A prism has a refractive index sqrt{3/2} and | vedclass

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(C) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given: $\mu = \sqrt{3/2}$,$A = 90^o$.

Vedclass · Accepted Answer

$30$

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(C) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given: $\mu = \sqrt{3/2}$,$A = 90^o$.Substituting the values: $\sqrt{3/2} = \frac{\sin((90^o + \delta_m)/2)}{\sin(90^o/2)}$.$\sqrt{3/2} = \frac{\sin(45^o + \delta_m/2)}{\sin(45^o)}$.Since $\sin(45^o) = 1/\sqrt{2}$,we have $\sqrt{3/2} = \frac{\sin(45^o + \delta_m/2)}{1/\sqrt{2}}$.Multiplying both sides by $1/\sqrt{2}$: $\sqrt{3/2} 	imes \frac{1}{\sqrt{2}} = \sin(45^o + \delta_m/2)$.$\sqrt{3/4} = \sin(45^o + \delta_m/2) \implies \frac{\sqrt{3}}{2} = \sin(45^o + \delta_m/2)$.We know that $\sin(60^o) = \frac{\sqrt{3}}{2}$,so $45^o + \delta_m/2 = 60^o$.$\delta_m/2 = 15^o \implies \delta_m = 30^o$.

$A$ prism has a refractive index $\sqrt{3/2}$ and a refracting angle of $90^o$. Find the minimum deviation produced by the prism in degrees.

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