Que 1: Define the terms:
(i) Interference : When light from two sources move in a direction then the light wave trains from there sources superimpose upon each other resulting in the modification of distribution of intensity i.e Position Maximum Intensity. This modification of light energy due to super position of two or more wave trains are called Interference.
(ii) Diffraction : The bending of waves around an obstacle and deviation from a rectilinear path is called Diffraction.
(iii) Stimulated Emission : It is a process in which there is an emission of a photon whenever an atom transits from a higher energy state to a lower energy under the influence of an external agency, i.e., an inducing photon.
(iv) Spontaneous Emission : It is a process in which there is an emission of a photon whenever an atom transits from a higher energy state to a lower energy state without the aid of any external agency.
(v) Absorption : Atom be initially in the lower state E1, If a photon of energy hν is incident on the atom in the lower state, the atom absorbs the incident photon and gets excited to the higher energy state E2. Where, B12 is the proportionality constant, N1 is population of the lower energy level, R12 is Rate of absorption.
(vi) LASER : Light Amplification by Stimulated Emission of Radiation Laser is a device which can produce a high intense, highly coherent, more directional and highly monochromatic beam.
(vii) Fringe width : The spacing between any two consecutive bright and dark fringes is called the fringe width(β).
Que 2: Define Interference phenomena of light. and Explain Young’s double slit experiment of Interference phenomenon of light using necessary diagram.
Ans: When two waves of the same frequency having a constant initial phase difference traverse simultaneously in a medium superpose each other, the resultant intensity of light is not distributed uniformly in space. This modification in intensity is called interference.
On the basis of wave theory of light, the formation of dark and bright fringes can be explained. The cylindrical wavefronts starting from S falls on S1 and S2. According to Huygens principle, S1 and S2 become center’s of secondary wavelets, i.e., two cylindrical wave fronts emerge out from S1 and S2.
Their radii increase as they move away from S1 and S2 so that they superimpose more and more on each other. At points where crest (or trough) due to one falls on a crest (or trough) due to the other, the resultant amplitude is the sum of the amplitudes due to each wave separately.
The intensity, which is proportional to the square of the amplitude, at these points is therefore a maximum. This is the case of constructive interference.
At points where a crest due to one falls on a trough due to the other, the resultant amplitude is the difference of the amplitudes due to separate waves and the resultant intensity is a minimum. This is the case of destructive interference.
Que 3: Discuss the types of interference phenomenon of light.
Ans: When two waves of the same frequency having a constant initial phase difference traverse simultaneously in a medium superpose each other, the resultant intensity of light is not distributed uniformly in space. This modification in intensity is called interference.
There are two type of interference. Constructive interference and Destructive interference.
When the amplitudes of the two wave trains and there by increasing the intensity of resultant wave It is called Constructive interference.
when the amplitude of the two wave trains and there by a decreasing the intensity of resultant wove It is called destructive interference.
Que 4: Obtain an expression for fringe width in case of Interference phenomena of light. Prove that in this case of interference dark and bright bands are of equal width.
Ans:
Let S be a narrow slit illuminated by monochromatic light and S, and S, be two parallel slits very close together and equidistant from S. The light waves from S, an S, produce an interference pattern on a screen XY placed parallel to S, and S, as shown in.
