Prism
A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle.1,ABQPandACRPare the two refracting faces. Angle A between them is the refracting angle or angle of prism.
The line AP where the two refracting faces meet is called the refracting edge of the prism. A section ABC of the prism made by a plane at right angles to the refracting edge of the prism is called Principle section of the prism,
Refraction through a prism
A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path.
(a) Calculation of angle of deviation
In fig. 2ABCis principle section of a prism with angle of prism= A
A ray of lightKLis incident on the faceABof the prism at∠i1. It bends towards the normalN1Oand is refracted along LM at∠r1. The refracted ray LM is incident at∠r2on face AC of the prism. It bends away from normalN2Oand emerges alongMNat∠i2. In passing through the prism, ray KL suffers two refractions and has turned through an∠QPN = δ, which is the angle of deviation.
In? PLM, δ = ∠PLM + ∠PML
δ = (i1– r1) + (i2– r2)
δ = (i1 + i2) – (r1 + r2) (1)
In? OLM, ∠O + r1+ r2= 180°(2)
In quadrilateralALOM,
As∠L + ∠M = 180°(because each angle is90°)
Therefore,A + ∠O = 180°(because sum of four angles of a quad. is360°)
Using eqn. (2),∠O + r1+ r2= A + ∠O
r1+ r2= A (3)
Put in (1),
δ = (i1+ i2) – A (4)
Ifµis refractive index of the material of the prism, then according to Snell’s law,
µ= sin i1/sin r1= i1, r1(when angles are small)
Therefore,i1= µr1; similarly,i2= µr2
Putting in (4), we get
δ = (µr1+ µr2) – A
δ = µ (r1+ r2) – A
Using (3), we obtain,
δ = µA – A
δ = (µ - 1) A (5)
This is the angle through which a ray deviates on passing through a thin prism of small refracting angle A.
(b) Derivation of prism formula
From (4) we find, that angle of deviation depends upon angle of prism, angle of incidence and also on nature of material of the prism. Fig. 3 shows the variation of angle of deviation(δ)with angle if incidence(i). Wheniis increased,δdecreases, reaches a minimum and increases again. For one value ofδ, there are two angles of incidencei1andi2. However, at minimum deviationδ = δm, i1= i2i.e. the incident ray and the emergent ray are symmetrical with respect to the refracting faces. The refracted ray in the prism, in that case will be parallel to the base, fig. 4.
Instead of referring to the graph, we can obtain the same result mathematically:
From (4),δ = (i1+ i2) – A
= (√i1)2+ (√i2)2– A
= (√i1)2+ (√i2)2- 2√i1i2+ 2√i1i2– A
δ = (√i1- (√i2)2+ 2√i1i2– A
δ will be minimum, when whole square on R.H.S. is minimum =0 i.e.
i1+ i2= I, say
Asi1+ i2, therefore,r1= r2= r, say
From (3),r + r = Atherefore,r = A/2
From (4),δm = i + i – A
或A + δm = 2i
或,I = A + δm/2
If µ is refractive index of the material of the prism, then according to Snell’s law,
µ = sin i/sin r
Therefore,µ = sin [(A + δm)/2]/sinA/2 (6)
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