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One face AC of an equilateral prism ABC is silvered as shown in fig. The angle of incidence of a light ray in order that it eventually leaves the prism in the opposite direction from base of prism is n ×15º where ‘n’ is $(μ = \sqrt{2})$

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answer is 4.

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Detailed Solution

From figure,

$A = r_{1} + c; C = r_{2} + c$

$∵ A = C = 60^{o} \Rightarrow r_{1} = r_{2} = r$

$\Rightarrow A = r + c$

From Snell's Law,

$\sin i_{1} = μ \sin r = \sin i_{2}$

$\Rightarrow i_{1} = i_{2} = i$

As light ray opposite to incident ray deviation = 180º ;

$δ_{net} = 180^{o}$

$i_{1} - r_{1} + 180^{o} - 2 c + i_{2} - r_{2} = 180^{o};$

$2 i = 2 r + 2 c \Rightarrow i = r + c = A$

$i = 60^{\circ} = n \times 15^{o} \Rightarrow n = 4$

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