The cross-section of a prism is an equilatera | vedclass

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(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Given that minimum deviation occurs when the angle of incidence $i = A = 60^{\circ}$.The

Vedclass · Accepted Answer

$5$

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(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Given that minimum deviation occurs when the angle of incidence $i = A = 60^{\circ}$.The refractive index $\mu$ of the prism is given by $\mu = \frac{\sin(\frac{\delta_{min} + A}{2})}{\sin(\frac{A}{2})}$.At minimum deviation,$i = e = 60^{\circ}$,so $\delta_{min} = 2i - A = 2(60^{\circ}) - 60^{\circ} = 60^{\circ}$.Thus,$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.The speed of light in the prism is $v = \frac{c}{\mu} = \frac{3 	imes 10^8}{\sqrt{3}} \, m/s$.The distance $AP$ is the height of the equilateral triangle with side $a = 10 \, cm = 0.1 \, m$.$AP = a \sin 60^{\circ} = 0.1 	imes \frac{\sqrt{3}}{2} = 0.05\sqrt{3} \, m$.The time taken $t = \frac{AP}{v} = \frac{0.05\sqrt{3}}{3 	imes 10^8 / \sqrt{3}} = \frac{0.05 	imes 3}{3 	imes 10^8} = 0.05 	imes 10^{-8} = 5 	imes 10^{-10} \, s$.Therefore,the value is $5$.

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