Question:
A ray of light entering from air into a denser medium of refractive index \(\dfrac{4}{3}\), 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 ray of light entering from air into a denser medium of refractive index \(\dfrac{4}{3}\), 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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The angle of incidence at the vertical face is 90 degrees minus the refraction angle at the top face; it must be at least the critical angle.
Updated On: Nov 4, 2026
- \(\sin^{-1}\frac{\sqrt{7}}{3}\)
- \(\sin^{-1}\frac{\sqrt{5}}{4}\)
- \(\sin^{-1}\frac{\sqrt{7}}{4}\)
- \(\sin^{-1}\frac{\sqrt{5}}{3}\)
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The Correct Option is A
Solution and Explanation
At the top face: \(\sin\theta = \dfrac43\sin\theta'\).
The ray strikes the vertical face with angle of incidence \(90^\circ - \theta'\). For TIR this must be at least the critical angle \(C\), where \(\sin C = \dfrac{1}{\mu} = \dfrac34\).
\(\sin(90^\circ-\theta') = \cos\theta' \ge \dfrac34\) \(\Rightarrow\) \(\sin\theta' \le \sqrt{1-\dfrac{9}{16}} = \dfrac{\sqrt7}{4}\).
\(\sin\theta = \dfrac43\sin\theta' \le \dfrac43\cdot\dfrac{\sqrt7}{4} = \dfrac{\sqrt7}{3}\).
Maximum \(\theta = \sin^{-1}\dfrac{\sqrt7}{3}\).
Answer: (1) \(\sin^{-1}\frac{\sqrt{7}}{3}\)
The ray strikes the vertical face with angle of incidence \(90^\circ - \theta'\). For TIR this must be at least the critical angle \(C\), where \(\sin C = \dfrac{1}{\mu} = \dfrac34\).
\(\sin(90^\circ-\theta') = \cos\theta' \ge \dfrac34\) \(\Rightarrow\) \(\sin\theta' \le \sqrt{1-\dfrac{9}{16}} = \dfrac{\sqrt7}{4}\).
\(\sin\theta = \dfrac43\sin\theta' \le \dfrac43\cdot\dfrac{\sqrt7}{4} = \dfrac{\sqrt7}{3}\).
Maximum \(\theta = \sin^{-1}\dfrac{\sqrt7}{3}\).
Answer: (1) \(\sin^{-1}\frac{\sqrt{7}}{3}\)
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