Interference ,Diffraction and Polarisation

Principle of Superposition of light

When two or more progressive wave travels simultaneously in a medium without affecting the motion of each other, the resultant displacement of each particle of the medium at any instant of time is equal to the vector sum of the displacement produced by the two waves separately. This is the principle of superposition of light

Time in x-axis

Displacement in y axis

Y=y1 +y2

The superposition principle holds good for both mechanical waves and electromagnetic waves.

The principle of superposition of waves can be explained as:

a. When two waves having same frequency travel in same direction meet each other, constructive interference is formed.

b. If there are number of waves, they travel independently and characteristics of the wave remain unchanged.

c.The resultant intensity at any point in the medium is not equal to sum of intensity of thetwo waves. In constructive interference, the total intensity is greater than sum of intensities of individual waves whereasin destructiveinterference the total intensities is less than the sum of individual intensities.

d. Beats are formed when two waves differing in frequency superimposed.

e. Energy distribution follows the law of conservation of energy.

Coherent sources

Two light sources are said to be coherent sources if they emit continuous light wave of same frequency or wavelength, nearly same amplitude and in phase or constant phase different.

The characteristic of the coherent are given below.

Coherent source have same frequency, wavelength and amplitude

Coherent source have same phase difference or constant phase difference.

They can be realized in practice in:

a. Lloyd’s single mirror (real narrow source and virtual image produced by reflection)

b.Fresenlbiprism(two virtual image produced by the same sourceby reflection)

c.Michelson’sinterferometer (dividing amplitude portion of the wave front into two parts by reflection and refraction

Interference of light

Interference of light is process in which the light energy is redistributed in the medium on account of superposition of light wave from two coherent sources. If the wave of two coherent sources traveling in a medium interferes (superpose) at a point in a way that the crest or tough of one wave fall exactly on the crest or trough of other wave then it is called constructive interference. But when crest of one wave falls on the trough of another wave or vice-versa, it is called destructive interference.

The conditions for interference of the light are given below:

The two sources should be coherent.

The two sources must be very closer to each other.

The two sources must emit continuous light.

The two sources must be travel in same direction

Two independent sources of light cannot produce interference because for producing interference pattern, there is required of two coherent source vibrating in same phase which not possible by two independent sources of light. Two independent sources cannot be coherent because light is emitted by individual atoms. Even a very small source consists of millions of atoms, and emission of light by them takes place independently. An atom emits an unbroken wave of above 10-8 second due to its transition from a higher energy state to a lower state. So, the phase difference between two independent light sources changes after 10-8second. The rapid changes cannot be seen by our eyes and almost a uniform illumination on the screen is observed. So, we cannot use two independent sources to produce interference pattern.

Let us consider the monochromatic light of wavelength λpasses through the two slits P and Q, which act as coherent sources .The light reaching at the point M has the phase difference ϕ.

Let y­1and y2 are the displacement of light eave coming from P and Q respectively.

By superposition principle Y =y1 +y2

=asinwt + asin (wt +ϕ)

= asinwt (1+ cosϕ) + acoswtsinϕ

Put, A cosϕ = a (1+ cosϕ)…………..1

A sinϕ = a sinϕ ………………2

Then,

Y = Acosϕsinwt + Asinϕcoswt

Y = Asin(wt +ϕ)

Squaring 1 and 2

A2 = 4 A20$A_0^2$cosϕ22$\frac{{\varphi }^2}{2}$

Since Intensity at a point is given by square of the amplitude.

I=4I0cosϕ22$\frac{{\varphi }^2}{2}$ and Io=A20$A_0^2$

Case. 1

Condition for constructive interference

For the constructive interference I=maximum as if cosϕ22$\frac{{\varphi }^2}{2}$=1

i.e. ɸ=2∏nwhere n=0, 1, 2, 3, 4, ……..etc.

Since the phase difference of 2∏ cross ponds to path difference of λ, the constructive path difference is , x=nλ where n=0, 1, 2, 3, 4, ……etc.

Hence the constructive path difference is integral multiple of 2∏ or λ.

On the other hand if the fringth width is equal to the integral multiple of the wavelength of the light i.e. β=(xnd)/D=n λ where n=0, 1, 2, 3, …….n

Case. 2

Conditionfor destructive interference

For destructive, I=minimum

i.e. cos2=0$\frac{}{2}=0$

or, ɸ=(2n-1)∏where n= 1,2, 3, 4, ………………….etc.

and hence destructive path difference is the integral multiple of ∏ or λ/2.

On the other hand we also have if the fringe width is equal to the odd multiple of half wavelength of light i.e. β=(xnd)/D={(2n+1)}λ/2 where n=0, 1, 2,3, ……….n.

As we see the intensity of the lightat maximum i.e. 4A2and minimum is 0 so, the variation of interference lies between 0 to 4A2. A/c to the law of conservation of the light energy i.e. the energy neither can be created nor created but is transferred fromplace to another in this case the point of minimum intensity to the point of maximum intensity. Thus the energy which apparently disseapar (dark) at minima has actually been transferred to the maximum (bright) where the intensity is greater than that produce by two beams acting sepreatly. So, interference follows the law of the conservation.

Young’s double slit experiment

LetA and B be the two narrow silt illuminated by monochromatic from the source S so the light vibrations issues from the two silts are in the phase and they behave as coherent sources. Let the distance between A and B be the d and the distance of the screen XY placed perpendicular to the SO be D. the point is equidistance from A and B, hence the wave originating from A and B will reinforce each other on reaching O. thus O will be the centre of the bright fringe.

