
Chapter 20 – dispersion and Spectra
Page No 442:
Question 8:
Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are μ_{r}, μ_{y} and μ_{v} respectively and those for the flint glass are μ’_{r}, μ’_{y} and μ’_{v} respectively. Find the ratio A‘/A for which (a) there is no net angular dispersion, and (b) there is no net deviation in the yellow ray.
Figure
Answer:
For the crown glass, we have:
Refractive index for red rays = μ_{r}
Refractive index for yellow rays = μ_{y}
Refractive index for violet rays = μ_{v}
For the flint glass, we have:
Refractive index for red rays = μ’_{r}
Refractive index for yellow rays = μ’_{y}
Refractive index for violet rays = μ’_{v}
Let δ_{cy} and δ_{fy} be the angles of deviation produced by the crown and flint prisms for the yellow light.
Total deviation produced by the prism combination for yellow rays:
δ_{y} = δ_{cy} − δ_{fy}
= 2δ_{cy} − δ_{fy}
=2(μ_{cy} + 1)A − (μ_{fy} − 1)A’
Angular dispersion produced by the combination is given by
δ_{v} − δ_{r} = [(μ_{vc} − 1)A − (μ_{vf} − 1)A’ + (μ_{v}_{c} − 1)A −
μrc1Aμrf1A’+μrc1AHere,
μ_{vc}_{ = }Refractive index for the violet colour of the crown glass
μ_{vf} = Refractive index for the violet colour of the flint glass
μrc= Refractive index for the red colour of the crown glass
μrf= Refractive index for the red colour of the flint glass
On solving, we get:
δ_{v} − δ_{r} = 2(μ_{vc} −1)A − (μ_{vf} − 1)A’
(a) For zero angular dispersion, we have:
δ_{t} − δ_{t} = 0 = 2(μ_{vc} −1)A − (μ_{vf} − 1)A’
⇒ A’A=2(μvf1)(μvc1)=2(μrμr)(μrμ)(b) For zero deviation in the yellow ray, δ_{y} = 0.
⇒ 2(μ_{cy} − 1)A = (μ_{fy} − 1)A
⇒ A’A=2(μcy1)(μfy1)=2(μy1)(μ’y1)
Page No 441:
Question 1:
The equation
ω=μuμrμ1was derived for a prism having small refracting angle. Is it also valid for a prism of large refracting angle? Is it also valid for a glass slab or a glass sphere?
Answer:
Dispersive power depends on angular deviation, and angular deviation is valid only for a small refracting angle and a small angle of incidence. Therefore, dispersive power is not valid for a prism of large refracting angle. It is also not valid for a glass slab or a glass sphere, as it has a large refracting angle.
Question 2:
Can the dispersive power
ω=μuμrμ1be negative? What is the sign of ω if a hollow prism is immersed into water?
Answer:
No, it cannot be negative, as the refractive index for violet light is always greater than that for red light. Also, refractive index is inversely proportional to
λ2. The sign of ω will be positive, as
μis still greater than 1 and as
μv>μr.
Question 3:
If three identical prisms are combined, is it possible to pass a beam that emerges undeviated? Undispersed?
Answer:
No, it is not possible even when prisms are be combined with their refractive angle reversed with respect to each other. There will be at least a net deviation and dispersion equal to the dispersion and deviation produced by a single prism.
Question 4:
“Monochromatic light should be used to produce pure spectrum”. Comment on this statement.
Answer:
No, monochromatic light cannot be used to produce a pure spectrum. A spectrum is produced when a light of different wavelengths is deviated through different angles and gets separated. Monochromatic light, on the other hand, has a single wavelength.
Question 5:
Does focal length of a lens depend on the colour of the light used? Does focal length of a mirror depend on the colour?
Answer:
Yes, the focal length of a lens depends on the colour of light.
According to lensmaker’s formula,
1f=(μ1)(1R11R2)Here, f is the focal length, μ is the refractive index, R is the radius of curvature of lens.
