If the critical angle for total internal reflection from a medium to…

The correct answer is Option (4) → \( 1.5\times10^{8}\,\mathrm{m\,s^{-1}} \)

The critical angle \( i_c \) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \( 90^\circ \). Snell’s law then gives the refractive index of the medium with respect to vacuum as \( n=1/\sin i_c \). The refractive index is also the ratio of the speed of light in vacuum to its speed in the medium, \( n=c/v \), with \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \).

    \( \begin{aligned} n &= \frac{1}{\sin i_c} \\ &= \frac{1}{\sin30^\circ} \\ &= 2 \end{aligned} \)

    \( \begin{aligned} v &= \frac{c}{n} \\ &= \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{2} \\ &= 1.5\times10^{8}\,\mathrm{m\,s^{-1}} \end{aligned} \)

NCERT (2026-27):
Total Internal Reflection – “The angle of incidence corresponding to an angle of refraction 90°, … is called the critical angle (i_c) for the given pair of media. … The refractive index of denser medium 1 with respect to rarer medium 2 will be n₁₂ = 1/sin i_c.”