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.

When two thin lenses are kept in contact, prove that their combined or effective focal length F is given by: 1 F = 1 F 1 + 1 F 2 where the terms have their usual meaning.

Let a point object O is placed on the principal axis By lens formula for image formation by lens L 1 1 v ' - 1 u = 1 f 1 -------------------(1) I' serves as a virtual object for the second lens L 2 which forms a final image I at a distance ' v ' . Then for L 2 1 v - 1 v ' = 1 f 2 -----------------(2) Adding equation (1) and (2) 1 v - 1 u = 1 f 1 + 1 f 2 ----------- (3) If we replace these two lenses by a single lens such that if an object is kept at u image is formed at v then its focal length f is given by 1 v - 1 u = 1 f ------------- (4) From equation (3) and (4) we get 1 f = 1 f 1 + 1 f 2

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A convex lens is cut into two parts in different ways that are arranged in four manners, as shown. Which arrangement will give maximum optical power?

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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 8

The refractive index of material of a Plano-convex lens, if the radius of curvature of the plane surface

(R_{1})

is infinite while the radius of curvature of convex surface

(R_{2})

20 c m

and focal length of the lens is

30 c m

, will be:

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 10

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.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 11

A thin convergent glass lens μ=1.5 has a power of $+ 5 D$ . When this lens is immersed in a liquid of refractive index $n$ it acts as a diverging lens of focal length $100 cm$ . The value of $n$ must be :

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 12

A concave lens of focal length

20 cm

product an image half in size of the real object. The distance of the real object is

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 13

The given lens is broken into four parts and rearranged as shown in the figure. If the initial focal length is $f$ then after rearrangement the equivalent focal length is

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 14

An object is placed at a certain distance from a convex lens of focal length $20 cm$ . Find the distance of the object if the image obtained is. magnified. $4$ times.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 15

The focal length of a concave lens is

30 cm

. Where should an object be placed so that its image is

\frac{1}{10}

of its size?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Refraction of Light at Spherical Surfaces : Lenses>Exercise>Q 16

A point object is placed

20 cm

left of a convex lens of focal length

f = 5 cm

(see the figure). The lens is made to oscillate with a small amplitude

A

along the horizontal axis. The image of the object will also oscillate along the axis with-
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A convex lens is cut into two parts in different ways that are arranged in four manners, as shown. Which arrangement will give maximum optical power?

Important Questions on Refraction of Light at Spherical Surfaces : Lenses

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