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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

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answer is C.

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Detailed Solution

$\begin{array}{l} {AD}_{1} = \frac{t_{1}}{μ}, {AD}_{2} = \frac{t_{2}}{μ}; A \cdot D_{1} + {AD}_{2} = \frac{t_{1}}{μ} + \frac{t_{2}}{μ} \\ (5 + 3) 1.5 = t_{1} + t_{2}; t_{1} + t_{2} = 12 cm \end{array}$

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