Question

A thin prism of mu =1.5 deviates a ray by a minimum angle of 5 what is the angle of prism

Answer 1

Refraction through a PrismA prism is a wedge-shaped body made from a refracting medium bounded by two plane faces inclined to each other at some angle. The two plane faces are called are the refracting faces and the angle included between these two faces is called the angle of prism or the refracting angle.In the below figure (1), ABC represents the principal section of a glass-prism having ∠A as its refracting angle.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)