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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2021
  • Hard
  • MCQ

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 -

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Answer & Solution

Correct Answer

(B) \(\sin ^{-1} \frac{\sqrt{7}}{3}\)

Step-by-step Solution

Detailed explanation

At maximum angle \(\theta\) ray at point \(B\) goes in gazing emergence, at all less values of \(\theta, \operatorname{TIR}\) occurs.

At point B

\(\frac{4}{3} \times \sin \theta^{\prime \prime}=1 \times \sin 90^{\circ}\)

\(\theta^{\prime \prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)

\(\theta^{\prime}=\left(\frac{\pi}{2}-\theta^{\prime \prime}\right)\)

At point \(A\)

\(1 \times \sin \theta=\frac{4}{3} \times \sin \theta^{\prime}\)

\(\sin \theta=\frac{4}{3} \times \sin \left(\frac{\pi}{2}-\theta^{\prime \prime}\right)\)

\(\sin \theta=\frac{4}{3} \cos \left[\cos ^{-1} \frac{\sqrt{7}}{4}\right]\)

\(\sin \theta=\frac{4}{3} \times \frac{\sqrt{7}}{4}\)

\(\theta=\sin ^{-1}\left(\frac{\sqrt{7}}{3}\right)\)

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