NEET  >  Principle of Superposition & Interference

# Principle of Superposition & Interference - Notes | Study Physics Class 11 - NEET

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Principle of superposition :

When two or more waves simultaneously pass through a point, the disturbance of the point is given by the sum of the disturbances each wave would produce in absence of the other wave(s). In case of wave on string disturbance means displacement, in case of sound wave it means pressure change, in case of electro-magnetic wave (E.M.W) it is electric field or magnetic field. Superposition of two light travelling in almost same direction results in modification in the distribution of intensity of light in the region of superposition. This phenomenon is called interference.

Superposition of two sinusoidal waves :

Consider superposition of two sinusoidal wave (having same frequency), at a particular point.

Let,    X1(t) = a1 sin ωt
and,    x2(t) = a2 sin (ωt + φ)
x = A sin (ωt + φ0)

where    ...(1.1) {Refer topic: combination of SHM}

and

Ex.1 If i1 = 3 sin wt and i2 = 4 cos wt, find i3 which is given by i3 = i1 i2

Sol.

Ex.2 S1 and S2 are two source of light which produce individually disturbance at point P given by E1 = 3 sin wt, E2 = 4 cos wt. Assuming  to be along the same line, find the result of their superposition.

Sol.

Interference :

Interference implies super position of waves. Whenever two or more than two waves superimpose each other they give sum of their individual displacement.

Let the two waves coming from sources S1 & S2 be

Due to superposition

v = y1 + y2
ynet = A1 sin (ωt + kx1) + A2 sin (ωt + kx2)
Phase difference between y1 & y2 = k(x2 - x1)
i.e., Δφ = k(x2 - x1)

As   (where Δx = path difference &. Δφ = phase difference)

Anet =

When the two displacements are in phase, then the resultant amplitude will be sum of the two amplitude & Inet will be maximum, this is known of constructive interference.

For Inet to be maximum

= 2nπ where n = {0,1,2,3,4,5..}

For constructive interference

When superposing waves are in opposite phase, the resultant amplitude is the difference of two amplitudes & Inet is minimum; this is known as destructive interference.

For Inet to be minimum,

For destructive interference

Ex.3 Light from two source, each of same frequency and travelling in same direction, but with intensity in the ratio 4 : 1 interfere. Find ratio of maximum to minimum intensity.

Sol.

=  = = 9 : 1

Ex.4 Find the maximum intensity in case of interference of n identical waves each of intensity I0 if the interference is (a) coherent and (b) incoherent.

Sol. The resultant intensity is given by

(a) The sources are said to be coherent if they have constant phase difference between them. Then intensity will be maximum when f = 2np; the sources are in same phase.

Thus Imax = I1 I2 2 =

Similarly, for n identical waves,

(b) The incoherent sources have phase difference that varies randomly with time

Hence for n identical waves,

YOUNG'S DOUBLE SLIT EXPERIMENT (Y.D.S.E.) :

In 1802 Thomas Young devised a method to produce a stationary interference pattern. This was based upon division of a single wavefront into two ; these two wavefronts acted as if they illuminated from two sources having a fixed phase relationship. Hence when they were allowed to interfere, stationary interference pattern was observed.

Figure : Young's Arrangement to produce stationary interference pattern by division of wave front S0 into S1 and S2

Figure : In Young's interference experiment, light diffracted from pinhole S0 encounters pinholes S1 and S2 in screen B. Light diffracted from these two pinholes overlaps in the region between screen B and viewing screen C, producing an interference pattern on screen C.

The geometry of experiment is simple Parallel wavefront of a monochromatic wave are incident on two identical narrow slits, each of width a separated by a distance d. The slit widths & their separation are of the order of the wavelength of the incident monochromatic light. Monochromatic light after passing through two slits S1 & S2 acts as coherent sources of light waves that interfere constructively & destructively at different point on the screen to produce a interference pattern.

Analysis of Interference Pattern :

We have insured in the above arrangement that the light wave passing through S1 is in phase with that passing through S2. However the wave reaching P from S2 may not be in phase with the wave reaching P from S1, because the latter must travel a longer path to reach P than the former. We have already discussed the phase-difference arising due to path difference. if the path difference is equal to zero or is an integral multiple of wavelengths, the arriving waves are exactly in phase and undergo constructive interference. If the path difference is an odd multiple of half a wavelength, the arriving waves are out of phase and undergo fully destructive interference. Thus, it is the path difference Dx, which determines the intensity at a point P.

