Derive an expression for the refractive index of the material of a prism.

Here ABC represents the triangular base of prism

PQ = incident ray A = angle of prism
OR = refracted ray

RS = emergent ray

d = angle of deviation

From triangle EQR,

d = i₁ – r₁ + i₂ – r₂

d = (i_{1 +} i₂) – (r₁ + r₂ ) … (1)

From triangle AQR,

A +(90^o – r₁ ) + (90^o – r₂ ) = 180^o

A = r₁ + r_{2 …} (2)

From (1) & (2)

d = (i_{1 +} i₂) – A

d +A = (i_{1 +} i₂) … (3)

Using Snell’s law, n₁ sin I = n₂ sin r

Using Snell’s Law At point M,

n₁ = refractive index of air = 1

n₂ = refractive index of material of prism = n

i = i₁ & r = r₁

sin i₁ = n sin i₂ …. (4)

Using Snell’s law at N,

n₁ = n

n₂ = 1

i = r₂ & r = i₂

n sin r₂ = sin i₂ …(5)

When i₁ = i_2, QR becomes parallel to BC and d becomes angle of minimum deviation (D).

Then from equation (3)

D +A = (i_{1 +} i₂) = 2i₁

i₁ = D+A/2

When i₁ = i₂, then r₁ = r₂

From equation (2)

2r₁ = A

r₁ = A/2

Substituting the values in equation (4),

Sin{(A+D)/2} = n sin (A/2)