Question

Draw a graph to show variation of angle of deviation D with that of angle of incidence i for a monochromatic ray of light passing through a prism of reflecting angle A. Hence deduce the relation.

\(\mu = \frac{sin(\frac{d_m+A}{2})}{sin A/2}\)

Answer 1

(a) The graph between D and i is as shown in Fig. 

(b) From the graph, we find that D is the same for two angles of incidence i₁ and i₂ but at minimum deviation D, there is only one angle of incidence.

It is found that at minimum deviation positions,

i.e. at D = D_m

e = i and r₁ = r₂ = r (say)

Since D = i + e - A

∴ At minimum deviation position,

we have

D_m = i + e - A = i + i - A

or D_m = 2i - A

or 2i = A D_m

or i = \(\frac{A+D_m}{2}\) .........(1)

Also at minimum deviation position

A = r₁ + r₂ = r + r = 2r

or  r = \(\frac{A}{2}\) ...........(2)

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. (1) and (2), we get

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