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 

(3) 0.32

(4) 0.35 

Answer 1

Correct option: (1) 0.27

Explanation: 

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