$$\small\begin{align}(S_{2}P)^{2} & = (S_{2}M_{2})^{2} + (PM_{2})^{2}\\
& = D ^ {2} + (x + d) ^{2}\end{align}\normalsize$$
$$= D ^ {2}[1 +\frac{(x + d)^{2}}{D^{2}}]$$
$$S_{2}P = D [1 + \frac{(x + d) ^{2}}{D^{2}}]^{1/2}$$
Since D>> (x+d), the binomial expansion up to two terms will give
$$S_{2}P = D [1 + \frac{(x + d) ^{2}}{D^{2}}]^{1/2}$$
$$= D[1 + \frac{1}{2} \frac{(x + d) ^{2}}{D^{2}}]$$
$$= D + \frac{(x + d)^{2}}{2D}$$
$$S_{1}P = D + \frac{(x – d) ^{2}}{2D}$$
$$S_{2}P – S_{1}P = \frac{2xd}{D}$$
Now for maxima or bright fringes, the path difference in given by (S2 P – S1 P ) = nλ where n = 0, 1, 2
$$\frac{2xd}{D} = nλ$$
$$x = n \frac{Dλ}{2d}$$
and For dark fringes, i.e, for minima the path difference in given by
$$S_{2}P – S_{1}P = (2n + 1)\frac{λ}{2} ,where n = 0, 1, 2$$
$$\frac{2xd}{D} = (2n + 1)\frac{λ}{2}$$
$$x =\frac{D}{2d} (2n + 1)\frac{λ}{2}$$
Now let xn and xn+1 denote the distances of nth and (n + 1)th bright fringes, then
$$x_{n} = n\frac{Dλ}{2d}$$
$$x_{n+1} =(n+1)\frac{Dλ}{2d}$$
Spacing between nth and (n + 1)th bright fringe or successive bright fringes is
$$x_{n+1} – x_{n} = (n+1)\frac{Dλ}{2d} – n\frac{Dλ}{2d}$$
It is independent of n. Hence, spacing between any two consecutive bright fringes is same. Similarly, it can be shown that spacing between two consecutive dark fringes will also be Dλ/2d.
The spacing between any two consecutive bright and dark fringes is called the fringe width (β)
$$β = \frac{Dλ}{2d}$$
Que 5: Derive the condition for Maxima and minima intensity of light in case of Interference phenomenon of light.
Ans: Condition for maximum intensity:-
Intensity I is maximum when cosδ = +1 ,
δ = 2nπ, n = 0, 1, 2 ,3_____
Path difference =(S2 P – S1 P ) = nλ
and Imax = a12+a22+2a1a2 = (a1+a2)2
Imax>I1+I2
This shows that the resultant intensity is greater then the sum of two separate intensities.
Condition for minimum intensity:-
Intensity I is maximum when cosδ = -1 ,
δ = (2n+1)π, n = 0, 1, 2 ,3_____ $$\small\begin{align}Path Difference & = (S_{2}P – S_{1}P )\\ & = \frac{(2n+1)πλ}{2}\normalsize\end{align}$$
and Imin = a12+a22-2a1a2 = (a1+a2)2
Imin<I1+I2
This, the resultant intensity is less then the sum of two separate intensities.
Que 6: Define Diffraction. Explain the types of diffraction phenomena of light with examples.
Ans: The bending of waves around an obstacle and deviation from a rectilinear path is called Diffraction.
It is matter matter of common experience that the path of light entering a dark room through hole in the window illuminated a by sunlight is straight But it has been observed that when a beam of light passes through a small opening it spreads to some extent into the region of the geometrical shadow also.
When Waves pass near an obstacle (barrier), they bend to bend around the edges of the obstacle.
The diffraction are divided into two classes, (i) Fresnel diffraction (ii) Fraunhofer diffraction
Fresnel diffraction: In this class of diffraction, the source of light and the screen are at finite distance from the diffracting aperture or obstacle having sharp edge. The wave front incident on the aperture or obstacle is either spherical or cylindrical.
Fraunhofer diffraction: In this class of diffraction the source of light and the screen are at finite distance from the diffraction aperture or obstacle having sharp edge. This can be achieved by placing the light source at the focal plane of the convex lens and placing the screen at the focal plane of another convex lens. In this case the wave front incident on the aperture or obstacle is a plane wave front.
Que 7: Differentiate Interference and Diffraction phenomena of light.
Ans:
| Interference | Diffraction |
| The Interference takes place Phenomenon, the interaction between two sperate wave fronts originating from two coherent sources. | The Phenomenon of Diffraction, the interaction takes place between the secondary wavelets originating from different points of the exposed parts of the same. wave front. |
| In Interference pattern the region of minimum intensity usually dark perfectly. | it is not so in diffraction pattern. |
| The widths of the fringes in interference may as may not be equal or uniform. | The widths of the fringes in diffraction never equal. |
| In an interference pattern all the maxima are of same intensity. | In diffraction pattern all the maxima are they are of varying intensity. |
Que 8: State different characteristics of Laser.
Ans: Coherence
The wave trains which are identical in phase and direction are called coherent waves. Since all the constituent photons of laser beam possess the same energy, momentum and in same direction, the laser beam is said to be highly coherent
High Intensity
Due to the coherent nature of laser, it has the ability to focus over a small area of 10-6 cm2, extremely high concentration of its energy over a small area.
