For a prism with refractive index mu = 1.5,th | vedclass

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(C) The condition for minimum deviation is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum devi

Vedclass · Accepted Answer

$82$

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(C) The condition for minimum deviation is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m$ is equal to the angle of incidence $i$,and for minimum deviation $i = (A + \delta_m)/2$,we have $\delta_m = A$.Substituting $\delta_m = A$ into the formula:$\mu = \frac{\sin((A + A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.Using the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$,we get:$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.Given $\mu = 1.5$,we have $1.5 = 2 \cos(A/2)$.$\cos(A/2) = 1.5 / 2 = 0.75$.Since $\cos 41^o = 0.75$,we have $A/2 = 41^o$.Therefore,$A = 82^o$.

For a prism with refractive index $\mu = 1.5$,the angle of minimum deviation is equal to the angle of incidence. Find the prism angle $A$ in degrees. (Given: $\cos 41^o = 0.75$)

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