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`        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

```							Here,A = 2030’ = π/72n = 1.58OP = 0.2 m For small angle AAssuming 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 α1sin α1  ≈  α1  => h  =  OP α1  δ = α1 + α2α1 ≈ h/OPα2 ≈ 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 refractionWe have2r = A=> r = A/2  ---2.Again, i + e = A + δi = α1=> α1 + e = A + δ  ---3.Sin(e)/sin(r) = μAssuming e and r to be smalle/r = μ=> e = μr=> e = ½μABy 3α1 +  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 mOC = 075 + 0.2 = 0.95 m RegardsArun (askIITians forum expert)
```
2 years ago
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