A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{0}$ . It travels a distance of 5cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

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$5 \sqrt{3}$ cm

$\frac{5}{2}$ cm

$5 \sqrt{3 / 2}$ cm

5 cm

answer is B.

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Detailed Solution

$\frac{\sin 60^{0}}{\sin r_{1}} = \sqrt{3} O r \sin r_{1} = \frac{\sin 60^{0}}{\sqrt{3}}$

Or $\sin r_{1} = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} = \frac{1}{2}$
Or $r_{1} = 30^{0}$
Now, $\sin (i_{1} - r_{1}) = \frac{d}{5}$
Or $d = 5 \sin (i_{1} - r_{1})$
Or $d = 5 \sin (60^{0} - 30^{0}) = 5 \sin 30^{0} = \frac{5}{2}$

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