Question

Derive the expression for refractive index of the prism in terms of angle of the prism and angle of minimum deviation

Answer 1

Answer:

A=Prism angle, δ=Angle of deviation, i1 =Angle of incidence, i2=Angle of emergent.

In the case of minimum deviation,∠r1 =∠r2 =∠r

A=∠r1 +∠r2

​SO, A=∠r+∠r=∠2r

∠r= A/2

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Now, again

A+δ=i1 +i2 (∵ In the case of minimum deviation  i1 = i2 = i and δ=δm)

So, A+δm =i +i = 2i

Now, i = A+δm/2

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Now, from snell's rule,

μ= sin i/sin r

μ= sin A+δm/2/ sin A/2

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