State the importance of coherent sources in the phenomenon of interference.
In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence, deduce the expression for the fringe width.
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
If coherent sources are not taken, the phase difference between two interfering waves, will change continuously and a sustained interference pattern will not be obtained. Thus, coherent sources provide sustained interference pattern.
Let S1 and S2 be two coherent sources separated by a distance d.
Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.
Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 
The path difference between two waves reaching at P from S1 and S2 is given by, 
Expression for Fringe Width:
Distance between any two consecutive bright fringes or any two consecutive dark fringes are called the fringe width. It is denoted by 
.
Let, 
be the distance of two consecutive fringes. Then, we have
So, fringe width = 
Fringe width is same for both bright and dark fringe.
When the entire apparatus is immersed in water, the fringe width decreases.
Expression for fringe width in Young's Double Slit Experiment
Let S1 and S2 be two slits separated by a distance d.
GG' is the screen at a distance D from the slits S1 and S2.
Point C is equidistant from both of the slits.
The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits is given by,
S1 = S2 P - S1 P
Now, S1 S2 = d, EF = d, and S2 F = D
In 
S2P = 
For constructive interference, the path difference is an integral multiple of wavelengths, that is, path difference is n
.
Graph of intensity distribution in young's double slit experiment is, 
ii)
Three distinguishing features observed in Young's Double Slit experiment as compared to single slit diffraction pattern is,
1. In the interference pattern, all the bright fringes have the same intensity. The bright fringes are not of the same intensity in a diffraction pattern.
2. In interference pattern, the dark fringes have zero or small intensity so that the bright and dark fringes can be easily distinguished. While in diffraction pattern, all the dark fringes are not of zero intensity.
3. In interference pattern, the width of all fringes are almost the same, whereas in diffraction pattern, the fringes are of different widths.
(a) How does an unpolarized light incident on a Polaroid get polarized?
Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
(b) Two Polaroid’s ‘A’ and ‘B’ are kept in crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A?a) When an unpolarized light falls on a polaroid, only those electric vectors that are oscillating along a direction perpendicular to the aligned molecules will pass through. Thus, incident light gets linearly polarized.
Electric vectors which are along the direction of the aligned molecules gets absorbed.
Whenever unpolarized light is incident on the boundary between two transparent media, the reflected light gets partially or completely polarized. When reflected light is perpendicular to the refracted light, the reflected light is a completely polarized light.
b) Let the angle between A and C be 
.
Intensity of light passing through A = Io/2
Intensity of light passing through C = Io/2 cos2 
Intensity of light passing through B = Io/2 cos2 
. cos2 (90 - 
)
Given that, 
Thus, the third polaroid is placed at an angle of 45o.
a) In Young’s double slit experiment, derive the condition for (i) constructive interference and
(ii) Destructive interference at a point on the screen.
(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
a)
Conditions of constructive interference and destructive interference.
Consider two coherent waves travelling in the same direction along a straight line.
Frequency of each wave is given by 
,
Amplitude of electric field vectors are a1 and a2 rspectively.
Wave equation is represented by, 
Intensity of the wave is proportional to the amplitude of the wave.
Thus, Intensity of the resultant wave is given by, 
Constructive interference: For maximum intensity at any point, cos = +1
So, maximum intensity is, 
Path difference is, 
Constructive interference is obtained when the path difference between the waves is an integral multiple of 
Destructive Interference: For minimum intensity at any point, cos = -1
Phase difference is given by, 
Path difference is, 
In destructive interference, path difference is odd multiple of 
.
b)
Given,
d = 0.28 mm = 0.28 x 10-3 m
As, 
If n1 = n then, n2 = n+1
(a) State Huygens principle. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe.
(c) If a monochromatic source of light is replaced by white light, what change would you observe in the diffraction pattern?
Huygen’s principle:
(i) Every point on a given wavefront may be regarded as a source of new disturbance.
(ii) The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets.
(iii) The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time.
a) Diffraction of light at a Single slit
A single narrow slit is illuminated by a monochromatic source of light. The diffraction pattern is obtained on the screen placed in front of the slits. There is a central bright region called as central maximum. All the waves reaching this region are in phase hence the intensity is maximum. On both side of central maximum, there are alternate dark and bright regions, the intensity becoming weaker away from the center. The intensity at any point P on the screen depends on the path difference between the waves arising from different parts of the wave-front at the slit.
According to the figure, the path difference (BP – AP) between the two edges of the slit can be calculated.
Path difference, BP – AP = NQ = a sin 
Angle is zero at the central point C on the screen, therefore all path difference are zero and hence all the parts of slit are in phase. Due to this, the intensity at C is maximum. If this path difference is 
, (the wavelength of light used), then P will be point of minimum intensity. This is because the whole wave-front can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is 
, then the path difference between the secondary waves from A and C reaching P will be 
/2, and path difference between the secondary waves from B and C reaching P will again be 
/2. Also, for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is 
/2. Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum.
b)
Central bright lies between 
Therefore, Angular width of central bright fringe = 
So, 1st diffraction fringe lies between 
and 
Therefore,
Angular width of first diffraction fringe is, 
So, 
c) When monochromatic light is replaced by white light, each diffraction band splits into a number of colored bands. Angular width of violet being the least and that of red light is maximum.
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