Question

A ray of light is refracted by a glass prism. Obtain an expression for the refractive index of the glass in terms of the angle of prism A and the angle of minimum deviation δ_m.

Answer 1

In the given diagram,

OP is the incident ray, which makes the angle i₁ normal, and ∠N'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.

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

In the case of minimum deviation,

`∠r_1 = ∠r_2 = ∠r`

A = `∠r_1 + ∠r_2`

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 = δ_m)

So, A + δ_m = i + i = 2i

Now, `i = ((A + δ_m))/2`

Now, from Snell's rule,

`mu = (sini)/(sinr)`

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