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Two immiscible liquids of refractive indices $$\frac{8}{5}$$ and $$\frac{3}{2}$$ respectively are put in a beaker as shown in the figure. The height of each column is $$6$$ cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $$\frac{\alpha}{4}$$ cm. The value of $$\alpha$$ is _______.
Correct Answer: 31
$$d' = \sum \frac{d_i}{\mu_i}$$
$$d' = \frac{6}{\frac{8}{5}} + \frac{6}{\frac{3}{2}} = \frac{30}{8} + \frac{12}{3} = \frac{15}{4} + 4 = \frac{31}{4}\text{ cm}$$
$$\frac{\alpha}{4} = \frac{31}{4} \implies \alpha = 31$$
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