A light ray is incident on an equilateral prism of refractive index 5/3. What should be the angle of incidence such that the ray gazes on the second surface?

s i n^{- 1} (\frac{1}{3})

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s i n^{- 1} (\frac{2}{3})

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s i n^{- 1} (\frac{2}{5})

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s i n^{- 1} (\frac{4}{5})

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Solution

The correct option is B $s i n^{- 1} (\frac{2}{3})$

For the first surface,
$\frac{s i n θ}{s i n r_{1}} = \frac{5}{3} . . (1)$
For the second surface,
$\frac{s i n r_{2}}{s i n 90} = \frac{3}{5}$
$\Rightarrow s i n r_{2} = \frac{3}{5}$
$\Rightarrow r_{2} = 37^{\circ}$
$\Rightarrow r_{1} = 60^{\circ} - 37^{\circ} = 23^{\circ}$
$\Rightarrow \frac{s i n θ}{s i n 23^{\circ}} = \frac{5}{3}$
$s i n θ = \frac{5}{3} \times s i n 23^{\circ}$
$= \frac{5}{3} \times . 4$
$= \frac{2}{3}$
$\Rightarrow θ = s i n^{- 1} (\frac{2}{3})$

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