JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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 -

- A \(\sin ^{-1} \frac{\sqrt{5}}{3}\)
- B \(\sin ^{-1} \frac{\sqrt{7}}{3}\)
- C \(\sin ^{-1} \frac{\sqrt{7}}{4}\)
- D \(\sin ^{-1} \frac{\sqrt{5}}{4}\)
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 \(\frac{4}{3} \times \sin \theta^{\prime \prime}=1 \times \sin 90^{\circ}\)…
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