Question

Draw a graph to show the angle of deviation d with the variation of angle of incidence i for a monochro-matic ray of light passing through a prism of refracting angle A. Deduce the relation 

n = \(\frac{sin\frac{A+\delta_m}{2}}{sin\ \frac{A}{2}}\)

Answer 1

(a) Prism: A prism is a wedge-shaped body made from a refracting medium bounded by two plane faces inclined to each other at some angle. The two plane faces are called the refracting faces. The angle included between these two faces is called the angle of prism or refracting angle. ABC represents the principal section of a glass prism having ∠A as refracting angle.

(b) A ray EF is incident on the face AB at the point F where N₁FO is the normal and so i is the angle of incidence. The refracted ray FG bends towards the normal such that r₁ is the angle of refraction. The ray FG suffers refraction from face AC at G bends away from the normal and goes along GH such that e is the angle of emergence. 

In the absence of the prism, the incident ray EF would have proceeded straight but due to refraction through the prism, it changes its path along the direction DGH. Thus ∠KDH gives the angle of deviation i.e. the angle through which the incident ray gets deviated in passing through the prism.

Now in ∆ DFG, external angle,

δ = ∠DFG + ∠DGF

or δ = i - r₁ + e - r₂

or δ = i + e - (r₁ + r₂) ...........(1)

In quadrilateral AFOG, we have

∠AFO + A + ∠AGO + ∠O = 360°

or 90° + A + 90° + ∠O = 360°

or A + ∠O = 180°

Also, in triangle FGO,

r₁ + r₂ + ∠O = 180°

∴ A + ∠O = r₁ + r₂ + ∠O

or A = r₁ + r₂ ..........(2)

Substituting in equation (1), we have

δ = i + e - A ..........(3)

or A + δ = i + e ..........(4)

Here δ is the angle through which a ray deviates on passing through a small angled prism, and the angle of deviation depends not only upon the material of the prism but also upon the angle of incidence and of prism.

(c) It is found that if the angle of incidence is increased gradually, there is variation in the angle of deviation, it decreases first, attains a minimum value dm and then it starts increasing. So minimum deviation is the angle at which the deviation of ray of light is minimum and the prism is said to be in minimum deviation position.

If we plot a graph between angle of incidence and angle of deviation, the graph is as shown in Fig. From the graph, we find that δ is same for two angles of incidence i₁ and i₂ but at minimum deviation δ_m, there is only one angle of incidence.
It is found that at minimum deviation position.

i.e. at δ = δ_m

e = i and r₁ = r₂ = r (say)

Since δ = i + e - A

∴ At minimum deviation position,

we have

δ_m = i + e - A = i + i - A

or δ_m = 2i - A

or 2i = A + δ_m

or i = \(\frac{A+\delta_m}{2}\) ............(5)

Also at minimum deviation position

A = r₁ + r₂ = r + r = 2r

or r = \(\frac{A}{2}\) ..........(6)

If µ is the refractive index of the prism, then from Snell's law at surface AB, we have

µ = \(\frac{sin\ i}{\sin r}\)

Using Eqs. (5) and (6), we get

µ = \(\frac{sin\frac{A+\delta_m}{2}}{sin\ A/2}\)