A thin convex lens of focal length 15 cm is placed – Class 12 Physics | ConceptFirst

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Q. A thin convex lens of focal length 15 cm is placed in contact with a thin concave lens of focal length 15 cm. What is the power of the combination? (A) 0 D (B) \infty D (C) 3 D (D) 6 D

Haryana BsehBSEH Intermediate Board 2020MCQ· 1mImportance★★★★★

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Equal and opposite focal lengths (convex +15 cm, concave −15 cm) cancel, giving zero net power.

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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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