Refraction At A Plane Surface

Question

A ray of light travelling through air strikes the smooth surface of a glass block of refractive index 1.5 at an angle of incidence of 60 degrees. Find the angle of refraction inside the glass.

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Refraction at a plane boundary is governed by Snell's law, $n_1\sin	heta_1=n_2\sin	heta_2$, where the ray bends toward the normal on entering a denser medium. With air $n_1=1$, glass $n_2=1.5$ and incidence $	heta_1=60°$, we solve $\sin	heta_2=\frac{n_1\sin	heta_1}{n_2}=\frac{1	imes\sin60°}{1.5}=\frac{0.866}{1.5}=0.577$. Hence $	heta_2=\sin^{-1}(0.577)\approx35.3°$, smaller than the incidence angle as expected for light entering glass. The bending occurs because light slows on entering the optically denser medium, and to keep the wavefronts continuous across the boundary the portion of each wavefront that enters first lags behind, swinging the ray toward the normal. The refractive index is precisely the ratio of the speed of light in vacuum to that in the medium, so a larger index means a greater slowing and a sharper bend. The value 30° is wrong because it would require $\sin	heta_2=0.5$, corresponding to a different index. The value 41.8° is wrong as it is the critical angle for glass, relevant only for light leaving the glass. The value 60° is wrong since it simply repeats the incidence angle, implying no bending. This is the standard NCERT application of Snell's law at a single interface. A direction check confirms the refracted ray bends toward the normal, consistent with entry into a denser medium.

Refraction At A Plane Surface

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