Question #ea4f0

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Question #ea4f0

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Answer 1 · 2017-08-10T15:12:00

We know the following relation of refractive index ( $mu$ ) of the prism with the angle of prism (A) and minimum deviation ( $del_m$ )

$mu=sin((del_m+A)/2)/sin(A/2)$

As per given condition we have $del_m=A$

So we have

$mu=sin((A+A)/2)/sin(A/2)$

$=>mu=sinA/sin(A/2)$

$=>mu=(2sin(A/2)cos(A/2))/sin(A/2)$

$=>cos(A/2)=mu/2$

$=>A=2cos^-1(mu/2)$

If we take for glass $mu=1.5$

Then $A=2cos^-1(1.5/2)=82.8^@$

If we think otherwise

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

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

Hence $" "A_"max"=2cos^-1(1/2)=2xx60^@=120^@$

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