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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
=> 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)
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