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An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is:
A
12
B
16
C
8
D
10
Explanation
Real depths from both faces add up to the thickness.
Detailed Solution
$H_{app} = \frac{H_R}{\mu_g} \Rightarrow H_R = \mu_gH_{app}$
Thickness $= H_{R_1} + H_{R_2} = \mu_g[H_{app_1} + H_{app_2}] = \frac{3}{2}[5 + 3] = 12$ cm