The illustration at any point P, distant x from O on the screen can be calculated by finding the path difference (BP-AP).joint AP and BP and draw AC perpendicular on BP then,

BP2=BF2+FP2=D2+(x+d2)2${\left(x+\frac{d}{2}\right)}^2$……………..1

Also, AP2=AE2+EP2=D2+(x−d2)2${\left(x-\frac{d}{2}\right)}^2$………………..2

On subtracting 2 from 1 we get

BP2-AP2=2dx

Or, (BP-AP) (BP+AP)=2dx…………3

As P lies very close to O hence to first approximation BP=AP=D

Or, path difference (BP-AP)=BC=dxD$BC=\frac{dx}{D}$……………………4

Condition for bright fringe: for P to lie at the centre of bright fringe from equation 4

We have dxD$\frac{dx}{D}$=nλ

or, x=Dnλd$\frac{Dn\lambda }{d}$

Let xn be the distance of nth bright fringe then xn= Dnλd…………..5$\frac{Dn\lambda }{d}\dots \dots \dots \dots ..5$

ann also let xn-1 be the distance of (n-1)th bright fringe then xn= (n−1)Dλd…………..6$\frac{\left(n-1\right)D\lambda }{d}\dots \dots \dots \dots ..6$

from 5 and 6 we get β=Dλd$\frac{D\lambda }{d}$

Condition for dark fringe:

for P to lie at the centre of bright fringe from equation 4

we havedxD$\frac{dx}{D}$=(2n+1)λ/2

or, x=D(2n+1)λ2d$\frac{D\left(2n+1\right)\lambda }{2d}$

letxn be the distance of nth dark fringe then xn= D(2n+1)λ2d…………..7$\frac{D\left(2n+1\right)\lambda }{2d}\dots \dots \dots \dots ..7$

ann also let xn-1 be the distance of (n-1)th bright fringe then xn= D{2(n−1)+1}λ2d…………..8$\frac{D\left\{2\left(n-1\right)+1\right\}\lambda }{2d}\dots \dots \dots \dots ..8$

from 7 and 8 we get β=Dλd$\frac{D\lambda }{d}$

Newton’s ring:

When the planoconcavex lens of the longer focal length is placed on the glass plate, a thin film of air is enclosed between the lowersurface of the plate. Thickness of the air film is negligible at the point of contact and it gradually increase from the centre toward periphery. When the air film is illuminated by a monochromatic light, the concentric dark and bright circular fringes with a dark centre are observed in the interference pattern. These dark and bright circular fringes are called Newton’s ring.

Experimental arrangement

Newton ringare formed when a plano-convex lens P of the larger radius of curvature placed on a sheet of plane glass AB illuminated from the top with monochromatic light. The combination from a thin circular air film of variable thickness in all direction around the pont of contact of the lens and the glass plate. The locus of all the point corresponding to the specific thickness of air film fall on the circle whose centre is at O. consequently, interference fringe observe in the form of concentric ring with their centre at O. Newton originally observed these concentric circular fringes and hence they are called Newton’s ring.

Figure

Monochromatic light from extended source S is regarded parallel by a lens L. it is incident on a glass plate inclined at 450to the horizontal, and is reflected normally drawn on to a planoconvex lens replaced on a glass plate. The part of the light incident on the system is reflected from the glass-to-air boundary say from D. the remainder of the light is transmitted through the air film. It is again reflected from the air-to-glass boundary. Say from J. the two rays reflected from the top and bottom of the air film are derived through the division of the amplitude from the same incident ray CD and are therefore coherent. The ray 1 and 2 are closed to each other and interfere to produce to darkness or brightness. The condition of brightness or darkness depends on the path difference between the two reflected light ray, which in turn depend on the thickness of the air film at the point of incidence.

Condition for bright or constructive interference:

For the constructive interference the ray which get reflected from the upper surface of the planoconvex have 2μt cosɵ-λ/2=nλ (when the light get reflected from surface of denser medium) therefore we have

2μt cosɵ= (2n+1)λ2$\frac{\left(2n+1\right)\lambda }{2}$=(2n+1)λ2$\frac{\lambda }{2}$

Since the film is thin so we have cosɵ=1and μ=1

Then we have 2t=(2n+1)λ2$\frac{\lambda }{2}$……………1

Hence the path difference for constructiveinterference or bright fringe is odd multiple of λ/2 in reflection case.

Let r be the radius of curvature of nth bright ringand D be the diameter as shown in figure and R be the radius of the curvature of lens then by geometry of circle.

r♢r=t(2R−t)$r\mathrm{♢}r=t\left(2R-t\right)$

or,$or,$r2=2Rt- t2

or,$or,$r2=2Rt ( since t is the thickness of the film which is very very thin and the square too much small so we neglecting )

Since D be the diameter of the ring then r=D/2 then from the above expression we have

D2=8Rt

Or, D2/4R=2t………….2

From 1 and 2 we have

D2=2(2n+1)λR……………………..3

From equation 3 we also have

We also have D2n$D_n^2$=2λR(2n+1)…………4

Equation 4 representsthe diameter of nth bright ring

Similarly we also have that the diameter of the (n+m)th bright ring is

D2n+m$D_{n+m}^2$=2λR {2(n+m)+1}…….5

Subtracting equation 4 by 5 we get

D2n+m−D2n=2λR{2n+2m−(2n+1)}$D_{n+m}^2-D_n^2=2\lambda R\left\{2n+2m-\left(2n+1\right)\right\}$

or,λ={D2n+m−D2n}4mR$or,\lambda =\frac{\{D_{n+m}^2-D_n^2\}}{4mR}$

This is the equation to determine the wavelength of the light.