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A cylindrical beam of diameter ( $d = R$ ) is incident on a glass hemisphere of radius $R$ and refractive index $(μ = \sqrt{2})$ on the curved surface symmetrically as shown in the figure. The radius of circular bright spot formed on the flat surface of the hemisphere is: (using snell’s Law)

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$\frac{R}{4 \sqrt{2}} [2 \sqrt{2} - \sqrt{7} + \sqrt{3}]$

$\frac{R}{2 \sqrt{2}} [2 \sqrt{2} - \sqrt{7} + \sqrt{3}]$

$\frac{R}{4 \sqrt{2}} [2 \sqrt{2} + \sqrt{7} - \sqrt{3}]$

$\frac{R}{10} [\sqrt{21} - 1]$

answer is D.

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Detailed Solution

Using Snell’s law: $\sin i = \frac{R}{2 R} \Rightarrow i = 30^{0}, \sin i = μ \sin r, \sin r = \frac{1}{2 \sqrt{2}}$

$\frac{x}{\sin r} = \frac{R}{\sin (120 - r)}$

$x = \frac{R}{10} (\sqrt{21} - 1)$

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