Question

How are angle of incidence and angle of prism related ?​

Answer 1

Answer:

When a ray of light suffers minimum deviation through a prism. Angle of incidence is equal to the angle of emergence. Ray of light passing through the prism is parallel to the base of the prism. Angle of refraction inside the material of prism is equal to half the angle of prism.

Explanation:

The angle of deviation of a ray of light in passing through a prism not only depends upon its material but also upon the angle of incidence. The above figure (2) shows the nature of variation of the angle of deviation with the angle of incidence. It is clear that an angle of deviation has the minimum value ‘δm’ for only one value of the angle of incidence. The minimum value of the angle of deviation when a ray of light passes through a prism is called the angle of minimum deviation.

The figure (3) shows the prism ABC, placed in the minimum deviation position. If a plane mirror M is placed normally in the path of the emergent ray MN the ray will retrace its original path in the opposite direction NMLK so as to suffer the same minimum deviation dm.

In the minimum deviation position, ∠i1 = ∠i2

and so ∠r1 = ∠r2 = ∠r    (say)

Obviously, ∠ALM = ∠LMA = 90º – ∠r

Thus, AL = LM

and so LM l l BC

A Prism, Placed in the Minimum Deviation Position.Hence, the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.

Since for a prism,

∠A = ∠r1 + ∠r2

So, A = 2r      (Since, for the prism in minimum deviation position, ∠r1 = ∠r2 = ∠r)

or r = A/2        …...(5)

Again, i1 + i2 = A + δ

or i1 + i1 = A + δm         (Since, for the prism in minimum deviation position, i1 = i2 and δ = δm)

2i1 = A + δm

or i1 = (A + δm) / 2           …... (6)

Now  µ = sin i1/sin r1 = sin i1/sin r

µ = sin [(A + δm) / 2] / sin (A/2)               …... (7)