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For a prism prove that i + e = A + δ where the symbols have their usual meanings.

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i. Consider a principal section ABC of a prism of absolute refractive index n kept in air as shown in figure.

ii. Let A be the refracting angle of prism and surface BC be the base.

iii. A monochromatic ray PQ obliquely strikes first reflecting surface AB such that, angle of incidence ∠PQM at Q is i.


iv. After refraction at Q, the ray deviates towards the normal and strikes second refracting surface AC at R which is the point of emergence.

v. Angles of refraction at Q (∠NQR) and at R (∠QRN) are r1 and r2 respectively.

vi. After R. the ray deviates away from normal and finally emerges along RS making e as the angle of emergence.

vii. Emergent ray RS meets an extended incident ray QT at X if traced backward. In this case, ∠TXS gives the angle of deviation.

viii. From figure,

∠AQN = ∠ARN = 90°

∴ From quadrilateral AQNR,

A + ∠QNR = 180° ………. (1)

From ∆ QNR,

r1 + r2 + ∠QNR = 180° ………. (2)

∴ A = r1 + r2 ……… (3)

ix. Angle δ forms an exterior angle for ∆ XQR.

∴ ∠XQR + ∠XRQ = δ

∴ (i – r1) + (e – r2) = δ

∴ (i + e) – (r1 + r2) = δ

From equation (3),

δ = i + e – A 

∴ i + .e = A + δ

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