Question

Derive the equation for an angle of deviation produced by a prism and thus obtain the equation for the refractive index of the material of the prism.

Answer 1

1. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
2. The deviation d₁ at the surface AB is,
   angle ∠RQM = d₁ = i₁ – r₁
3. The deviation d₂ at the surface AC is, angle ∠QRM = d₂ = i₂ – r₂
   
4. Total angle of deviation d produced is
   d = d₁ + d₂
   Substituting for d₁ and d₂,
   d = (i₁ – r₁) + (i₂ – r₂)
   After rearranging,
5. d = (i₁ + i₂) – (r₁ + r₂)
   In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
6. ∠A + ∠QNR = 180°
   From the triangle ∆ QNR,
   r₁ + r₂ + ∠QNR = 180°
   Comparing these two equations and we get,
   r₁ + r₂ = A
   substituting in equation for angle of deviation,
   d = i₁ + i₂ – A
   
7. A graph plotted between the angle of incidence and angle of deviation.
8. The minimum value of angle of deviation is called the angle of minimum deviation D. At minimum deviation,
   
   1. the angle of incidence is equal to the angle of emergence, i₁ = i₂
   2. the angle of refraction at face one and face two are equal, r₁ = r₂
      
   3. the incident ray and emergent ray are symmetrical with respect to the prism.
   4. the refracted ray inside the prism is parallel to its base of the prism.
      The case of the angle of minimum deviation is shown in figure.

9. Refractive index of the material of the prism
   At minimum deviation, i₁ = i₂ = i and r₁ = r₂ = r
   D = i₁ + i₂ – A = 2i – A (or)
   i = `("A + D")/2`
   r₁ + r₂ = A ⇒ 2r = A (or)
   r = `"A"/2`
   n = `(sin "i")/(sin "r")`

   n = `(sin (("A + D")/2))/(sin ("A"/2))`