A thin prism of angle 2degree 30 min and of index 1.58 for sodium light. What is the distance between the slit and its image formed by the prism?

For small angle A

Assuming the prism to be set in min deviation position,

δm = (μ -1)A

  = (1.58 - 1) π/72

  = 0.025

  Again, sin α1 = h/OB

  ≈ h/OP  [considering paraxial rays]

  h = OP sin α1

sin α1  ≈  α1

  => h  =  OP α1

  δ = α1 + α2

α1 ≈ h/OP

α2 ≈ h/PC

=> h/OP + h/PC =  δ

=> h( 1/OP + 1/PC) = δ

=>  OP α1 ( 1/OP + 1/PC) = δ

=> α1 ( 1 + OP/PC) = δ  ----1.

Let r be the angle of refraction

We have

2r = A

=> r = A/2  ---2.

Again, i + e = A + δ

i = α1

=> α1 + e = A + δ  ---3.

Sin(e)/sin(r) = μ

Assuming e and r to be small

e/r = μ

=> e = μr

=> e = ½μA

By 3

α1 +  ½μA = A + δ 

=> α1  = A(1 - ½μ) + δ  ---4.

By 1.

{A(1 - ½μ) + δ} ( 1 + OP/PC) = δ

=>  {π/72( 1 – 0.5×1.58)  + 0.025}{ 1 + 0.2/PC} =  0.025

=> PC ≈ 0.75 m

OC = 075 + 0.2 = 0.95 m

Regards

Arun (askIITians forum expert)