The refractive index (μ) depends on the inverse of square of wavelength.
The focal length of a mirror is independent of the colour of light.
Question 6:
Suggest a method to produce a rainbow in your house.
Answer:
A rainbow can be produced using a prism. Another way of producing a rainbow is to dip a mirror inside water, keeping it inclined along the wall of a tumbler. The light coming from water after reflecting from the mirror will give a rainbow.
Question 1:
The angular dispersion produced by a prism
(a) increases if the average refractive index increases
(b) increases if the average refractive index decreases
(c) remains constant whether the average refractive index increases or decreases
(d) has no relation with average refractive index
Answer:
(a) increases if the average refractive index increases
If μ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by
δ=(μ1)A.
Page No 442:
Question 1:
A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the mean ray is zero. The refractive index of flint and crown glasses for the mean ray are 1.620 and 1.518 respectively. If the refracting angle of the flint prism is 6.0°, what would be the refracting angle of the crown prism?
Answer:
Given:
Refractive index of the flint glass, μ_{f} = 1.620
Refractive index of the crown glass, μ_{c} = 1.518
Refractive angle of the flint prism, A_{f} = 6°
Now,
Let the refractive angle of the crown prism be A_{c}.
For the net deviation of the mean ray to be zero,
Deviation by the flint prism = Deviation by the crown prism
i.e., (μ_{f} − 1)A_{f} = (μ_{c} − 1)A_{e}
⇒Ac=μf1μe1Af
⇒Ac=1.62011.5181×6.0°=7.2°Thus, the refracting angle of the crown prism is 7.2
°.
Question 2:
A certain material has refractive indices 1.56, 1.60 and 1.68 rfor red, yellow and violet lightespectively. (a) Calculate the dispersive power. (b) Find the angular dispersion produced by a thin prism of angle 6° made of this material.
Answer:
Given:
Refractive index for red light, μ_{r} = 1.56
Refractive index for yellow light, μ_{y} = 1.60
Refractive index for violet light, μ_{v} = 1.68
Angle of prism, A = 6°
(a) Dispersive power
ωis given by
ω=μvurμy1On substituting the values in the above formula, we get:ω=1.681.561.601
=0.120.60=0.2(b) Angular dispersion = (μ_{v} − μ_{r})A
=(0.12) × 6° = 0.72°
Thus, the angular dispersion produced by the thin prism is 0.72°.
Question 3:
The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersive power of the material of the lens.
Answer:
Focal lengths of the convex lens:
For red rays,
fr=100 cmFor yellow rays,
fy=98 cmFor violet rays,
fv=96 cmLet:
μr= Refractive index for the red colour
μy= Refractive index for the yellow colour
μv= Refractive index for the violet colour
Focal length of a lens
fis given by
1f=μ11R11R2Here,
μis the refractive index and R_{1} and R_{2} are the radii of curvatures of the lens.
Thus, we have:
μ1=1f×11R11R2⇒ μ1=kf k=11R11R2For red rays,
μr1=k100For yellow rays,
μy1=k98For violet rays,
μv1=k96Dispersive power (ω) is given by
ω=μvμrμy1Or,ω=(μv1)(μr1)(μy1)Substituting the values, we get:
ω=k96k100k98=98×49600⇒ω=0.0408Thus, the dispersive power of the material of the lens is 0.048.
Question 4:
The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material.
Answer:
Given:
Difference in the refractive indices of violet and red lights = 0.014
Let μ_{v} and μ_{r} be the refractive indices of violet and red colours.
Thus, we have:
μ_{v} − μ_{r} = 0.014
Now,
Real depth of the newspaper = 2.00 cm
Apparent depth of the newspaper = 1.32 cm
Refractive index=Real depthApparent depth∴Refractive index for yellow light μy is given byμy=2.001.32=1.515
Also,Dispersive power, ω=μvμrμy1 =0.0141.5151 Or, ω=0.0140.515=0.027Thus, the dispersive power of the material is 0.027.