High Directionality
An ordinary light source emits light in all possible directions. But, since laser travels as a parallel beam it can travel over a long distance without spreading.
The angular spread of a laser beam is 1 mm/meter. This reveals the directionality of the laser beam
High Monochromaticity
The light from a normal monochromatic source spreads over a range of wavelength of the order 100 nm. But, the spread is of the order of 1 nm for laser. Hence, laser is highly monochromatic, i.e. if can emit light of single wavelength.
Que 9: Discuss the differences between Spontaneous emission and Stimulated emission.
Ans:
| Stimulated emission | Spontaneous emission |
| Emission of a light photon takes place through an inducement by an external photon. | Emission of a light photon takes place immediately without any inducement. |
| It is not a random process. | It is a random process. |
| The photons get multiplied through chain reaction. | The photons do not get multiplied through chain reaction. |
| It is a controllable process. | It is an uncontrollable process. |
| More intense. | Less intense. |
| Monochromatic radiation. | Polychromatic radiation. |
Que 10: What are Einstein Coefficients? Derive the relation between them for LASER.
Ans: Einstein obtained a mathematical expression for the existence of two different kinds of processes Spontaneous emission and stimulated emission.
Since the transition between the atomic energy states is a statistical process, it is not possible to predict which particular atom will make a transition from one state to another at a particular instant.
Let us assume that the atomic system is in equilibrium with electromagnetic radiation. Hence at thermal equilibrium, the number of upward transitions is equal to the number of downward transitions per unit volume per second.
The rate of absorption = The rate of emission
B12 N1ρ= A21 N2+ B21 N2 ρ
From the above equation,
(B12 N1 – B21 N2)ρ = A21 N2
$$ρ= \frac{A_{21} N_{2}}{B_{12} N_{1} – B_{21} N_{2}}$$
$$ρ= \frac{A_{21}}{B_{12} \frac{N_{1}}{N_{2}} – B_{21}}—– (1)$$
Under thermal equilibrium, the number of atoms N1 and N2 in the energy states E1 and E2 at a temperature T is given by Boltzmann distribution law. Hence, we have
$$N_{1} = N_{0} e^{-E_{1}/k_{B}T}$$
$$N_{2} = N_{0} e^{-E_{2}/k_{B}T}$$
$$\frac{N_{2}}{N_{1}} = \frac{e^{-E_{2}/k_{B}T}}{e^{-E_{1}/k_{B}T}} = e^{-(E_{2} -E_{1})/k_{B}T}$$
$$\frac{N_{2}}{N_{1}} = e^{-hv/k_{B}T}$$
$$\frac{N_{1}}{N_{2}} = e^{hv/k_{B}T}——(2)$$
Substituting equation (1) and (2)
$$ρ= \frac{A_{21}}{B_{12} e^{hv/k_{B}T} – B_{21}}$$
$$ρ= \frac{A_{21}}{B_{21}}\frac{1}{\frac{B_{12}}{B_{21}} e^{hv/k_{B}T} – 1}—–(3)$$
But from Planck’s black body theory of radiation
$$ρ = \frac{8πhv^{3}}{c^{3}}\frac{1}{e^{hv/k_{B}T}-1}—–(4)$$
Hence, comparing equations (3) and (4), we get
$$\frac{A_{21}}{B_{21}} = \frac{8πhv^{3}}{c^{3}}$$
B12 = B21
Take B12 = B21 = B and A21 = A. The constants A and B are called Einstein’s coefficients Equation shows that the probability of absorption is equal to the probability of stimulated emmision.
Que 11: List out the types of LASER. Describe the construction and working principle of solid state Laser (Nd-YAG Laser) along with necessary diagrams.
Ans: Types of LASER
Solid State Lasers ( Nd-YAG laser, Ruby laser)
Liquid Lasers
Gaseous Lasers ( He-Ne laser, CO2 laser)
Dye lasers
Semiconductor Laser
Construction:
This laser system has two absorption band (0.73 m and 0.8 m)
Optical pumping mechanism is employed.
Laser transition takes place between two laser levels at 1.06 mm.
