Question

derive expression for relative index of prism

Answer 1

HEY MATE HERE'S YOUR ANSWER;

there is derivation for refractive index in two ways. wavefront method. (light as a wave) and light traveling in a rectilinear motion as a ray.
The first method is shown in the first diagram.

Assume that light wave is incident on the boundary between two media at P. The wavefront is PQ. Angle of incidence is i. Let us consider the situation after a time interval t. The disturbance (wave) at P travels to Q' with velocity v' in medium2. The wave at point Q travels to point P' with velocity v in medium 1.

Using trigonometry we have sin i / sin r = μ = a constant = speed of light in medium 1 / speed of light in medium 2.
Please see diagram for proof.
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Another method for refractive index is by using a glass prism.

The angle of deviation D = angle bent by emerging ray with respect to incident ray. D is the exterior angle of the triangle PP'Q'. It is sum of two interior angles.
See derivation in the diagram.

A + Q +90 + 90 = 360 and Q+r1+r2 = 180 
So A = r1 + r2

D = (i-r1) + ( e-r2) = (i  + e) - (r1 +r2)
D = i + e - A

Vary i and then e varies. At one point D is minimum and at that point i = e.

When D is minimum and = Dm, we see that i = e and r1 = r2 = r

D = 2 i - A
i = (D+A)/2
r = A/2          as r1+r2 = 2 r  = A.

μ = sin i / sin r  = [ sin (A+Dm)/2 ]  / (  sin A/2 )

we can do an experimental setup to measure A, Dm and create the conditions for minimum deviation of incident light. We can calculate the refractive index for the material that the prism is made of.

Refractive index is not property of Prism. it is of the material.