When a glass prism of refracting angle 60^cir | vedclass

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(B) Given: Refracting angle $A = 60^\circ$,Angle of minimum deviation $\delta_m = 30^\circ$.The refractive index $\mu$ of the prism with respect to th

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$45$

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(B) Given: Refracting angle $A = 60^\circ$,Angle of minimum deviation $\delta_m = 30^\circ$.The refractive index $\mu$ of the prism with respect to the liquid is given by the formula:$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$Substituting the values:$\mu = \frac{\sin((60^\circ + 30^\circ)/2)}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$The relation between refractive index and critical angle $C$ is $\mu = 1/\sin(C)$.Therefore,$\sin(C) = 1/\mu = 1/\sqrt{2}$.This gives $C = \sin^{-1}(1/\sqrt{2}) = 45^\circ$.

When a glass prism of refracting angle $60^\circ$ is immersed in a liquid,its angle of minimum deviation is $30^\circ$. The critical angle of glass with respect to the liquid medium is......$^\circ$

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