The angle of a prism is 60^o and its refracti | vedclass

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(B) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given: Prism angle $A = 60^o$,refract

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$30$

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(B) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given: Prism angle $A = 60^o$,refractive index $\mu = \sqrt{2}$.Substituting the values: $\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(60^o/2)}$.$\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(30^o)}$.Since $\sin(30^o) = 0.5$,we have $\sqrt{2} 	imes 0.5 = \sin((60^o + \delta_m)/2)$.$\frac{1}{\sqrt{2}} = \sin((60^o + \delta_m)/2)$.We know that $\sin(45^o) = \frac{1}{\sqrt{2}}$,therefore $(60^o + \delta_m)/2 = 45^o$.$60^o + \delta_m = 90^o$.$\delta_m = 30^o$.

The angle of a prism is $60^o$ and its refractive index is $\sqrt{2}$. The angle of minimum deviation suffered by a ray of light in passing through it is......$^o$

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