Path difference Dp = S1P - S2P =  ...(1)

Approximation I :

For D >> d, we can approximate rays and as being approximately parallel, at angle q to the principle axis.

Now, S1P - S2P = S1A = S1 S2 sinθ

⇒ path difference = d sinθ ...(2)

Approximation II :

further if q is small, i.e., y << D, sinθ  tanθ =

and hence, path difference =  ...(3)

for maxima (constructive interference)

Dp =  = nl

⇒ y = , n = 0, ± 1, ± 2, ±3 ...(4)

Here n = 0 corresponds to the central maxima

n = ± 1 correspond to the 1 st maxima

n = ± 2 correspond to the 2nd maxima and so on.

for minima (destructive interference).

Dp = ± , ± ±

consequently,

...(5)

Here n = ± 1 corresponds to first minima,

n = ± 2 corresponds to second minima and so on.

Fringe width :

It is the distance between two maxima of successive order on one side of the central maxima. This is also equal to distance between two successive minima.

fringe width b =

Notice that it is directly proportional to wavelength and inversely proportional to the distance between the two slits.

As vertical distance y is related to q by q = so Dq =  which is referred as angular fringe width

Bq =

Ex.5 In a YDSE performed with wavelength l = 5890 Å the angular fringe width is 0.40°. What is the angular fringe width if the entire set-up is immersed in water ?

Sol. Angular fringe width is given by

Ex.6 A beam of light consisting of two wavelengths 6500Å and 5200Å is used to obtain interference fringes in a Young's double slit experiment. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of slits and the screen is 120 cm.

Sol. The position of nth bright fringe on the screen is

yn =

Let the nth bright fringe of 6500Å and the nth bright fringe of 5200Å coincide; then

=

=  =

Thus the minimum values of m and n are 4 and 5 respectively.

Hence y =  = 0.156 cm = 1.56 mm

Ex.7  In a YDSE, D = 1m, d = 1mm and l =1/2 mm

(i) Find a distance between the first and central maxima on the screen.

(ii) Find the no of maxima and minima obtained on the screen.

Sol. D >> d

Hence DP = d sinθ

,

clearly, n <<  is not possible for any value of n.

Hence Dp =  cannot be used

for Ist maxima,

Dp = d sinθ = l

(ii) Maximum path difference
ΔPmax = d = 1 mm
⇒ Highest order maxima,
and highest order minima

Total no. of maxima =  *(central maxima)
Total no. of minima =

Ex.8 Monochromatic light of wavelength 5000 A° is used in Y.D.S.E., with slit-width, d = 1mm, distance between screen and slits, D = 1m. If intensity at the two slits are I1 = 4I0 , I2 = I0, find

(i) fringe width b

(ii) distance of 5th minima from the central maxima on the screen
(iii) Intensity at y  mm

(iv) Distance of the 1000th maxima

(v) Distance of the 5000th maxima

Sol. (i)  =  = 0.5 mm

(ii) y = (2n - 1) , n = 5 ⇒ y = 2.25 mm

(iii) At y = mm, y << D

Now resultant intensity

(iv)

n = 1000 is not << 2000

Hence now Δp = d sin θ must be used
Hence, d sin θ = nλ = 1000λ

(v) Highest order maxima

nmax =  = 2000

Hence, n = 5000 is not possible.

Ex.9 A beam of light consisting of wavelength 6000 Å and 4500 Å is used in YDSE with D = 1 m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.

Sol.

Let n1 th maxima of λand n2 th maxima of λ2 coincide at a position y. the, y =    = n1P1 = n2P2 = LCM of β1 and β2
⇒ y = LCM of 0.6 cm and 0.45 mm
y=1.8mm A.
At this point 3rd maxima for 6000 Å & 4th maxima for 4500 Å coincide

GEOMETRICAL PATH & OPTICAL PATH

Actual distance travelled by light in a medium is called geometrical path (Dx). Consider a light wave given by the equation

If the light travels by Dx, its phase changes by kDx = , where w, the frequency of light does not depend on the medium, but v, the speed of light depends on the medium as v =

Consequently, change in phase,

Df = kDx = (mDx)

It is clear that a wave travelling a distance Dx in a medium of refractive index m suffers the same phase change as when it travels a distance mDx in vacuum. i.e. a path length of Dx in medium of refractive index m is equivalent to a path length of mDx in vacuum.

