Question

Derive an expression relating angle of prism, angle of incidence, angle of emergence and angle of deviation when light is refracted by a prism

Answer 1

A ray KL is incident on the face AB at the point F where N1LO is the normal and ∠i1 is the angle of incidence. Since the refraction takes place from air to glass, therefore, the refracted ray LM bends toward the normal such that ∠r1 is the angle of refraction. If µ be the refractive index of glass with respect to air, thenµ = sin i/sin r     (By Snell’s law)The refracted ray LM is incident on the face AC at the point M where N2MO is the normal and ∠r2 is the angle of incidence. Since the refraction now takes place from denser to rarer medium, therefore, the emergent ray MN such that ∠i2 is the angle of emergence.In the absence of the prism, the incident ray KL would have proceeded straight, but due to refraction through the prism, it changes its path along the direction PMN. Thus, ∠QPN gives the angle of deviation ‘δ’, i.e., the angle through which the incident ray gets deviated in passing through the prism. Thus,   δ = i1 – r1 + i2 -r2     ….... (1) δ = i1 + i2 – (r1 + r2 )Again, in quadrilateral ALOM,∠ALO + ∠AMO = 2rt∠s                [Since, ∠ALO = ∠AMO = 90º]So, ∠LAM +∠LOM = 2rt∠s           [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s]    ….... (2)Also in ?LOM,∠r1 +∠r2 + ∠LOM = 2rt∠s         …... (3)Comparing (2) and (3), we get∠LAM = ∠r1 +∠r2 A = ∠r1 +∠r2 Using this value of ∠A, equation (1) becomes,δ = i1 + i2 - Aor i1 + i2 = A + δ                   …... (4)

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