A converging lens of focal length $f_1$ is placed coaxially in contact with a diverging lens of focal length $f_2$ ($f_1 > f_2$). Determine the power and nature of the combination in terms of $f_1$ and $f_2$. OR How is the resolving power of a compound microscope affected if (a) wavelength of light used is decreased, and (b) the diameter of its objective lens is increased ? Justify your answers.

A converging lens of focal length f₁ is placed coaxially in contact with a diverging lens of focal length f₂ (

Question

f_1

(OR)

How is the resolving power of a compound microscope affected if

(a) wavelength of light used is decreased, and

(b) the diameter of its objective lens is increased ? Justify your answers.

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Power of a Lens and Combination of Lenses

The power of a lens is a measure of how strongly it converges or diverges light, defined as $P=1/f$ (SI unit dioptre, D; $1\,\text{D}=1\,\text{m}^{-1}$ ), positive for a converging lens and negative for a diverging one; from the lens maker's formula, $P=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ , so a higher refractive index or more sharply curved (smaller-radius, 'bulkier') surfaces give greater power (and hence a shorter focal length), while gently curved ('skinny') lenses have low power. For two thin lenses of focal lengths $f_1,f_2$ placed in contact (a common optical centre), writing and adding the lens equation for each in turn (the first lens's image serving as the second lens's object) gives the effective focal length $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ , extendable to any number of lenses as $\dfrac{1}{F}=\sum\dfrac1{f_i}$ , or equivalently as an algebraic sum of powers $P=P_1+P_2+\ldots$ (some positive for convex, some negative for concave); the total magnification of the combination is the product of the individual magnifications, $m=m_1\times m_2\times\cdots$ . Two thin lenses can even combine to give zero net power ( $P_1=-P_2$ ) if a converging and a diverging lens of exactly matching magnitudes are used together. When two thin lenses of focal length $f_1,f_2$ are separated by a distance $d$ and only paralle …

Part (b)Concept understanding — Resolving Power of Microscope

The Core Problem: When Two Points Become One Blur

Imagine you are looking at two tiny dots drawn very close together on a piece of paper. From far away, they look like a single dot. As you bring the paper closer, at some point your eye suddenly sees two separate dots. That moment — the threshold where your eye (or a microscope) can just barely tell that there are two objects instead of one — is the heart of resolving power.

A microscope's job is to show you fine detail. But no matter how good the lenses are, there is a fundamental limit: light itself behaves like a wave. When light passes through the circular opening of a lens, it does not travel in perfect straight lines. It spreads out and forms a pattern called an Airy disk — a bright central spot surrounded by faint rings. Every point in your specimen becomes a tiny blurry disk in the image, not a perfect point.

If two points in the specimen are very close, their Airy disks overlap. When they overlap too much, your eye cannot tell them apart — they merge into one blob. The resolving power of a microscope is its ability to show two closely spaced points as distinct.

Note

Resolving power is not about magnification. You can magnify a blurry image as much as you like — it only becomes a bigger blur. Resolution is about separating detail, not enlarging it.

The Precise Criterion: Lord Rayleigh's Condition

Lord Rayleigh proposed a practical rule: two points are just resolved when the centre of one Airy disk falls exactly on the first dark ring of the other. At that point, the combined intensity has a small dip between the two peaks — your eye can just detect that there are two sources.

For a microscope, the smallest distance $d$ between two points that can just be resolved is given by:

d = \frac{1.22 \lambda}{2 n \sin \beta}

where:

$\lambda$ is the wavelength of light used
$n$ is the refractive index of the medium between the specimen and the objective lens
$\beta$ is the half-angle of the cone of light entering the objective

The quantity $n \sin \beta$ is called the numerical aperture (NA) of the objective lens. So the formula is often written as:

d = \frac{1.22 \lambda}{2 \cdot \text{NA}}

\text{Resolving power} = \frac{1}{d} = \frac{2 \cdot \text{NA}}{1.22 \lambda}

What This Tells Us: Two Levers for Better Resolution

1. Shorter wavelength $\lambda$ — Blue light resolves better than red light. Ultraviolet light resolves even better, which is why electron microscopes (using much shorter "wavelengths" of electrons) can see atoms.

