Question

A transparent oil, of refractive index 7/5, is poured over a layer of water (refractive index = 4/3) in a glass container. The ‘depths’ of the oil and water layers, are in the inverse ratio of their refractive indices. 

A tiny shining metal ball, lying at the bottom of the container, is viewed from above. Its ‘apparent height’, above the bottom of the container equals, a fraction, f, of the total depth of the two liquids in the container. The fraction, f, is (nearly) equal to

(1) 0.27

(2) 0.30

(2) 0.30

(4) 0.35

Answer 1

(1) 0.27

The depths, of the two liquids, are in the ratio 4/3 : 7/5 ; ie, 20:21. We can therefore, take the total depth, of the two liquids, as [(20+21)x] units, i.e.; (41x) units.

Apparent depth of the top layer = (20x ÷ 7/5) = 100x/7 units

Apparent depth of the bottom layer = (21x ÷ 4/3) = 63x/4 units

∴ Total apparent depth = (841/28 x)units

∴ Apparent height of the ball = (41x - 841/28 x)units

= 307/28 x units