The critical angle for a denser rarer interface is 45°. The speed of light in rarer medium is $3 \times 10^{8} m / s$ . The speed of light in the denser medium is

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$3.12 \times 10^{7} m / s$

$\sqrt{2} \times 10^{8} m / s$

$5 \times 10^{7} m / s$

$2.12 \times 10^{8} m / s$

answer is C.

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Detailed Solution

$i_{c} = 45^{\circ}, V_{rarer} = 3 \times 10^{8} \frac{m}{s}$

$V_{denser} = ?$

${sini}_{c} = \frac{V_{denser}}{V_{rarer}}$

$\Rightarrow V_{denser} = \frac{1}{\sqrt{2}} \times 3 \times 10^{8} = 2.12 \times 10^{8} \frac{m}{s}$

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