Question

If prism has 1.5 refractive index and emission angle is double of incidence angle, what will be the angle of minimum deviation?

Answer 1

In case of minimum deviation, the angle of deviation D is equal to incidence angle i

$$D=i (1)$$

Now by Snell’s Law applying at interface of prism:

$$\frac { \sin i } {\sin r } = 1.5 (2)$$

In our case,

$$r=2i (3)$$

We put (3) in (2)

$$\frac { \sin i } {\sin 2i } = \frac { \sin i } {2 \sin i \cos i } = \frac { 1} {\cos i } = 1.5 (4)$$

Using (4) we get:

$$\cos i = 0.67 (5)$$

$$i=48° (6)$$

We put (6) in (1)

$$D=48°$$

The angle of minimum deviation is equal to 48°