The quantity mDx is called the optical path length of light, Dxopt. And in terms of optical path length, phase difference would be given by,

Df = =  .....(1)

where l0 = wavelength of light in vacuum

However in terms of the geometrical path length Dx,

where l = wavelength of light in the medium (l = ).

Displacement of fringe :

on introduction of a glass slab in the path of the light coming out of the slits -

On introduction of the thin glass-slab of thickness t and refractive index m, the optical path of the ray S1P increases by t(m - 1). Now the path difference between waves coming form S1 and S2 at any point P is

for central bright fringe;

The whole fringe pattern gets shifted by the same distance

Notice that this shift is in the direction of the slit before which the glass slab is placed. It happens so because S2 compensates the path difference (arised due to optical path length covered by S1) by covering more geometrical path length. If the glass slab is placed before the upper slit, the fringe pattern gets shifted upwards and if the glass slab is placed before the lower slit the fringe pattern gets shifted downwards.

Ex.10 In a YDSE with d = 1 mm and D = 1 m, slabs of (t = 1 mm, m = 3) and (t = 0.5 mm, m = 2) are introduced in front of upper and lower slit respectively. Find the shift in the fringe pattern.

Sol. Optical path for light coming from upper slit S1 is

Similarly optical path for light coming from S2 is

Path difference :

for central bright fringe Δp = 0

The whole pattern is shifted by 1.5 mm upwards

Ex.11 Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000 Å. The separation between the two slits is 2 mm. The distance between the slits and screen is 10 cm. When a transparent plate of thickness 0.5 mm is placed over one of the slits, the fringe pattern is displaced by 5 mm. Find the refractive index of the material of the plate.

Sol.

Here d = 2 cm = 2 x 10-3 m, D = 10 cm = 0.10 m, t = 0.5 mm = 0.5 x 10-3 m, Δx = 5 mm = 5 x 10-3 m, λ, = 6 x 10-7 m

Ex.12 In a YDSE light of wavelength l = 5000 Å is used, which emerges in phase from two slits a distance d = 3 × 10-7 apart. A transparent sheet of thickness t = 1.5 × 10-7 m, refractive index n = 1.17, is placed over one of the slits. Where does the central maxima of the interference now appear?

Sol. The path difference introduced due to introduction of transparent sheet is given by Dx = (m - 1)t.

If the central maxima occupies position of nth fringe,

Hence is angular position of central maxima is

For small angles sin

Shift of central maxima is   This formula can be used if D is given.

YDSE WITH OBLIQUE INCIDENCE :

In YDSE, ray is incident on the slit at an inclination of q0 to the axis of symmetry of the experimental set-up for points above the central point on screen, (say for P1)

For point O, Δp = dsinθ0 (because S2O = S1O) and for points below O on the screen, (say for P2)

(if d << D)

We obtain central maxima at a point where, Δp = 0

or θ2 = θ0,

This corresponds to the point O' in the diagram

Hence we have finally for path difference,

Ex.13 In YDSE with D = 1m, d = 1mm light of wavelength 500 nm is incident at an angle of 0.57° w.r.t the axis of symmetry of the experimental set up. If centre of symmetry of screen is O as shown.

(i) find the position of central maxima

(ii) Intensity at point O in terms of intensity of central maxima I0

(iii) Number of maxima lying between O and the central maxima.

Sol.

(ii) for point 0, θ = 0
Hence, Δp = d sin θ0, dθ0 = 1 mm x (10-2 rad)
= 10,000 nm = 20 x (500 nm)
⇒ Δp = 20 λ
Hence point O corresponds to 20th maxima
⇒ intensity at O = I0

(iii) 19 maxima lie between central maxima and O, excluding maxima at O and central maxima.

Shape of Interference Pattern :

1. Shape of the Pattern when the interference takes place due to waves produced by two slits.

2. Shape of the Pattern when the interference takes place due to waves produced by two point sources(where the line of sources is perpendicular to the screen).

3. Shape of the Pattern when the interference takes place due to waves produced by two point sources(where the line of sources is parellel to the screen).

YDSE with white light

Central Maxima position (where phase difference = 0) is independent of the wavelength of light.

White light is used to find out the central maxima position of YDSE set up because at this position only, all the wavelength show constructive interference i.e. why we get white spot at that position.

However slightly below or above the position of central maxima fringes will be coloured.

