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Grade: 12
        
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?
2 years ago

Answers : (1)

Arun
23006 Points
							

Here,

A = 2030’ = π/72

n = 1.58

OP = 0.2 m

 

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 

=> α = 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)

2 years ago
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