A ray of light travelling through rarer medium is incident at a very small angle 'i' on a glass slab and after refraction its velocity is reduced by 20% . The angle of deviation is

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

Since the velocity of light is reduced by $20%$ after refraction, the velocity in the glass slab is:

$v_{2} = v_{1} - 0.20 v_{1} = 0.80 v_{1}$

The refractive index of the glass slab with respect to the rarer medium is:

$μ = \frac{v _{1}}{v _{2}} = \frac{v _{1}}{0.80 v _{1}} = \frac{10}{8} = \frac{5}{4}$

According to Snell's law, $μ = \frac{sin i}{sin r}$ . For very small angles of incidence and refraction, $sin i \approx i$ and $sin r \approx r$ .

$μ = \frac{i}{r} \Rightarrow r = \frac{i}{μ}$

The angle of deviation $δ$ is given by:

$δ = i - r$

Substituting $r = \frac{i}{μ}$ :

$δ = i - \frac{i}{μ} = i (1 - \frac{1}{μ})$

Substituting $μ = \frac{5}{4}$ :

$δ = i (1 - \frac{4}{5}) = i (\frac{1}{5}) = \frac{i}{5}$

Answer: $\frac{i}{5}$

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