A ray of light travelling through a rarer medium is incident at very small angle ' i ' on a glass slab and after refraction its velocity is reduced by 25% . The angle of deviation is

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Let the velocity of light in the rarer medium be $v_{1}$ .

The velocity of light in the glass slab is reduced by $25%$ , so $v_{2} = v_{1} - 0.25 v_{1} = \frac{3}{4} v_{1}$ .

The refractive index of the glass slab with respect to the rarer medium is $μ = \frac{v _{1}}{v _{2}} = \frac{v _{1}}{\frac{3}{4} v _{1}} = \frac{4}{3}$ .

From Snell's law for small angles of incidence and refraction, we have $i = μ r$ .

$r = \frac{i}{μ} = \frac{i}{\frac{4}{3}} = \frac{3 i}{4}$

The angle of deviation $δ$ is given by $δ = i - r$ .

$δ = i - \frac{3 i}{4} = \frac{i}{4}$

Answer: $\frac{i}{4}$

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