A ray of light suffers minimum deviation when | vedclass

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(B) For a prism in the condition of minimum deviation,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\

Vedclass · Accepted Answer

$45$

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(B) For a prism in the condition of minimum deviation,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Also,at minimum deviation,the angle of incidence $i$ is related to the prism angle $A$ and the angle of minimum deviation $\delta_m$ as $i = \frac{A + \delta_m}{2}$.Substituting this into the refractive index formula,we get: $\mu = \frac{\sin i}{\sin(A/2)}$.Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.Substituting the values: $\sqrt{2} = \frac{\sin i}{\sin(60^{\circ}/2)}$.$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.Since $\sin 30^{\circ} = 0.5$,we have: $\sin i = \sqrt{2} 	imes 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = \arcsin(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

$A$ ray of light suffers minimum deviation when incident on a prism having an angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is . . . . . . .

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