The critical angle of medium P with respect to air is \(\sin^{-1}\left(\frac{3}{5}\right)\). That of medium Q is \(\sin^{-1}\left(\frac{4}{5}\right)\). For light going from P to Q, the critical angle at their boundary is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sin^{-1}\left(\frac{3}{4}\right)\)
For light going from a medium into air, \( n=1/\sin i_c \). This gives the refractive index of each medium. The medium with the larger index is the denser one. At the boundary, \( \sin i_c \) equals the index of the rarer medium divided by that of the denser medium.
\( \begin{aligned} n_P &= \frac{1}{3/5} \\ &= \frac{5}{3} \end{aligned} \)
\( \begin{aligned} n_Q &= \frac{1}{4/5} \\ &= \frac{5}{4} \end{aligned} \)
Since \( n_P > n_Q \), P is the denser medium, and light going from P to Q can be totally reflected.
\( \begin{aligned} \sin i_c &= \frac{n_Q}{n_P} \\ &= \frac{5/4}{5/3} \\ &= \frac{3}{4} \end{aligned} \)
Hence, the critical angle at the P–Q boundary is \( \sin^{-1}(3/4) \).
NCERT (2026-27):
Total Internal Reflection – “… there is an upper limit to the value of sin i for which the law can be satisfied, that is, i = ic such that”
\( \displaystyle \sin i_c=n_{21} \)\( \qquad (9.12) \)
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