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Q. The refractive index of the material of an equilateral prism is $1.6$. The angle of minimum deviation due to the prism would be

AIIMSAIIMS 2019Ray Optics and Optical Instruments

Solution:

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$A=60^{\circ}$
$\mu=1.6$
$\mu=\frac{\sin \left(\frac{A+\delta \min }{2}\right)}{\sin A / 2} \delta \min$
$1.6=\frac{\sin \left(\frac{60+8 \min}{2}\right)}{\sin 30^{\circ}}$
$\sin \left(\frac{60+8 \min}{2}\right)=1.6 \times \frac{1}{2}=0.8$
$\sin ^{-1}(0.8)=53.10^{\circ}$
$\frac{60+\delta \min }{2}=53.10^{\circ}$
$\delta \min =\frac{53.10\times 2}{60} = 1.77^{\circ}$