A thin layer of oil of refractive index \(\sqrt{2}\) is spread on the flat top surface of a glass block of refractive index 1.5. A ray of light in air falls on the oil at an angle of incidence of \(45^\circ\). The angle of refraction in the glass is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sin^{-1}(\sqrt{2}/3)\)
The ray is refracted twice, first at the air-oil surface and then at the oil-glass surface. For the second surface we need the refractive index of glass with respect to oil, \( n_{go}=n_{ga}\times n_{ao} \).
\( i=45^\circ \)
\( n_{oa}=\sqrt{2} \)
\( n_{ga}=1.5 \)
\( \begin{aligned} \sin r_o &= \frac{\sin i}{n_{oa}} \\ &= \frac{1/\sqrt{2}}{\sqrt{2}} \\ &= \frac{1}{2} \end{aligned} \)
\( \begin{aligned} n_{go} &= n_{ga}\times n_{ao} \\ &= \frac{1.5}{\sqrt{2}} \end{aligned} \)
\( \begin{aligned} \sin r_g &= \frac{\sin r_o}{n_{go}} \\ &= \frac{1}{2}\times\frac{\sqrt{2}}{1.5} \\ &= \frac{\sqrt{2}}{3} \end{aligned} \)
The oil index does not appear in the result. The ray in glass is the same as with no oil. Hence, the angle of refraction in the glass is \( \sin^{-1}(\sqrt{2}/3) \).
NCERT (2026-27):
Refraction – “(ii) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. … We have”
\( \displaystyle \frac{\sin i}{\sin r}=n_{21} \)\( \qquad (9.10) \)
“where n21 is a constant, called the refractive index of the second medium with respect to the first medium.”
Refraction – “… if n32 is the refractive index of medium 3 with respect to medium 2 then n32 = n31 × n12, where n31 is the refractive index of medium 3 with respect to medium 1.”