Derive the relation, μ = sin A + δ m 2 sin A 2 (symbols have their usual meanings) and show that i = A + δ m 2 , r = A 2 , for minimum angle fo deviation.

In the given diagram, O P is the incidence ray, which is making the angle i 1 with normal, and Q R is the angle of emergence, which is represented by i 2 . A is the prism angle and μ is the refractive index of the prism. Now, we know that, A = Prism angle, δ = Angle of deviation, i 1 = Angle of incidence, i 2 = Angle of emergent. In the case of minimum deviation, ∠ r 1 = ∠ r 2 = ∠ r A = ∠ r 1 + ∠ r 2 So, A = ∠ r = ∠ r = ∠ 2 r ∠ r = A 2 Now, again A + δ = i 1 + i 2 ( ∵ In the case of minimum deviation i 1 = i 2 = i and δ = δ m ) So, A + δ m = i + i = 2 i Now, i = A + δ m 2 Now, from snell's rule: μ = sin i sin r μ = sin A + δ m 2 sin A 2

Derive the relation, μ = sin A + δ m 2 sin A 2 (symbols have their usual meanings) and show that i = A + δ m 2 , r = A 2 , for minimum angle fo deviation.

In the given diagram, O P is the incidence ray, which is making the angle i 1 with normal, and Q R is the angle of emergence, which is represented by i 2 . A is the prism angle and μ is the refractive index of the prism. Now, we know that, A = Prism angle, δ = Angle of deviation, i 1 = Angle of incidence, i 2 = Angle of emergent. In the case of minimum deviation, ∠ r 1 = ∠ r 2 = ∠ r A = ∠ r 1 + ∠ r 2 So, A = ∠ r = ∠ r = ∠ 2 r ∠ r = A 2 Now, again A + δ = i 1 + i 2 ( ∵ In the case of minimum deviation i 1 = i 2 = i and δ = δ m ) So, A + δ m = i + i = 2 i Now, i = A + δ m 2 Now, from snell's rule: μ = sin i sin r μ = sin A + δ m 2 sin A 2

The refractive index of a prism of angle 60 ° is 1 . 62 for sodium light. Find the angle of minimum deviation. sin 54 . 1 ° = 0 . 81 .

Given, Angle of prism, A = 60 ° Refractive index of material of prism, μ = 1 . 62 We have, μ = sin A + δ m 2 sin A 2 ⇒ sin A + δ m 2 = μ sin A 2 sin 60 ° + δ m 2 = 1 .62 × sin 30 ° = 0 .81 60 ° + δ m 2 = sin − 1 0 .81 δ m = 2 sin − 1 0 .81 − 60 ° δ m = 2 × 54 . 1 ° − 60 ° = 48 . 2 °

HARD

12th ICSE

IMPORTANT

Earn 100

Derive the relation, $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{(\sin \frac{A}{2})}$ (symbols have their usual meanings) and show that $i = \frac{A + δ_{m}}{2}, r = \frac{A}{2}$ , for minimum angle fo deviation.

Important Questions on Refraction and Dispersion of Light through a Prism

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 4

What do you understand by the term scattering of light? Why sky appears blue and dangerous signals are of red colour? Write Rayleigh's concept of scattering and relate it to above mentioned colours.

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 5

Define angular dispersion and dispersive power of an optical medium. Explain why deviation is maximum for violet colour and minimum for red colour also explain the dependence of deviation on the refractive index.

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12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 1

A ray of light is incident on a

60 °

prism of glass of refractive index

1.5

at an angle of

40 °

. Find the angle of emergence of the ray.

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 2

The angle of a prism is

60 °

and the angle of minimum deviation is

39 °

. What is the refractive index of the material of the prism?

(\sin 49.5 ° = 0.76)

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 3

Calculate the refractive index of the material of an equilateral prism for which the angle of minimum deviation is

60 °

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12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 4

The refractive index of a prism of angle

60 °

1.62

for sodium light. Find the angle of minimum deviation.

(\sin 54.1 ° = 0.81)

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 5

Calculate the angle of minimum deviation for a glass prism of angle

60 °

, the refractive index of glass being

1.658

. What is the corresponding angle of incidence? Given:

\sin 56 ° = 0.829

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 4 - Refraction and Dispersion of Light through a Prism>QUESTIONS>Q 6

A regular

60 °

glass prism has refractive index

\sqrt{2}

for a given wavelength. Calculate the angle of minimum deviation

δ_{m}

. At what angle of incidence should a monochromatic light of this wavelength fall on the face of the prism so as to get a minimum deviation? (use

\sin 30 ° = \frac{1}{2}

and

\sin 45 ° = \frac{1}{\sqrt{2}}

Derive the relation, μ=sinA+δm2sinA2 (symbols have their usual meanings) and show that i=A+δm2, r=A2, for minimum angle fo deviation.

Important Questions on Refraction and Dispersion of Light through a Prism

Learn and Prepare for any exam you want !

Derive the relation, $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{(\sin \frac{A}{2})}$ (symbols have their usual meanings) and show that $i = \frac{A + δ_{m}}{2}, r = \frac{A}{2}$ , for minimum angle fo deviation.