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(C) Given: Refractive index $\mu = \sqrt{3}$ and angle of minimum deviation $\delta_m = A$.The formula for the refractive index of a prism is given by

Vedclass · Accepted Answer

$60$

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(C) Given: Refractive index $\mu = \sqrt{3}$ and angle of minimum deviation $\delta_m = A$.The formula for the refractive index of a prism is given by:$\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$Substituting the given values:$\sqrt{3} = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})}$$\sqrt{3} = \frac{\sin(A)}{\sin(\frac{A}{2})}$Using the trigonometric identity $\sin(A) = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$:$\sqrt{3} = \frac{2 \sin(\frac{A}{2}) \cos(\frac{A}{2})}{\sin(\frac{A}{2})}$$\sqrt{3} = 2 \cos(\frac{A}{2})$$\cos(\frac{A}{2}) = \frac{\sqrt{3}}{2}$Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$, we have:$\frac{A}{2} = 30^{\circ}$$A = 60^{\circ}$

$A$ prism is made up of a material with a refractive index of $\sqrt{3}$. The angle of the prism is $A$. If the angle of minimum deviation is equal to the angle of the prism, then the value of $A$ is: (in $^{\circ}$)

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