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Refraction Of Light By Spherical Lenses (Convex Lens And Concave Lens)

A thin convex lens L (refractive index =1.5 ) is placed on a p

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Solving time: 4 mins

A thin convex lens $L$ (refractive index $= 1.5$ ) is placed on a plane mirror $M$ . When a pin is placed at $A$ , such that $O A = 18 cm$ , its real inverted image is formed at $A$ itself, as shown in the figure. When a liquid of refractive index $μ_{l}$ is put between the lens and the mirror, the pin has to be moved to $A^{'}$ , such that $O A^{'} = 27 cm$ , to get its inverted real image at $A^{'}$ itself. The value of $μ_{l}$ will be
$Refraction Of Light By Spherical Lenses (Convex Lens And Concave Lens)$

Name: Video solution 1: A thin convex lens L (refractive index =1.5 ) is placed on a plane mirror M. When a pin is placed at A, such that OA=18 cm, its real inverted image is formed at A itself, as shown in the figure. When a liquid of refractive index μl is put between the lens and the mirror, the pin has to be moved to A′, such that OA′=27 cm, to get its inverted real image at A′ itself. The value of μl will be
Uploaded: 2022-11-06T12:08:20.000Z
Duration: 18 min

3

2

\frac{4}{3}

\frac{3}{2}

Views: 5,392 students

Updated on: Apr 28, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 3 tutors discussing this question

Evelyn Discussed

A thin convex lens

L

(refractive index

= 1.5

) is placed on a plane mirror

M

. When a pin is placed at

A

, such that

O A = 18 cm

, its real inverted image is formed at

A

itself, as shown in the figure. When a liquid of refractive index

μ_{l}

is put between the lens and the mirror, the pin has to be moved to

A^{'}

, such that

O A^{'} = 27 cm

, to get its inverted real image at

A^{'}

itself. The value of

μ_{l}

will be

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Light from the plane mirror is reflected back on its path, so that image of

A

coincides with

A

itself.

This would happen when rays refracted by the convex lens fall normally on the plane mirror, i.e. the refracted rays form a beam parallel to the principal axis of the lens. Hence, the object would then be considered at the focus of the convex lens.

∴

Focal length of curvature of a convex lens is,

f = 18 cm

With liquid between the lens and mirror, the image again coincides with the object, so the second measurement is the focal length of the combination of the liquid lens and convex lens.

∴ \frac{1}{t _{1}} + \frac{1}{t _{2}} = \frac{1}{t _{q}} \Rightarrow \frac{1}{18} + \frac{1}{t _{2}} = \frac{1}{27} \Rightarrow t_{2} = - 54 cm

For convex lens by lens maker's formula, we have

\frac{1}{f} = (μ - 1) (\frac{2}{R}) \Rightarrow \frac{1}{18} = 0.5 \times \frac{2}{R} \Rightarrow R = 18 cm

and for the plano-convex liquid lens, we have

\frac{1}{f} = (μ_{I} - 1) (\frac{- 1}{R}) \Rightarrow - \frac{1}{54} \Rightarrow μ_{1} = (μ_{1} - 1) (\frac{- 1}{18}) = 1 + \frac{1}{3} = \frac{4}{3}

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Question Text	A thin convex lens $L$ (refractive index $= 1.5$ ) is placed on a plane mirror $M$ . When a pin is placed at $A$ , such that $O A = 18 cm$ , its real inverted image is formed at $A$ itself, as shown in the figure. When a liquid of refractive index $μ_{l}$ is put between the lens and the mirror, the pin has to be moved to $A^{'}$ , such that $O A^{'} = 27 cm$ , to get its inverted real image at $A^{'}$ itself. The value of $μ_{l}$ will be
Updated On	Apr 28, 2023
Topic	Ray Optics and Optical Instruments
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	321
Avg. Video Duration	16 min

Text solutionVerified

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