Proof for isosceles prism ray inside it passes parallel to base
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PRISM:-Refraction through a Prism:-A 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 therefracting 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)The angle of deviation of a ray of light in passing through a prism not only depends upon its material but also upon the angle of incidence. The above figure (2) shows the nature of variation of the angle of deviation with the angle of incidence. It is clear that an angle of deviation has the minimum value ‘δm’ for only one value of the angle of incidence. The minimum value of the angle of deviation when a ray of light passes through a prism is called the angle of minimum deviation.The figure (3) shows the prism ABC, placed in the minimum deviation position. If a plane mirror M is placed normally in the path of the emergent ray MN the ray will retrace its original path in the opposite direction NMLK so as to suffer the same minimum deviation dm.In the minimum deviation position, ∠i1 = ∠i2and so ∠r1 = ∠r2 = ∠r (say)Obviously, ∠ALM = ∠LMA = 90º – ∠rThus, AL = LMand so LM l l BCHence, the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.Since for a prism,∠A = ∠r1 + ∠r2So, A = 2r (Since, for the prism in minimum deviation position, ∠r1 = ∠r2 = ∠r)or r = A/2 …...(5)Again, i1 + i2 = A + δor i1 + i1 = A + δm (Since, for the prism in minimum deviation position, i1 = i2 and δ = δm)2i1 = A + δmor i1 = (A + δm) / 2 …... (6)Now µ = sin i1/sin r1 = sin i1/sin r µ = sin [(A + δm) / 2] / sin (A/2) …... (7)Equation (7) gives the relation between the refractive index of the material of the prism and the minimum deviation.Grazing Incidence:-When i = 90°, the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass - air. This is known as grazing incidence.Grazing Emergence:-When e = 90°, the emergent ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for glass - air. This is known as grazing emergence.Maximum Deviation:-The angle of deviation is same for both the above cases (grazing incidence & grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection.
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