm}.

Now the beam appears to come from $420\ \text{cm}$ .

Conclusion. The two answers ( $220\ \text{cm}$ and $420\ \text{cm}$ ) are different, so the result depends on the side of incidence; the pair cannot be replaced by one thin lens and the notion of a single effective focal length is not useful here. (The algebraic combination $\tfrac{1}{F}=\tfrac{1}{f_1}+\tfrac{1}{f_2}-\tfrac{d}{f_1 f_2}=-\tfrac{1}{300}$ gives $F=-300\ \text{cm}$ , but this is referred to principal planes that themselves shift with the side of incidence.)

(b) Magnification and image size

Object height $h_o=1.5\ \text{cm}$ , placed $40\ \text{cm}$ before the convex lens. …

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Method: Sequential Image Formation (Lens-by-Lens Analysis)

This is the standard method for compound lens systems — treat each lens separately, using the image from the first lens as the object for the second lens.

Steps

Step 1: Identify the given data

From Exercise 9.10 (standard NCERT reference):

Convex lens: $f_1 = +30\ \text{cm}$
Concave lens: $f_2 = -20\ \text{cm}$
Separation between lenses: $d = 8.0\ \text{cm}$

Step 2: For part (a) — Effective focal length

Use the formula for the effective focal length $F$ of two thin lenses separated by distance $d$ :

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

Substitute:

\frac{1}{F} = \frac{1}{30} + \frac{1}{(-20)} - \frac{8.0}{(30)(-20)}

\frac{1}{F} = \frac{1}{30} - \frac{1}{20} + \frac{8}{600}

\frac{1}{F} = \frac{20 - 30 + 8}{600} = \frac{-2}{600} = -\frac{1}{300}

Therefore:

\boxed{F = -300\ \text{cm}}

Step 3: Does the answer depend on which side the light is incident?

Yes, the effective focal length formula assumes a specific order of lenses.
If parallel light is incident from the convex side first, the calculation above holds.
If incident from the concave side first, the roles of $f_1$ and $f_2$ swap, giving a different $F$ .
Conclusion: The notion of effective focal length is not very useful here because the system is not a simple equivalent lens — the image position depends on which lens the light hits first.

Step 4: For part (b) — Magnification and image size

Object distance from convex lens: $u_1 = -40\ \text{cm}$ (sign convention: object to left of lens)
Lens 1 (convex): $f_1 = +30\ \text{cm}$

Using lens formula:

\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}

\frac{1}{v_1} = \frac{1}{30} + \frac{1}{-40} = \frac{4 - 3}{120} = \frac{1}{120}

v_1 = +120\ \text{cm} \quad (\text{real image, to the right of convex lens})

Magnification by lens 1:

m_1 = \frac{v_1}{u_1} = \frac{120}{-40} = -3

Step 5: Image from lens 1 becomes object for lens 2

Distance between lenses = $8.0\ \text{cm}$
So, $v_1 = 120\ \text{cm}$ from lens 1 means it is $120 - 8 = 112\ \text{cm}$ to the right of lens 2.
For lens 2 (concave), object distance: $u_2 = +112\ \text{cm}$ (object is on the right side — virtual object for lens 2)

Step 6: Lens 2 (concave): $f_2 = -20\ \text{cm}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Treating the two-lens system as a single thin lens with a simple formula

The error: Students try to use the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ directly, ignoring the separation between lenses.

Why it's wrong: That formula works only for lenses in contact (separation $d = 0$ ). Here $d = 8.0\ \text{cm}$ , so you must use the lens combination formula:

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

How to avoid: Always check if $d = 0$ before using the simple formula. If $d \neq 0$ , use the full formula above.

Mistake 2: Forgetting that effective focal length depends on the side of incidence

The error: Students assume $F$ is the same regardless of which side the light enters from.

Why it's wrong: The effective focal length is different for light entering from the left vs. the right when $d \neq 0$ . The formula above gives $F$ for light incident from the left (first lens = $f_1$ ). For light from the right, swap $f_1$ and $f_2$ in the formula.

How to avoid:

For part (a), compute $F$ for both directions explicitly.
The answer does depend on which side the light is incident — state this clearly.

Mistake 3: Thinking the "effective focal length" concept is always useful

The error: Students assume that once $F$ is found, it can be used like a single lens for any object position.

Why it's wrong: The effective focal length is only meaningful for parallel incident light (object at infinity). For a finite object distance, you cannot use $F$ directly — you must trace the image through each lens step-by-step.

How to avoid: For part (b), do not use $F$ . Instead:

Find the image from the first lens using $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$
Use that image as the object for the second lens (accounting for separation $d$ )
Find the final image position and magnification

Mistake 4: Sign convention errors in the two-lens calculation

The error: Students forget to adjust the object distance for the second lens properly.

Why it's wrong: If the first image forms at distance $v_1$ from lens 1, and the lenses are $d$ apart, then the object distance for lens 2 is:

u_2 = d - v_1

(using the Cartesian sign convention consistently)

How to avoid: Draw a clear ray diagram. Label distances from each lens. Always check:

Is $u_2$ positive or negative?
Is the object for lens 2 real or virtual?

Mistake 5: Confusing total magnification with individual magnifications

The error: Students add magnifications or forget to multiply them. …

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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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(a) Effective focal length

(b) Magnification and image size

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