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derive an expression for the refractive index of the material of a prism in terms of angle of minimum deviation and angle of prism

derive an expression for the refractive index of the material of a prism in terms of angle of minimum deviation and angle of prism

Grade:12

1 Answers

Arun
25750 Points
5 years ago
Dear Raza
 
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.

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