Dispersion without Deviation: Definition, Formula, Derivation, Solved Examples

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Dispersion without deviation is a phenomenon in which deviation produced by one prism is canceled by deviation produced by the prism made up of another glass.

The splitting of white light into its constituent color component is called dispersion of light.
The bending of light rays at some angle from their original path is called deviation.
To achieve dispersion without deviation, flint glass prisms are used.
Deviation produced by crown and flint glass prisms is equal but opposite in direction.
To observe this, the crown and flint glass prism are joined together such that the base of one prism opposes the angle of another prism.

Table of Content

Dispersion without Deviation
Dispersion without Deviation - Formula
Derivation of Dispersion without Deviation
Things to Remember
Sample Questions

Key Terms: Dispersion without Deviation, Dispersion, Deviation, Refractive Index, Refracting Angle, Prism

Dispersion without Deviation

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When two prisms of different angles and refractive index are combined in the opposite order, then the emergent ray may be dispersed but suffers no deviation.

Dispersion without deviation can be observed when crown and flint glass prisms are kept in the opposite order.
Dispersion is the process in which white light splits into its constituent's colors while passing through a prism.
For example, deviation can be seen in the formation of a rainbow.
Deviation of light occurs when a beam of light traveling through one medium enters into a second medium.
Deviation can be seen in the phenomenon of refraction of light, where light traveling through a medium deviated from its original path while entering another medium with a different refractive index.

Read Also: Some Natural Phenomena due to Sunlight

Dispersion without Deviation - Formula

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The formula for Dispersion without Deviation is given by:

(μ_y - 1)A = – (μ’_y - 1)A’

A’/A = – [(μ_y - 1)/(μ’_y - 1)]

Where

μy is the refractive index of crown glass prism
A is the refracting angle of crown glass prism
μ’y is the refractive index of flint glass prism
A’ is the refracting angle of flint glass prism

Read More: Ray Optics and Optical Instruments

Derivation of Dispersion without Deviation

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Consider a white light striking a crown glass prism, it disperses into multiple colors and deviates.

The deviation of a white light produced by a crown prism for mean color is equal and opposite in direction to the deviation produced by the flint glass prism.
To see the dispersion without deviation, two prisms with different glasses (different refractive indices) are placed in contact (as shown below).

Dispersion without deviation

Dispersion without deviation

The deviation produced by a crown glass prism and a flint glass prism is equal.

From the given figure, it is given

The angle of a crown glass prism is A.
A’ be the angle of the flint glass prism.

Let the refractive indices of the crown and flint glass prism be μ_y and μ’_y respectively for the mean ray.

The mean ray deviation produced by the crown glass prism is given as

δ_y = (μ_y - 1)A …(1)

The mean ray deviation produced by the crown glass prism is given as

δ’_y = (μ’_y - 1)A’ …(2)

Since in this combination dispersion takes place without deviation thus, the net deviation will be zero.

δ_y + δ’_y = 0

⇒ δ_y = - δ’_y

Substitute eq (1) and eq (2) in the above expression, we get

(μ_y - 1)A = – (μ’_y - 1)A’

Also,

A’/A = – (μ_y - 1)]/(μ’_y - 1)

Also, Let

The refractive indices of a crown glass prism for violet and red colors rays are μ_V and μ_R respectively.
The refractive indices of a flint glass prism for violet and red color rays are μ’_V and μ’_R respectively.

For crown glass prism, the angular dispersion is

θ₁ = ( μ_V - μ_R)A …(3)

For flint glass prism, the angular dispersion is

θ₂ = ( μ’_V - μ’_R)A’ …(4)

The net angular dispersion is given by

⇒ θ = θ₂ - θ₁

Substitute eq (3) and eq (4)

⇒ θ = (μ’_V - μ’_R)A’ - ( μ_V - μ_R)A

⇒ θ = (μ’_V - μ’_R)[-(μ_y - 1)]/(μ’_y - 1)]A - ( μ_V - μ_R)A

⇒ θ = (μ_y - 1)A [(μ’_V - μ’_R)/(μ’_y - 1)] - [( μ_V - μ_R)/(μ_y - 1)]

⇒ θ = (μ_y - 1)A (ω’ - ω)

Here, ω’ and ω are the dispersive powers of flint and crown prism respectively.

Hence,

ω = [( μ_V - μ_R)/(μ_y - 1)]

and,

ω’ = [(μ’_V - μ’_R)/(μ’_y - 1)]

Also Read:

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Snell’s Law	Refraction Through a Prism	Dispersion of Light by Prism

Things to Remember

Dispersion without deviation is observed when deviation produced by both the prism annulled each other.
Deviation produced by crown and flint prism for a mean ray is equal and opposite in direction.
Dispersion without deviation is given by: A’/A = – (μ_y - 1)]/(μ’_y - 1).
The angular dispersion of a prism is given by: θ = ( μ_V - μ_R)A.
There is some dispersion left, hence this combination of prisms is used in direct vision spectroscopes.

Read More: Optics

Sample Questions

Ques. The angle of the crown and flint glass prisms are 30^o and 60^o respectively. Find the relation between the refractive indices of prisms. (3 Marks)

Ans. Let the refractive indices of the crown and flint glass prism be μ_y and μ’_y respectively.

