Question

For any prism, obtain a relation between the angle of the prism (A), the angle of minimum deviation (δ_m) and the refractive index of its material (μ or n).

Answer 1

In the given diagram,

OP is the incident ray, which makes the angle i, with normal, and QR is the angle of emergence, which is represented by i₂. A is the prism angle, and it 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 = ∠2r

`=> ∠"r" = "A"/2`

Now again

A + δ = i₁ + i₂    ...(∵ In the case of minimum deviation)

i₁ = i₂ = i and δ = δ_m

So, A + δ_m = i + i = 2i

Now, `angle "i" = ("A" + delta_"m")/2`

For Snell's law:

`mu = (sin i)/(sin r)`

`=> mu = (sin  ("A" + delta_m)/2)/(sin  "A"/2)`