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Answer 1 · 2017-08-10T15:12:00

We know the following relation of refractive index ( $μ$ $mu$ ) of the prism with the angle of prism (A) and minimum deviation ( $\partial_{m}$ $del_m$ )

$μ = \frac{sin (\frac{\partial_{m} + A}{2})}{sin (\frac{A}{2})}$ $mu=sin((del_m+A)/2)/sin(A/2)$

As per given condition we have $\partial_{m} = A$ $del_m=A$

So we have

$μ = \frac{sin (\frac{A + A}{2})}{sin (\frac{A}{2})}$ $mu=sin((A+A)/2)/sin(A/2)$

$\Rightarrow μ = \frac{sin A}{sin (\frac{A}{2})}$ $=>mu=sinA/sin(A/2)$

$\Rightarrow μ = \frac{2 sin (\frac{A}{2}) cos (\frac{A}{2})}{sin (\frac{A}{2})}$ $=>mu=(2sin(A/2)cos(A/2))/sin(A/2)$

$\Rightarrow cos (\frac{A}{2}) = \frac{μ}{2}$ $=>cos(A/2)=mu/2$

$\Rightarrow A = 2 {cos}^{- 1} (\frac{μ}{2})$ $=>A=2cos^-1(mu/2)$

If we take for glass $μ = 1.5$ $mu=1.5$

Then $A = 2 {cos}^{- 1} (\frac{1.5}{2}) = {82.8}^{\circ}$ $A=2cos^-1(1.5/2)=82.8^@$

If we think otherwise

This relation reveals that the value of $A$ $A$ increases with decrease in value of $μ$ $mu$ and A will be maximum when $μ$ $mu$ is minimum.The medium of the material of the prism is optically denser than the surrounding medium. So the minimum value of the refractive ( $μ$ $mu$ ) of the medium with respect to the surrounding one may be taken as $\approx 1$ $~~1$ .

Imposing this condition we can have the maximum value of $A$ $A$ .

Hence $A_{max} = 2 {cos}^{- 1} (\frac{1}{2}) = 2 \times 60^{\circ} = 120^{\circ}$ $" "A_"max"=2cos^-1(1/2)=2xx60^@=120^@$

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