Question 5:
A thin prism is made of a material having refractive indices 1.61 and 1.65 for red and violet light. The dispersive power of the material is 0.07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4.0° in favourable conditions. Calculate the angle of the prism.
Answer:
The refractive indices for red and yellow lights are μ_{r} = 1.61 and μ_{y} = 1.65, respectively.
Dispersive power, ω = 0.07
Angle of minimum deviation, δ_{y} = 4°
Now, using the relation ω=μvμrμy1, we get:⇒0.07=1.651.61μy1
⇒μy1=0.040.07=47Let the angle of the prism be A.
Angle of minimum deviation, δ = (μ − 1)A
⇒A=δyμy1=447=7°Thus, the angle of the prism is 7
°.
Question 6:
The minimum deviations suffered by, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.
Answer:
Given:
Minimum deviations suffered by
Red beam, δ_{r} = 38.4°
Yellow beam, δ_{y} = 38.7°
Violet beam, δ_{v} = 39.2°
If A is the angle of prism having refractive index μ, then the angle of minimum deviation is given by
δ=(μ1)A
⇒ μ1=δA …(1)
Dispersive power
ωis given by
ω=μvμrμy1 =(μv1)(μr1)(μy1)From equation (1), we get:
ω=δvAδrAδyA
⇒ω=δvδrδy=(39.2)(38.4)(38.7)
⇒ω=(0.8)38.7=0.0206So, the dispersive power of the medium is 0.0206.
Question 7:
Two prisms of identical geometrical shape are combined with their refracting angles oppositely directed. The materials of the prisms have refractive indices 1.52 and 1.62 for violet light. A violet ray is deviated by 1.0° when passes symmetrically through this combination. What is the angle of the prisms?
Answer:
Let A be the angle of the prisms.
Refractive indices of the prisms for violet light, μ_{1} = 1.52 and μ_{2} = 1.62
Angle of deviation, δ = 1.0°
As the prisms are oppositely directed, the angle of deviation is given by
δ = (μ_{2} − 1)A − (μ_{1} − 1)A
δ = (μ_{2} −μ_{1} )A
A=δμ2μ1=1(1.62)(1.52)=10.1⇒A=10°So, the angle of the prisms is 10^{âˆ˜}.
Question 2:
If a glass prism is dipped in water, its dispersive power
(a) increases
(b) decreases
(c) does not change
(d) may increase or decrease depending on whether the angle of the prism is less than or greater than 60°
Answer:
(b) decreases
If μ is the refractive index and A is the angle of prism, then the angular dispersion produced by the prism will be given by
δ=(μ1)A.
Because the relative refractive index of glass with respect to water is small compared to the refractive of glass with respect to air, the dispersive power of the glass prism is more in air than that in water.
Question 3:
A prism can produce a minimum deviation δ in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is
(a) 0
(b) δ
(c) 2δ
(d) 3δ
Answer:
(b) δ
In combination (refractive angles of prisms reversed with respect to each other), the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.
Question 4:
Consider the following two statements :
(A) Line spectra contain information about atoms.
(B) Band spectra contain information about molecules.
(a) Both A and B are wrong
(b) A is correct but B is wrong
(c) B is correct but A is wrong
(d) Both A and B are correct
Answer:
(d) Both A and B are correct.
Because line spectra contain wavelengths that are absorbed by atoms and band spectra contain bunch wavelengths that are absorbed by molecules, both statements are correct.
Question 5:
The focal length of a converging lens are f_{v} and f_{r} for violet and red light respectively.
(a) f_{v} > f_{r}
(b) f_{v} = f_{r}
(c) f_{v} < f_{r}
(d) Any of the three is possible depending on the value of the average refractive index μ.