A Nd: YAG rod and a krypton flash lamp are enclosed inside an ellipsoidal reflector. In order to make the entire flash radiation to focus on the laser rod, the Nd: YAG rod is placed at one focal axis and the flash lamp at the other focal axis of the ellipsoidal reflector.
Working:
The flash lamp is switched on. The optical pumping excites the Nd3+ ions from the ground energy state E0 to the higher energy level E3 and E4 by absorbing radiations of wavelength 0.80 μm and 0.73 pm, respectively. The energy-level diagram is shown in Figure 6.5. The excited Nd3+ ions then make a transition from these energy levels. The transition from the energy level E4 to E2 is a non-radiative transition. The state E2 is the metastable state. Upon continuous excitation, population inversion of Nd3+ ions is achieved at the metastable state E2.
Any of the spontaneously emitted photon will make the excited Nd3+ ions to undergo a transition between E2→E1 state. Thus, during this transition the stimulated photon is generated. The photons travelling parallel to the resonator axis experience multiple reflections at the mirrors. As a result, the transition E2→E1 yields an intense and coherent laser beam of wavelength 1.064 um. These lasers give beam continuously. The Nd3+ ions then make a transition between E1→E0 which is a non-radiative transition
Que 12: Discuss the various applications of LASER in the field of science and engineering.
Ans: (1) It is used in fiber optic communication.
(2) Communication between planets is possible with laser.
(3) It is used in holography.
(4) It is used in underwater communication between submarines, as they are not easily absorbed by water.
(5) It is used to accelerate some chemical reactions.
(6) It is used to create new chemical compounds by destroying atomic bonds between molecules.
(7) It is used to drill minute holes in cell walls without damaging the cell itself.
Que 13: In a Young’s double slit experiment , the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe has been measured to be 1.2 cm. Determine the wavelength of light.
Ans: Distance between the slits, 2d = 0.28×10−3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n=4) fringe, x = 1.2×10−2 m
$$⇒x =\frac{nλD}{2d}$$
$$⇒λ=\frac{2dx}{nD}$$
$$⇒λ=\frac{0.28×10^{−3}× 1.2×10^{−2}}{1.4×4}$$
$$⇒λ=6×10^{−7} m = 600\,nm$$
Que 14: Two coherent sources are placed 1 mm apart and generate interference fringes on a screen 0.9 m away. The second dark fringe is formed at a distance of 0.9 mm from the central fringe. Determine the wavelength of the monochromatic light used.
Ans: $$x = (2n−1)\frac{λD}{2d}$$
$$λ = \frac{2xd}{(2n−1)D}$$
$$λ =\frac{2×0.9×10^{−3}×1}{(4-1)×0.9}$$
$$λ = 6.67 × 10^{-4}\,m$$
Que 15: Two coherent sources are placed 0.2 mm apart and the fringes are observed on the screen 1 m away. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 10 mm from the central fringe. Find the wavelength of light.
Ans: $$x_{n} = n\frac{Dλ}{2d}$$
$$λ = \frac{2dx_{n}}{D×n}$$
Here, xn= 10 mm = 1 cm , 2d = 0.2 mm = 0.02 cm , n=4, D = 1 m = 100 cm
$$ λ = \frac{1×0.02}{4×100}= 5×10^{-5} cm =5000 Å$$
Que 16: In Young’s double slit experiment, the angular width of a fringe found on a distant screen is 0.1°. The wavelength of light used is 6000 Å. What is the spacing between the slits?
Ans:
$$Angular \,fringe\, width: β =\frac{λ}{d}$$
$$d=\frac{6000×10^{-10}}{β}$$
$$∴ d =6×10^{−7}×{\frac{180}{0.1×π}}$$
$$d = 3.438×10^{−4}m$$
Que 17: Two coherent sources of monochromatic light of wavelength 6000 Å produce an interference pattern on a screen kept at a distance of 1 mm from them. The distance between two consecutive bright fringes on the screen is 0.5 mm. Find the distance between the two coherent sources.
Ans: The fringe width (β) is given by
$$β =\frac{Dλ}{2d}$$
The distance between the two coherent sources (2d) is
$$2d = \frac{Dλ}{β}$$
Given D = 1 m = 100 cm, λ = 6000 Å = 6000 × 10-8 cm and β = 0.5mm = 0.05 cm
$$2d = \frac{100×6000×10^{-8}}{0.05} = 0.12\,cm$$