Question

derive an expression for refractive index of the material of the prism and angle of minimum deviation
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Answer 1

In the given diagram,

OP is the incidence ray, which is making the angle i1 with normal, and QR is the angle of emergence, which is represented by i2 .A is the prism angle and μ is the refractive index of the prism.

Now, We know that,

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

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

Now, from snell's rule,

μ = sin i / sin r

μ = ( sin( ( A + δm ) / 2 ) / sin (A/2).

Hope this helps you ✌️✌️☘️☘️.