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Question

Physics Question on Optics

In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and a transparent material (μ=43).(μ = \frac{4}{3} ). A point object O is placed in air on the principal axis of the surface 20 cm from P. The distance of the image of O from P will be _____.Radius of curvature
Fill in the blank with the correct answer from the options given below

A

16 cm left of P in air

B

16 cm right of P in water

C

20 cm right of P in water

D

20 cm left of P in air

Answer

16 cm left of P in air

Explanation

Solution

The refraction at a spherical surface is governed by the formula

μ2vμ1u=μ2μ1R\frac{μ2}{v} −\frac{ μ1}{ u} = \frac{μ2 − μ1}{ R}

where, μ1=1 (for air), μ2=43, u=20 cm, R=10 cm\mu_1 = 1 \ (\text{for air}), \ \mu_2 = \frac{4}{3}, \ u = -20 \ \text{cm}, \ R = 10 \ \text{cm}

Substitute the values into the formula

43v120=43110\frac{4}{3v} - \frac{1}{-20} = \frac{\frac{4}{3} - 1}{10}

Simplifying, 43v+120=130\frac{4}{3v} + \frac{1}{20} = \frac{1}{30}

43v=130120=2360=160\frac{4}{3v} = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = \frac{-1}{60}

Thus, v=16 cmv = -16 \ \text{cm},
This means the image is 16 cm to the left of P in air.