Answer:
(c) f_{v} < f_{r}
Focal length is inversely proportional to refractive index and refractive index is inversely proportional to
λ2. So, keeping other parameters the same, we can say:
f∝1λ2 (∵ λr<λv)∴ f_{v} < f_{r}
Question 1:
A narrow beam of white light goes through a slab having parallel faces.
(a) The light never splits in different colours
(b) The emergent beam is white
(c) The light inside the slab is split into different colours
(d) The light inside the slab is white
Answer:
(b) The emergent beam is white.
(c) The light inside the slab is split into different colours.
White light will split into different colours inside the glass slab because the value of refractive index is different for different wavelengths of light; thus, they suffer different deviations. But the emergent light will be white light. As the faces of the glass slide are parallel, the emerging lights of different wavelengths will reunite after refraction.
Question 2:
By properly combining two prisms made of different materials, it is possible to
(a) have dispersion without average deviation
(b) have deviation without dispersion
(c) have both dispersion and average deviation
(d) have neither dispersion nor average deviation
Answer:
(a) have dispersion without average deviation
(b) have deviation without dispersion
(c) have both dispersion and average deviation
Consider the case of prisms combined such that the refractive angles are reversed w.r.t. each other. Then, the net deviation of the yellow ray will be
δy=(μy1)A(μy’1)A’And, the net angular dispersion will be
δyδr=μy1Aωω’Thus, by choosing appropriate conditions, we can have the above mentioned cases.
Question 3:
In producing a pure spectrum, the incident light is passed through a narrow slit placed in the focal plane of an achromatic lens because a narrow slit
(a) produces less diffraction
(b) increases intensity
(c) allows only one colour at a time
(d) allows a more parallel beam when it passes through the lens
Answer:
(d) allows a more parallel beam when it passes through the lens
To produce a pure spectrum, a parallel light beam is required to be incident on the dispersing element. So, the incident light is passed through a narrow slit placed in the focal plane of an achromatic lens.
Question 4:
Which of the following quantities related to a lens depend on the wavelength or wavelengths of the incident light?
(a) Power
(b) Focal length
(c) Chromatic aberration
(d) Radii of curvature
Answer:
(a) Power
(b) Focal length
(c) Chromatic aberration
The focal length, power and chromatic aberration are dependent on the refractive index of the lens, which itself is dependent on the wavelength of the light.
Question 5:
Which of the following quantities increase when wavelength is increased? Consider only the magnitudes.
(a) The power of a converging lens
(b) The focal length of a converging lens
(c) The power of a diverging lens
(d) The focal length of a diverging lens
Answer:
(b) The focal length of a converging lens
(d) The focal length of a diverging lens
The focal length of a lens is inversely proportional to the refractive index of the lens and the refractive index of the lens is inversely proportional to the square of wavelength. Therefore, the focal length is directly dependent on wavelength; it increases when the wavelength is increased.
Page No 443:
Question 9:
A thin prism of crown glass (μ_{r} = 1.515, μ_{v} = 1.525) and a thin prism of flint glass (μ_{r} = 1.612, μ_{v} = 1.632) are placed in contact with each other. Their refracting angles are 5.0° each and are similarly directed. Calculate the angular dispersion produced by the combination.
Answer:
For crown glass, we have:
Refractive index for red colour, μ_{cr} = 1.515
Refractive index for violet colour, μ_{cv} = 1.525
For flint glass, we have:
Refractive index for red colour, μ_{fr} = 1.612
Refractive index for violet colour,μ_{fv} = 1.632
Refracting angle, A = 5°
Let:
δ_{c} = Angle of deviation for crown glass
δ_{f}_{ }= Angle of deviation for flint glass
As prisms are similarly directed and placed in contact with each other, the total deviation produced
δis given by
δ = δ_{c} + δ_{f}
= (
μc– 1)A + (
μf– 1)A
= (
μc+
μf– 2)A
For violet light, δ_{v} = (μ_{cv} + μ_{fv} – 2)A
For red light, δ_{r} = (μ_{cr} + μ_{fr} – 2)A
Now, we have:
Angular dispersion of the combination:
δ_{v} – δ_{r} = (μ_{cv} + μ_{fv} – 2)A – (μ_{cr} + μ_{fr} – 2)A
= (μ_{cv} + μ_{fv} – μ_{cr} – μ_{fr}) A
= (1.525 + 1.632 – 1.515 – 1.612)5
= 0.15°
So, the angular dispersion produced by the combination is 0.15°.
