An equilateral prism has an angle of deviation 30° when the angle of incidence is 60°. The angle of deviation if a ray is incident normally on a surface is:

60°

For a prism, angle of deviation

$δ = l_1 + l_2 – A$ …(1)

and $r_1 + r_2 = A$ …(2)

Here $δ = 30°, l_1 = 60°, A = 60°$

Putting these values in (1) and (2), we get

$l_2 = 30°$ and $r_1 + r_2 = 60°$

Also from Snell’s Law

$μ = \frac{\sin l_1}{\sin r_1} = \frac{\sin l_2}{\sin r_2} ⇒ \frac{\sin 60}{\sin r_1}=\frac{\sin 30°}{\sin(60°-r_1)}$

Solving we get, $\tan r_1 = \frac{3}{2+\sqrt{3}}$

$⇒ \sin r_1 = \frac{3}{2\sqrt{4+\sqrt{3}}}$

Hence $μ = \frac{\sin 60°}{\sin r_1°}≈1.383$.

Now if the ray is incident normally on this prism angle of incidence at the Second surface = 60° (as shown)

But, $\sin 60° > \frac{1}{μ}$

∴ Total internal reflection will take place At the third surface this again incident normally (as shown).

Hence the ray suffer deviation only at the second surface, which is equal to 180° – 2 × 60° = 60°.