A small bulb is placed 12 cm below the surface of a liquid of refractive index 1.25. Light from the bulb comes out of the surface only through a circular patch. The radius of this patch is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 16 cm
Rays from the bulb that meet the surface at more than the critical angle are totally reflected back into the liquid. So light comes out only inside a circle. At the edge of this circle, the angle of incidence equals the critical angle.
\( \begin{aligned} \sin i_c &= \frac{1}{n} \\ &= \frac{1}{1.25} \\ &= \frac{4}{5} \end{aligned} \)
\( \tan i_c=\frac{4}{3} \)
For a bulb at depth \( h \), the radius \( r \) of the patch follows from \( \tan i_c=r/h \).
\( \begin{aligned} r &= h\tan i_c \\ &= 12\times\frac{4}{3} \\ &= 16\ \mathrm{cm} \end{aligned} \)
Hence, the radius of the circular patch is 16 cm.
NCERT (2026-27):
Total Internal Reflection – “The refractive index of denser medium 1 with respect to rarer medium 2 will be n12 = 1/sin ic. …”
Total Internal Reflection – “… For values of i larger than ic, Snell’s law of refraction cannot be satisfied, and hence no refraction is possible.”