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Question
A ray of light incident at an angle of incidence i1 passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges at an angle of emergence i2.
- How is the angle of emergence ‘i2’ related to the angle of incidence ‘i1’.
- What can you say about the angle of deviation in such a situation?
Solution
- ∠i1 = ∠i2
and refracted ray is parallel to base.
∠5 = ∠6 = 90°
∵ PQ || BC, ∠APQ = ∠AQP = 60°
∵ ΔABC is equilateral and ∠APQ and ∠AQP are corr. ∠K to ∠B and ∠C
∴ ∠2 = ∠4 ...[∵ 60° + ∠2 = 60° + ∠4]
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
∴ ∠i1 = ∠i2
∵ ∠i1 = ∠1 + ∠2; ∠i2 = ∠3 + ∠4
∴ Angle of incidence = Angle of emergence is the relation when PQ || Base BC -
The angle of deviation is at its smallest when i1 = i2 and the refracted beam inside the prism is parallel to the base.
δ = i + e - A = i1 + i2- A
∵ i1 = i2 = i
δ = δmin = 2 i - A
For a given prism and given colour of light δmin is unique.
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