Find the angle of minimum deviation for an eq | vedclass

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(D) Given: Angle of prism $A = 60^\circ$,Refractive index $\mu = 1.732 = \sqrt{3}$.The formula for the refractive index of a prism is $\mu = \frac{\si

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$60^\circ, 60^\circ$

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(D) Given: Angle of prism $A = 60^\circ$,Refractive index $\mu = 1.732 = \sqrt{3}$.The formula for the refractive index of a prism is $\mu = \frac{\sin(\frac{\delta_m + A}{2})}{\sin(A/2)}$.Substituting the values: $\sqrt{3} = \frac{\sin(\frac{\delta_m + 60^\circ}{2})}{\sin(30^\circ)}$.Since $\sin(30^\circ) = 0.5$,we have $\sqrt{3} 	imes 0.5 = \sin(\frac{\delta_m + 60^\circ}{2}) \Rightarrow \frac{\sqrt{3}}{2} = \sin(\frac{\delta_m + 60^\circ}{2})$.We know $\sin(60^\circ) = \frac{\sqrt{3}}{2}$,so $\frac{\delta_m + 60^\circ}{2} = 60^\circ$.$\delta_m + 60^\circ = 120^\circ \Rightarrow \delta_m = 60^\circ$.For minimum deviation,the angle of incidence $i$ is given by $i = \frac{\delta_m + A}{2}$.$i = \frac{60^\circ + 60^\circ}{2} = 60^\circ$.Thus,the angle of minimum deviation is $60^\circ$ and the angle of incidence is $60^\circ$.

Find the angle of minimum deviation for an equilateral prism made of a material of refractive index $1.732$. What is the angle of incidence for this deviation?

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