VIBGYOR

& as we known

As we move away from central maxima first maxima & minima are of violet colour but in the near by region of central maxima reddish colour will dominate because in this region intensity of violet colour decreases at a faster rate as compared to red colour.

In usual interference pattern with a monochromatic source, a large number of identical interference fringes are obtained & it is usually not possible to determine the position of central maxima Interference with white light is used to determine the position of central maxima in such case.

Ex.14 White light, with a uniform intensity across the visible wavelength range 430-690 nm, is perpendicularly incident on a water film, of index of refraction m = 1.33 and thickness d = 320 nm, that is suspended in air. At what wavelength l is the light reflected by the film brightest to an observer ?

Sol. This situation is like that of Figure shown, for which equation written below gives the interference maxima.

for constructive interference.

Solving for l and inserting the given data, we obtain

=  =

for m = 0, this give us l = 1700 nm, which is in the infrared region. For m = 1, we find I = 567 nm, which is yellow-green light, near the middle of the visible spectrum. For m = 2, l = 340 nm, which is the ultraviolet region. So the wavelength at which the light seen by the observer is brightest is

l = 567 nm. Ans.

Note : When a light gets reflected from a denser medium there is an abrupt phase change of p no phase change occurs when reflection takes place from rarer medium

Ex.15 Find the minimum value of x for which a maxima is obtained at P.

Sol. For maxima, Dx= l (because x should be minimum)

Path difference between the direct & reflected ray

= x +  (due to reflection, a phase change of p or path change of  takes place)

2x = ⇒ x =

Ex.16 Find the value of q for which a maxima is obtained at P.

For maxima at P, Δx = λ
Path difference between direct &. reflected ray

Fresenel's Biprism :

Fig shows the Fresnel's biprism experiment schematically. The thin prism P refracts light from the slit source S into two beams AC & BE. When a screen MN is placed as shown in the figure, the interference fringes are observed only in the region BC. If the screen MN is removed, the two beam will overlap over the whole region AE.

If A is the angle of refraction of thin prism & m is the refractive index of its medium, then the angle of deviation produced by the prism is

If l1 is the distance between the source & the prism, then the separation between virtual sources is

If l2 is the distance between the prism & the screen, then the distance between virtual sources & the screen is given by

D = l1 + l2

Thus, by using the result of young's experiment, the fringe width is given by

⇒

⇒

Fringes observed in the Fresnel's biprism experiment are vertical straight lines.

Ex.20 In a biprism experiment, the slit is illuminated with light of wavelength 4800 Å. The distance between the slit and diprism is 20 cm and that between biprism and eyepiece is 80 cm. If two virtual sources are 0.3 cm apart, determine the distance between the 5th bright band on one side of the central bright band and the 4th dark bank on the other side.

Sol. Here l = 4.8 × 10-7 m, d = 0.3 × 10-2 m,

D = 20 80 = 100 cm = 1 m

Distance of 5th bright from the central bright band is

x5 =

Ex.21 In a biprism experiment, fringe width is measured as 0.4 mm. When the eyepiece is moved away from the biprism through 30 cm, the fringe width increases by 50%. If the two virtual sources are 0.6 mm apart, find the wavelength of light used.

Sol.

Wavelength of light used,

Ex.22 Interference fringes are produced by a Fresnel's biprism in the focal plane of reading microscope which is 100 cm from the slit. A lens interposed between the biprism and the microscope gives two images of the slit in two positions. If the images of the slits are 4.05 mm apart in one case, 2.90 mm in the other and the wavelength of light used is 5893 Å, find the distance between two consecutive bands.

Sol.

Here d, = 4.05 mm = 0.405 cm, d2 = 2.09 mm = 0.209 cm

Distance between the two coherent sources will be

Huygen's Principle :

The various postulates are :

1. Each source of light is a centre of disturbance from which waves spread in all directions. All particles equidistant from the source & vibrating in same phase lie on the surface known as wavefront.

2. Wave propagates perpendicular to wavefront

3. Each ray take same time to reach from one wavefront to another wavefront

4. Every point on a wavefront is a source of new disturbance which produces secondary wavelets. These wavelets are spherical & travel with the speed of light in all directions in that medium.

5. Only forward envelope enclosing the tangents at the secondary wavelets at any instant gives the new position of wavefront.

There is no backward flow of energy when a wave travels in the forward direction.

Ex.23 For the given ray diagram, draw the wavefront

Sol.

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