For a prism with a prism angle of 60^circ and | vedclass

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(A) Given: Prism angle $A = 60^\circ$,Refractive index $\mu = \sqrt{2}$.The formula for the refractive index of a prism in terms of the angle of minim

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$30$

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(A) Given: Prism angle $A = 60^\circ$,Refractive index $\mu = \sqrt{2}$.The formula for the refractive index of a prism in terms of the angle of minimum deviation $\delta_m$ is:$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$Substituting the values:$\sqrt{2} = \frac{\sin((60^\circ + \delta_m)/2)}{\sin(60^\circ/2)}$$\sqrt{2} = \frac{\sin(30^\circ + \delta_m/2)}{\sin(30^\circ)}$Since $\sin(30^\circ) = 0.5$:$\sqrt{2} 	imes 0.5 = \sin(30^\circ + \delta_m/2)$$\frac{1}{\sqrt{2}} = \sin(30^\circ + \delta_m/2)$We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,therefore:$45^\circ = 30^\circ + \delta_m/2$$15^\circ = \delta_m/2$$\delta_m = 30^\circ$.

For a prism with a prism angle of $60^\circ$ and a refractive index of $\mu = \sqrt{2}$,the angle of minimum deviation is .......... $^\circ$.

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