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  • A ray of light entering from air into a denser medium of refractive index , as shown in figure. The light ray suffers total internal reflection at the adjac

A ray of light entering from air into a denser medium of refractive index \frac{4}{3}, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle \theta should be equal to :
Option: 1 \sin ^{-1} \frac{\sqrt{7}}{3}
Option: 2 \sin ^{-1} \frac{\sqrt{5}}{4}
Option: 3 \sin ^{-1} \frac{\sqrt{7}}{4}
Option: 4 \sin ^{-1} \frac{\sqrt{5}}{3}

Answers (1)

best_answer


\mu _{1}\sin \theta = \mu _{2}\sin {\theta }'
1\sin \theta= \frac{4}{3}\, {\sin \theta}'\rightarrow \left ( 1 \right )
For the maximum value of \theta,  {\theta}' is maximum and consequently {\theta }'' is the minimum for TIR,\left ( {\theta }'' \right )_{min}= critical\, angle\, \left ( \theta _{c} \right )
\sin {\theta }''= \sin \theta _{c}= \frac{1}{\mu}= \frac{3}{4}
\sin {\theta }''= \sin \left ( 90-{\theta }' \right )= \cos {\theta }'
\cos {\theta }'= \frac{3}{4}
\sin {\theta }'= \sqrt{1-\cos ^{2}{\theta }'}= {\frac{\sqrt7}{4}}
from Eqn (1)
1\times \sin \theta = \frac{4}{3}\times \frac{\sqrt{7}}{4}= \frac{\sqrt{7}}{3}
\rightarrow \sin \theta = \frac{\sqrt{7}}{3}
\theta =\sin^{-1} \left ( \frac{\sqrt{7}}{3} \right )
The correct option is (1)

Posted by

vishal kumar

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