2. Larger numerical aperture $n \sin \beta$ — You can increase $n$ by using oil between the slide and the objective (oil immersion). Air has $n \approx 1$ , but special oils have $n \approx 1.5$ . You can increase $\sin \beta$ by using a lens that collects light from a wider cone — a lens with a shorter focal length and larger diameter. …

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Part (a): net power $P=\dfrac{1}{f_1}-\dfrac{1}{f_2}$ ; with $f_1>f_2$ the combination is diverging ( $P<0$ ).

Part (b): microscope resolving power $\propto\dfrac{2\mu\sin\theta}{\lambda}$ , so it increases both when $\lambda$ decreases and when the objective diameter increases.

Part (a): Power and Nature of a Lens Combination

For two thin lenses placed coaxially in contact, powers add: $P=P_1+P_2$ , with $P=1/f$ ( $f$ in metres), converging positive and diverging negative.

Converging lens: $P_1=+\dfrac{1}{f_1}$ .
Diverging lens: focal length is $-f_2$ , so $P_2=-\dfrac{1}{f_2}$ .
Net power:

P=\frac{1}{f_1}-\frac{1}{f_2}

Nature: given $f_1>f_2\Rightarrow\dfrac{1}{f_1}<\dfrac{1}{f_2}\Rightarrow P<0$ . The combination behaves as a diverging lens. …

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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 34 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If power of a lens is +2 dioptre, then it is -$
(i) Concave lens of focal length 50 cm.
(ii) Convex lens of focal length 500 mm
(iii) Concave lens of focal length 0.25 m.
(iv) Convex lens of focal length 25 cm.
›Reveal solutionSolution

f = 1/P; positive power means a convex (converging) lens.

Power $P=+2$ D, so $f=\dfrac{1}{P}=\dfrac{1}{2}\text{ m}=0.5\text{ m}=50\text{ cm}=500\text{ mm}$ . Since the power is positive, the …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Focal length of field lens of power 2D is:$
(a) 50 cm
(b) –50 cm
(c) 200 cm
(d) –200 cm
›Reveal solutionSolution

Focal length is the reciprocal of power; for $P = 2\,D$ , $f = 0.5\,\text{m} = 50\,\text{cm}$ .

Power of a lens is $P = \dfrac{1}{f(\text{in metres})}$ . Given $P = 2\,D$ : $f = \dfrac{1}{P} = \dfrac{1}{2} = 0.5\,\text{m} = 50\,\text{cm}$ . Since the power is …

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CBSE 2025Set D1 markMCQ
$Q. If two converging lenses of equal focal length f are kept in contact then the focal length of the combination will be (A) f (B) 2f (C) f/2 (D) 3f$
›Reveal solutionSolution

For thin lenses in contact powers add: P = P₁ + P₂, giving F = f/2 for two equal lenses.

When two thin lenses are placed in contact, their combined power is the sum of the individual powers:

1/F = 1/f₁ + 1/f₂

With f₁ = f₂ = f:

…

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CBSE 2025Set D1 markMCQ
$Q. Power of a convex lens is 2 dioptre. Its focal length will be (A) 20 cm (B) 50 cm (C) 40 cm (D) 60 cm$
›Reveal solutionSolution

Focal length (in metres) is the reciprocal of power in dioptres: f = 1/P.

The power of a lens is P = 1/f, with f in metres. For P = 2 D:

f = 1/P = 1/2 = 0.5 m = 50 cm

…

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CBSE 2025Set A1 mark
$Q. Write answer in one sentence: What is power of a lens having +0.5 m focal length?$
›Reveal solutionSolution

A lens with focal length +0.5 m has power +2 dioptre.

The power of a lens is defined as the reciprocal of its focal length expressed in metres:

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

Here $f = +0.5$ m (positive, by sign convention for a converging/convex lens):

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two thin lenses of focal length f1 and f2 are placed in contact with each other, the effective focal length of the combination will be$
(a) f = f1 + f2
(b) f = f1 - f2
(c) f = (f1 + f2) / (f1 f2)
(d) f = (f1 f2) / (f1 + f2)
›Reveal solutionSolution

For thin lenses in contact, the powers (reciprocals of focal length) simply add, so the combined focal length is the product over the sum of the individual focal lengths.

