Question

Write a relation for the angle of deviation $\delta$ for a ray of light passing through an equilateral prism in terms of the angle of incidence (i_1), angle of emergence (i_2) and angle of prism (A).

Answer 1

In the below figure (1), ABC represents the principal section of a glass-prism having ∠A as its refracting angle.

A ray KL is an incident on the face AB at the point F where N1LO is the normal and ∠i1 is the angle of incidence. Since the refraction takes place from air to glass, therefore, the refracted ray LM bends toward the normal such that ∠r1 is the angle of refraction. If µ be the refractive index of glass with respect to air, then

$\mu =\frac{sini}{sinr}$ (By Snell’s law)

∠QPN gives the angle of deviation ‘δ

Thus, $\delta =i1–r_1+i_2-r_2$ ….... (1)

$\delta =i_1+i_2–(r_1+r_2)$

Again, in quadrilateral ALOM,

∠ALO + ∠AMO = 2 right angles [Since, ∠ALO = ∠AMO = 90º]

So, ∠LAM +∠LOM = 2 right angles [Since, Sum of four angles of a quadrilateral = 4 right angles] ….... (2)

Also in $△$ LOM,

$\angle r_1+\angle r_2$+ ∠LOM = 2 right angles …... (3)

Comparing (2) and (3), we get

∠LAM =$\angle r_1+\angle r_2$

A = $\angle r_1+\angle r_2$

Using this value of ∠A, equation (1) becomes,

$\delta =i_1+i_2-A$