Question

A lens has a power of +3 diopters in air. What will be the power of the lens if it is completely immersed in water? Given, μg = \(\frac {3}{2}\) and μw = \(\frac {4}{3}\).

(a) 5 D

(b) 3 D

(c) 1 D

(d) \(\frac {3}{4}\) D

I got this question at a job interview.

This is a very interesting question from Ray Optics in division Ray Optics and Optical Instruments of Physics – Class 12

Answer 1

Correct option is (c) 1 D

The explanation is: Pw=\(\frac {\mu_w}{f_w}\)=μw\( [ \frac {\mu_g}{\mu_w} -1 ] [ \frac {1}{R1} – \frac {1}{R2} ]\)……………………………..1

Pa=\(\frac {1}{f_a}\)=[μg-1]\([ \frac {1}{R1} – \frac {1}{R2} ]\)……………………………………2

Dividing 1 and 2

We get ➔ \(\frac {P_w}{P_a} = \frac {\mu_g-\mu_w}{μ_g-1}\)

Pw=\(\frac {\frac {3}{2}-\frac {4}{3}}{\frac {3}{2}-1}\) × 3

Pw=\(\frac {\frac {1}{6}}{\frac {1}{2}}\) × 3

Pw=\(\frac {2\times 3}{6}\)=1 D

Therefore, the power of the lens gets altered inside water.