For a single thin lens, the lens maker's relation for an object gives $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ . For two thin lenses of focal length $f_1, f_2$ placed in contact, the image formed by the first lens acts as the object for the second. Adding the two lens equations for the combination:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If the focal length of a lens is f metre, then the value of its power will be$
(a) f dioptre
(b) 1/f dioptre
(c) (1 - f) dioptre
(d) 100/f dioptre
›Reveal solutionSolution

The power of a lens is defined as P = 1/f, with f measured in metres, giving power in dioptres (D).

By definition, lens power P (in dioptres) is the reciprocal of the focal length f (in metres):

P = 1/f

…

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CBSE 2025Set ANNUAL1 mark
$Q. Two thin lenses of power + 4D and - 2D are in contact. The focal length of the combination is ______.$
›Reveal solutionSolution

Powers of thin lenses in contact simply add; here P = (+4 D) + (−2 D) = +2 D, giving f = 1/P = 0.5 m.

For two thin lenses of powers $P_1$ and $P_2$ placed in contact (coaxially, touching), the power of the combination is the algebraic sum:

$P = P_1 + P_2$

Here $P_1 = +4\ \text{D}$ and $P_2 = -2\ \text{D}$ , so:

$P = 4 + (-2) = +2\ \text{D}$

Since power and focal length are related by $P = \dfrac{1}{f}$ (f in metres): …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The tangent of the angle by which it converges or diverges a beam of light parallel to the principle axis falling at unit distance from optical centre is called:$
(a) Malus law
(b) Lens formula
(c) Power of a lens
(d) Snell's law.
›Reveal solutionSolution

The power of a lens is defined exactly as the tangent of the angle of convergence/divergence produced in a ray at unit distance from the optical centre.

If a parallel beam of light, on refraction through a lens, converges or diverges through an angle $\theta$ at unit distance from the optical centre, the power of the lens is $P = \tan\theta$ , which for small angles equals $1/f$ (f in metres), measured in dioptres (D). A convex (converging) lens has positive power, a conca …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two thin lenses of focal length f1 and f2 are in contact and coaxial. The power of the combination is -$
(a) (f1 + f2)/2
(b) (f1 + f2)/(f1 f2)
(c) sqrt(f1/f2)
(d) sqrt(f2/f1)
›Reveal solutionSolution

For thin lenses in contact, powers simply add.

For two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact and coaxially, the powers add algebraically:

…

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CBSE 2024Set IMPROVEMENT1 mark
$+2\ \text{D}$
›Reveal solutionSolution

Focal length is the reciprocal of power: $f = 1/P$ .

The power of a lens is related to its focal length (in metres) by $P=\dfrac{1}{f}$ . Given $P=+2\ \text{D}$ :

$f=\frac{1}{P}=\frac{1}{2}=0.5\ \text{m}=50\ \text{cm}$ …

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CBSE 2024Set A1 markMCQ
$Q. Powers of two lenses kept in contact, are P₁ and P₂. The power of equivalent lens will be (A) P₁/P₂ (B) P₂/P₁ (C) P₁ \times P₂ (D) P₁ + P₂$
›Reveal solutionSolution

Powers of thin lenses in contact add up: P = P₁ + P₂.

For two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact, the equivalent focal length is

$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}.$

Since power is the reciprocal of focal length, $P = 1/f$ (with f in metres), this becomes

$P = P_1 + P_2.$

…

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Q.A converging lens of focal length f1f_1f1​ is placed coaxially in contact with a diverging lens of focal length f2f_2f2​ (f1>f2f_1 > f_2f1​>f2​). Determine the power and nature of the combination in terms of f1f_1f1​ and f2f_2f2​.

Part (a): Power and Nature of a Lens Combination

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Q.A converging lens of focal length $f_1$ is placed coaxially in contact with a diverging lens of focal length $f_2$ ( $f_1 > f_2$ ). Determine the power and nature of the combination in terms of $f_1$ and $f_2$ .

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