SAQ : Deduce the relation δ = i1 + i2 – A for refraction through a prism

Question

Deduce the relation δ = i₁ + i₂ – A for refraction through a prism.

Answer

The angle of deviation (δ): The angle between the direction of the incident ray and the emergent ray is called the angle of deviation.

To prove δ = (i₁ + i₂) – A : 

In fig, ∠ LMQ = ∠ MPQ + ∠ MQP

∴   Angle of deviation (δ) = δ₁ + δ₂ —- (i)

Since ∠ MPN = i₁ and ∠ MQN = i₂

∴   ∠ MPQ = δ₁ = i₁ – r₁

and    ∠ MQP  = δ₂ = i₂ – r₂

∴  From eq (i), δ = (i₁ – r₁) – (i₂ – r₂)

or,       δ = (i₁ + i₂) – (r₁ + r₂)—- (ii)

Also from the quadrilateral APNQ in fig

∠ APN = ∠ AQN = 90^o

∴ ∠ PNQ + ∠ PAQ = 180^o

or,   ∠ PNQ = 180^o – A (∵ ∠ PAQ = A) —-(iii)

But in Δ PNQ,

∠ PNQ = 180^o – (r₁ + r₂)

∴ From eq (ii) and (iii),

A =  r₁ + r₂ —-(iv)

Hence from eq (i) and (iv)

δ = (i₁ + i₂) – A (Proved).