A = 30^o

A’ = 60^o

Then dispersion produced by the combination of crown and flint glass prisms is given by:

A’ = [-(μ_y - 1)]/(μ’_y - 1)]A …(i)

Substitute values

60^o = 30^o [-(μ_y - 1)]/(μ’_y - 1)]

2 = [-(μ_y - 1)]/(μ’_y - 1)]

⇒ 2 - 2μ’_y = μ_y - 1

⇒ μ_y + 2μ’_y = 3

Above equation is required relationship between the refractive indices of prisms.

Ques. Find the relation between refractive indices of crown and flint prisms if their angle is equal in the combination of dispersion without deviation. (3 Marks)

Ans. The dispersion without deviation is given as:

A’ = [-(μ_y - 1)]/(μ’_y - 1)]A …(i)

Let refractive indices of crown and flint glass prism be μ_y and μ’_y respectively.

Given,

A’ = A …(ii)

Substitute eq (ii) in eq (i)

[(μ_y - 1)]/(μ’_y - 1)] = -1

⇒ μ_y - 1 = 1 - μ’_y

⇒ μ_y + μ’_y = 2

Ques. The angles of a crown and flint prism are 30o and 60o respectively. The refractive index of a crown prism is 1.5. Find the refractive index of the flint prism if light undergoes dispersion without deviation through the combination. (3 Marks)

Ans. Then dispersion produced by the combination of crown and flint glass prisms is given by:

A’ = [-(μ_y - 1)]/(μ’_y - 1)]A …(i)

Given

A = 30, μ_y = 1.5

A’ = 60^o

Put these values in eq (i)

60^o = 30^o [-(1.5 - 1)]/(μ’_y - 1)]

⇒ -2 (μ’_y - 1) = 0.5

⇒ 1.5 = 2μ’_y

⇒ μ’_y = 0.75

∴ The refractive index of the flint prism is 0.75.

Ques. A prism made up of glass with a refractive index of 1.45 has a refracting angle of 10^o. It is now combined with another prism with a refractive index of 1.9. This combined structure produces dispersion without deviation. Find the refracting angle of the second prism. (3 Marks)

Ans. The dispersion produced without deviation is given as

A’ = [-(μ_y - 1)]/(μ’_y - 1)]A …(1)

A = 10^o

μ = 1.45

μ’ = 1.9

A’ = (refracting angle of the second prism)

Substitute these values in eq (1)

A’ = [-(1.45 - 1)]/(1.9 - 1)]10^o

⇒ A’ = (0.45/0.9)10^o

A’ = 5^o

Ques. What is the necessary condition for dispersion without deviation in the prism? (2 Marks)

Ans. The necessary condition to produce dispersion without deviation is that deviation produced by the crown prism is equal and opposite to the deviation produced by the flint prism.

Ques. What is the required condition for the occurrence of dispersion? (1 Mark)

Ans. Dispersion of light occurs when white light is separated into its seven constituent colors by the mean of refraction and Snell’s Law.

Ques. Will a monochromatic light undergo dispersion when it passes through a prism? Explain. (2 Marks)

Ans. Monochromatic light does not undergo dispersion. Because it does not have any components (unlike white light). Hence, it will not be dispersed through the prism, instead, it will deviate.

Ques. The refractive indices of the two prisms are 1.4 and 1.8 respectively. Find the ratio of their refracting angle if the light undergoes dispersion without deviation. (3 Marks)

Ans. For dispersion without deviation

A’ = [-(μ_y - 1)]/(μ’_y - 1)]A …(1)

Here,

μ_y = 1.4, μ’_y = 1.8

A and A’ are the refracting angle of the first and second prism respectively.

Substitute the values

A’ = [(0.4)/(0.8)]A

⇒ A/A’ = 2

Hence, the ratio of their refracting angles is 2^o.

Ques. The refracting angle of the crown glass prism is A and dispersion produced without deviation is A’. If the refracting angle is made double. Find the relation between A’ and new dispersion. (3 Marks)

Ans. The expression for dispersion without deviation is given as

A’ = -A[(μ_y - 1)/(μ’_y - 1)] …(1)

From eq (1) it is seen that

A’ (dispersion) ∝ A (refracting angle of a crown prism)

Thus, dispersion varies linearly with refracting angle.

Hence, on doubling the refracting angle, the dispersion will also be doubled. Thus, the new dispersion will be twice the previous dispersion.

Ques. A prism has refractive indices of 1.46, 1.52, and 1.60 for red, yellow, and violet light respectively. Calculate (3 Marks)
(a) Dispersive power
(b) Angular dispersion is produced by a prism of angle 10^o

Ans. (a) Dispersive power is given by

ω = [( μ_V - μ_R)/(μ_y - 1)] … (1)

Given,

μ_V = 1.60
μ_R = 1.46
μ_y = 1.52

Substitute these values in eq(1)

ω = (1.60 - 1.46)/(1.52 - 1)

⇒ ω = 0.27

(b) Angular dispersion (θ) = (μ_V - μ_R)A

θ = (1.60 - 1.46)10^o

θ = 1.4^o

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Dispersion without Deviation: Definition, Formula, Derivation, Solved Examples

Dispersion without Deviation

Dispersion without Deviation - Formula

Derivation of Dispersion without Deviation

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Dispersion without Deviation: Definition, Formula, Derivation, Solved Examples

Dispersion without Deviation

Dispersion without Deviation - Formula

Derivation of Dispersion without Deviation

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

3.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: