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Optical Physics - Prism


A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle.1, ABQP and ACRP are 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. 2 ABC is principle section of a prism with angle of prism = A

A ray of light KL is incident on the face AB of the prism at ∠i1. It bends towards the normal N1O and is refracted along LM at ∠r1. The refracted ray LM is incident at ∠r2 on face AC of the prism. It bends away from normal N2O and emerges along MN at ∠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 quadrilateral ALOM,

As ∠L + ∠M = 180° (because each angle is 90°)

Therefore, A + ∠O = 180° (because sum of four angles of a quad. is 360°)

Using eqn. (2), ∠O + r1 + r2 = A + ∠O

r1 + r2 = A                                                 (3)

Put in (1), 

δ = (i1 + i2) – A                                         (4)

If µ is 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). When is increased, δ decreases, reaches a minimum and increases again. For one value of δ, there are two angles of incidence i1 and i2. However, at minimum deviation δ = δm, i1 = i2 i.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)– 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

As i1 + i2, therefore, r1 = r2 = r, say

From (3), r + r = A therefore, r = A/2

From (4), δm = i + i – A 

Or A + δm = 2i

Or, 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) - Physics HelpOptical Physics Assignments, Prism Assignment Help, Prism Homework Help, Prism Assignment Tutors, Prism Solutions, Prism Answers, Optical Physics Assignment Tutors

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