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 = 90o
∴ ∠ PNQ + ∠ PAQ = 180o
or, ∠ PNQ = 180o – A (∵ ∠ PAQ = A) —-(iii)
But in Δ PNQ,
∠ PNQ = 180o – (r1 + r2)
∴ From eq (ii) and (iii),
A = r1 + r2 —-(iv)
Hence from eq (i) and (iv)
δ = (i1 + i2) – A (Proved).