Question

For any prism, show that refractive index ‘n’ of its material is given by:

n = `(sin((A + delta m)/2))/(sin(A/2))`

where the terms have their usual meaning.

Answer 1

A ray of light is passing through a triangular prism ABC as shown in the figure.

In ΔQLR,

Angle of deviation (δ) = ∠LQR + ∠LRQ

= (i₁ − r₁) + (i₂ − r₂)

= (i₁ + i₂) − (r₁ + r₂)    ...(i)

In quadrilateral AQL'R,

∠A + ∠L' = 180°    ...(ii)

and in ΔQL'R,

r₁ + r₂ + L' = 180°    ...(iii)

From equations (ii) and (iii),

r₁ + r₂ = A    ...(iv)

Putting the values of (r₁ + r₂) in eq (i),

δ = i₁ + i₂ − A    ...(v)

For minimum deviation,

i₁ = i₂ = i

and δ = δ_m

∴ From equation (v),

δ_m = 2i − A

⇒ i = `(delta_m + A)/2`

From equation (iv)    ...[∵ r₁ = r₂ = r]

and 2r = A

⇒ r = `A/2`

Using Snell’s law,

Refractive index of glass,

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

⇒ n = `(sin((delta_m + A)/2))/(sin(A/2))`