For a prism of angle pi/3,the angle of minimu | vedclass

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Question

(D) Given: Prism angle $A = \pi/3 = 60^{\circ}$,Angle of minimum deviation $\delta_m = \pi/6 = 30^{\circ}$,Velocity of light in vacuum $c = 3 	imes 1

Vedclass · Accepted Answer

$2.12 	imes 10^{8} 	ext{ m/s}$

Vedclass · Answer

(D) Given: Prism angle $A = \pi/3 = 60^{\circ}$,Angle of minimum deviation $\delta_m = \pi/6 = 30^{\circ}$,Velocity of light in vacuum $c = 3 	imes 10^{8} 	ext{ m/s}$.The refractive index $\mu$ of the prism is given by the formula:$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$Also,$\mu = c/v$,where $v$ is the velocity of light in the prism.Therefore,$v = c 	imes \frac{\sin(A/2)}{\sin((A + \delta_m)/2)}$.Substituting the values:$v = 3 	imes 10^{8} 	imes \frac{\sin(60^{\circ}/2)}{\sin((60^{\circ} + 30^{\circ})/2)}$$v = 3 	imes 10^{8} 	imes \frac{\sin(30^{\circ})}{\sin(45^{\circ})}$$v = 3 	imes 10^{8} 	imes \frac{0.5}{0.7071}$$v \approx 2.12 	imes 10^{8} 	ext{ m/s}$.

For a prism of angle $\pi/3$,the angle of minimum deviation is $\pi/6$. If the velocity of light in vacuum is $3 \times 10^{8} \text{ m/s}$,then the velocity of light in the material of the prism is:

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