The core of an optical fibre has a refractive index of 1.6. The critical angle at the boundary between the core and the cladding is \(60^\circ\). The refractive index of the cladding is about
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.39
Light in the core meets the cladding, which is the rarer medium. At this boundary, \( \sin i_c \) equals the refractive index of the cladding divided by that of the core.
\( \sin i_c=\frac{n_{\mathrm{clad}}}{n_{\mathrm{core}}} \)
\( \begin{aligned} n_{\mathrm{clad}} &= n_{\mathrm{core}}\sin 60^\circ \\ &= 1.6\times0.866 \\ &\approx 1.39 \end{aligned} \)
Hence, the refractive index of the cladding is about 1.39. It is lower than that of the core, so light in the core can be totally reflected.
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”
\( \displaystyle \sin i_c=n_{21} \)\( \qquad (9.12) \)
Total internal reflection in nature and its technological applications – “… The refractive index of the material of the core is higher than that of the cladding.”