QQuestion
A ray of light is incident normally on one of the faces of a prism of refracting angle A and refractive index μ. If δ = angle of deviation, then establish the relation between μ, A and δ.
✓Answer
The principal section ABC of a prism is shown in Fig. A ray of light OP is incident at P on the face AB at an angle of incidence i₁. The angle of refraction at P is r₁. The refracted ray moves along PQ and is incident on the face AC of the prism at an angle of incidence r₂ at Q, and emerges along QR at an angle of emergence i₂. So, ∠LMR = δ is the angle of deviation, i.e., the angle through which the emergent ray is deviated from the incident ray.
Now, from quadrilateral APNQ,
∠APN + ∠AQN = 90°
∴ ∠PNQ + ∠PAQ = 180°
But in ΔPNQ,
∠PNQ = 180° − (r₁ + r₂)
or, 180° − (r₁ + r₂) + A = 180°
∴ A = r₁ + r₂
Again, the angle of deviation δ is given by,
δ = ∠MPQ + ∠MQR
δ = (∠MPN − r₁) + (∠MQN − r₂)
δ = (i₁ − r₁) + (i₂ − r₂)
δ = (i₁ + i₂) − (r₁ + r₂)
δ = i₁ + i₂ − A
∴ δ = i₁ + i₂ − A
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