For two thin lenses kept in contact with each other, show that: 1 F = 1 f 1 + 1 f 2 where the terms have their usual meaning.

Let two lenses A and B kept in contact and the object is kept at the point O beyond the focus of the first lens A. The first lens produces an image at I 1 . Since image I 1 is real, it serves as a virtual object for the second lens B, producing the final image at I. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P. For the image formed by the first lens A, we get: 1 f 1 = 1 v 1 - 1 u .....(i) For the image formed by the second lens B, we get 1 f 1 = 1 v - 1 v 1 .....(ii) On adding (i) and(ii) we get: 1 f 1 + 1 f 2 = 1 v - 1 u .....(iii) If the two lens system is considered as a single lens of focal length F , we can write: 1 F = 1 v - 1 u ......(iv) From(iii) and(iv) we get: 1 F = 1 f 1 + 1 f 2 ....(v) v is the required expression.

For two thin lenses kept in contact with each other, show that: 1 F = 1 f 1 + 1 f 2 where the terms have their usual meaning.

Let two lenses A and B kept in contact and the object is kept at the point O beyond the focus of the first lens A. The first lens produces an image at I 1 . Since image I 1 is real, it serves as a virtual object for the second lens B, producing the final image at I. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P. For the image formed by the first lens A, we get: 1 f 1 = 1 v 1 - 1 u .....(i) For the image formed by the second lens B, we get 1 f 1 = 1 v - 1 v 1 .....(ii) On adding (i) and(ii) we get: 1 f 1 + 1 f 2 = 1 v - 1 u .....(iii) If the two lens system is considered as a single lens of focal length F , we can write: 1 F = 1 v - 1 u ......(iv) From(iii) and(iv) we get: 1 F = 1 f 1 + 1 f 2 ....(v) v is the required expression.

A biconvex lens made of glass (refractive index 1 . 5 ) has two spherical surfaces having radii 20 c m and 30 c m . Calculate its focal length.

Given Refractive index of glass μ = 1 . 5 Radius of curvature R 1 = 20 cm R 2 = - 30 cm Focal length of the lens f = ? Using lens maker formula : 1 f = μ - 1 1 R 1 - 1 R 2 Substituting, the values, we have, 1 f = 1 . 5 - 1 1 20 - 1 - 30 2 = 0 . 5 50 600 or f = 24 cm

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If the image formed by a thin convex lens of power $P$ has magnification $m$ then image distance $v$ is

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Important Questions on Refraction of Light at Spherical Surfaces : Lenses

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 28

For two thin lenses kept in contact with each other, show that:

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

where the terms have their usual meaning.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 29

How is focal length defined for a two lens system, separated by distance d?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 30

What is power of a lens?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 31

If two lenses of powers

P_{1} & P_{2}

are kept in contact with each other, what will be the effective power of the combination?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 32

When two thin lenses are kept in contact, prove that their combined or effective focal length F is given by:

\frac{1}{F} = \frac{1}{F_{1}} + \frac{1}{F_{2}}

where the terms have their usual meaning.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 33

A concave lens of focal length

24 cm

is kept in front of a concave mirror at a distance of

20 cm

on the axis. If a pin is kept at a distance of

36 cm

in front of the lens, the inverted image of the pin formed by the combination of lens and mirror coincides with the object. Calculate the focal length of the mirror.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 34

A biconvex lens of focal length

15 cm

is in front of a plane mirror. The distance between the lens and the mirror is

10 cm

. A small object is kept at a distance of

30 cm

from the lens. The final image is

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 35

A point object is placed at a distance of

12 cm

on the axis of a convex lens of focal length

10 cm

. On the other side of the lens, a convex mirror is placed at a distance

10 cm

from the lens such that image formed by the combination coincides with the object itself. What is the focal length of the convex mirror ?

If the image formed by a thin convex lens of power P has magnification m then image distance v is

Important Questions on Refraction of Light at Spherical Surfaces : Lenses

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If the image formed by a thin convex lens of power $P$ has magnification $m$ then image distance $v$ is