Sarthaks Test
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In the figure given below, PQ represents an incident ray falling in the side AB of a prism, when monochromatic light is used

1. Draw the refracted ray, emergant ray and mark the angle of deviation

2. Derive an equation for refractive index of the material of the prism in terms of angle of minimum deviation

3. Draw the incident ray and refracted ray, at the angle of minimum deviation

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2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the face AB at an angle i1 . QR is the refracted ray inside the prism, which makes two angles r1 and r2 (inside the prism). RS is the emergent ray at angle i2.

The angle between the emergent ray and incident ray is the deviation ‘d’.

In the quadrilateral AQMR,

∠A + ∠R = 180°

[since N1M are normal]

ie, ∠A + ∠M = 180° .......(1)

In the ∆ QMR,

∴ r1 + r2 + ∠M = 180° .......(2)

Comparing eq (1) and eq (2)

r1 + r2 = ∠A ....... (3)

From the ∆ QRT,

(i1 – r1) + (i2 – r2) = d

[since exterior angle equal sum of the opposite interior angles]

(i1 + i2) – (r1 + r2) = d
but, r1 + r2 = A
∴ (i1 + i2) – A = d
(i1 + i2) = d + A  .......(4)

It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’. At the minimum deviation position,

i1 = i2 =i, r1 = r2 = r and d = D

Hence eq (3) can be written as,

r + r = A

or r = \(\frac{A}{2}\) .....(5)

Similarly eq (4) can be written as,

i + i = A + D

n = \(\frac{A+D}{2}\)  ......(6)

Let n be the refractive index of the prism, then we can write, 

n = \(\frac{sin\,i}{sin\,r}\)  .....(7)

Substituting eq (5) and eq (6) in eq (7),

n = \(Sin\frac{A+D}{\frac{2}{Sin\frac{A}{2}}}\)

i – d curve:

It is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.

3. Refracted ray is parallel to base

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