Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle, the time taken by light to travel from P (midpoint of BC) to A is ____ $\times 10^{- 10} \sec .$ (Given, speed of light in vacuum = $3 \times 10^{8} m / s$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ )

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answer is 5.

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Detailed Solution

$PA = 0.1 \sin 60^{\circ} = \frac{0.1 \sqrt{3}}{2} [As, i = 60^{\circ}]$

Minimum deviation: $δ_{min} = i + e - A \Rightarrow δ_{min} = 2 i - A$

$\Rightarrow δ_{min} = 2 \times 60^{\circ} - 60^{\circ} = 60^{\circ}$

$μ = \frac{\sin (\frac{δ_{min} + A}{2})}{\sin \frac{A}{2}}$

Putting values of $δ_{min} and A$

$μ = \frac{\sin (\frac{60^{o} + 60^{o}}{2})}{\sin (\frac{60^{o}}{2})} = \sqrt{3}$

velocity of light in prism:

$v = \frac{C}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} m / s$

time taken to travel PA:

$t = \frac{PA}{V} = \frac{0.1 \sqrt{3} \times \sqrt{3}}{2 \times 3 \times 10^{8}} = 5 \times 10^{- 10} \sec$

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