For a prism of refractive index 1.732,the ang | vedclass

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(C) The refractive index of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$Given that the angle of min

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$60$

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(C) The refractive index of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$Given that the angle of minimum deviation $\delta_m$ is equal to the angle of the prism $A$,we substitute $\delta_m = A$ into the formula:$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$Given $\mu = 1.732 = \sqrt{3}$,we have:$\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 $\frac{A}{2} = 30^\circ$,which implies $A = 60^\circ$.

For a prism of refractive index $1.732$,the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is......$^o$

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