Question 10:
A thin prism of angle 6.0°, ω = 0.07 and μ_{y} = 1.50 is combined with another thin prism having ω = 0.08 and μ_{y} = 1.60. The combination produces no deviation in the mean ray. (a) Find the angle of the second prism. (b) Find the net angular dispersion produced by the combination when a beam of white light passes through it. (c) If the prisms are similarly directed, what will be the deviation in the mean ray? (d) Find the angular dispersion in the situation described in (c).
Answer:
Given:
For the first prism,
Angle of prism, A‘ = 6°
Angle of deviation, ω’ = 0.07
Refractive index for yellow colour, μ’_{y} = 1.50
For the second prism,
Angle of deviation, ω = 0.08
Refractive index for yellow colour, μ_{y}= 1.60
Let the angle of prism for the second prism be A.
The prism must be oppositely directed, as the combination produces no deviation in the mean ray.
(a) The deviation of the mean ray is zero.
Thus, we have:
δ_{y} = (μ_{y} – 1)A – (μ’_{y} – 1)A‘ = 0
∴(1.60 – 1)A = (1.50 – 1)A‘
⇒ A =
0.50×6°0.60=5°
(b) Net angular dispersion on passing a beam of white light:
(μ_{y} – 1)ωA – (μ_{y} – 1)ω’A‘
⇒ (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°)
⇒ 0.24° – 0.21° = 0.03°
(c) For the prisms directed similarly, the net deviation in the mean ray is given by
δ_{y} = (μ_{y} – 1)A + (μ_{y} – 1)A‘
= (1.60 – 1)5° + (1.50 – 1)6°
= 3° + 3° = 6°
(d) For the prisms directed similarly, angular dispersion is given by
δ_{v} – δ_{r} = (μ_{y} – 1)ωA – (μ_{y} – 1)ω’A‘
= 0.24° + 0.21°
= 0.45°
Question 11:
The refractive index of a material M1 changes by 0.014 and that of another material M2 changes by 0.024 as the colour of the light is changed from red to violet. Two thin prisms, one made of M1(A = 5.3°) and the other made of M2(A = 3.7°) are combined with their refracting angles oppositely directed. (a) Find the angular dispersion produced by the combination. (b) The prisms are now combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.
Answer:
If μ’_{v} and μ’_{r} are the refractive indices of material M_{1}, then we have:
μ’_{v} – μ’_{r} = 0.014
If μ_{v} and μ_{r} are the refractive indices of material M_{2}, then we have:
μ_{v} – μ_{r} = 0.024
Now,
Angle of prism for M_{1}_{, }A’ = 5.3°
Angle of prism for M_{2}, A = 3.7°
(a) When the prisms are oppositely directed, angular dispersion
δ1is given by
δ_{1} = (μ_{v} – μ_{r})A – (μ’_{v} – μ’_{r})A‘
On substituting the values, we get:
δ_{1} = 0.024 × 3.7° – 0.014 × 5.3°
= 0.0146°
So, the angular dispersion is 0.0146°.
(b) When the prisms are similarly directed, angular dispersion
δ2is given by
δ_{2} = (μ_{v} – μ_{r})A + (μ’_{v} – μ’_{r})A’
On substituting the values, we get:
δ_{2} = 0.024 × 3.7° + 0.014 × 5.3°
= 0.163°
So, the angular dispersion is 0.163°.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity
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