Nootan Solutions Interference of Light ISC Class-12 Physics Ch-20 Nageen Prakashan Numericals. Step by step Solutions of Kumar and Mittal ISC Physics Class-12 Nageen Prakashan Numericals Questions. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

## Nootan Solutions Interference of Light ISC Class-12 Physics Ch-20 Nageen Prakashan

 Board ISC Class 12 Subject Physics Publication Nageen Prakashan Writer Kumar and Mittal Vol 2nd Book Name Nootan Chapter-20 Interference of Light Topics Solution of Numericals Questions Page-Number 815, 816

### Nootan Solutions Interference of Light

ISC Class-12 Physics Ch-20 Nageen Prakashan Numericals of Kumar and Mittal

Interference is a natural phenomenon that happens at every place and at every moment. Yet we don’t see interference patterns everywhere. Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or same amplitude. The most commonly seen interference is the optical interference or light interference. This is because light waves are randomly generated every which way by most sources.

#### Principle of Super Position

Two, or more Progressive waves can travel simultaneously in a medium without affecting the motion of one another, Therefore, the resultant displacement of each particle of the medium at any instant is equal to the vector sum of the displacements produced by the two waves separately. This principle is called principle of superposition

#### Meaning of principle of superposition :

The principle of superposition means that if a number of waves are travelling in a medium, then each one travels independently as if the other waves were not present at all; the shape and other characteristics of any wave are not changed.

#### interference of two waves

When two waves of same frequency travel in a medium simultaneously in the same direction then, due to their superposition, the resultant intensity at any point of the medium is different from the sum of intensities of the two waves. At some points the intensity of the resulatant wave is very large while at some other points it is very small or zero. This phenomenon is called the interference of waves.

#### interference Type

• ##### Constructive interference:

Constructive interference takes place when the crest of one wave falls on the crest of another wave such that the amplitude is maximum. These waves will have the same displacement and are in the same phase.

• ##### Destructive interference:

In destructive interference the crest of one wave falls on the trough of another wave such that the amplitude is minimum. The displacement and phase of these waves are not the same.

### Conditions for Interference of Light Waves

For sustained interference of light to occur, the following conditions must be met:

1. Coherent sources of light are needed.
2. Amplitudes and intensities must be nearly equal to produce sufficient contrast between maxima and minima.
3. The source must be small enough that it can be considered a point source of light.
4. The interfering sources must be near enough to produce wide fringes.
5. The source and screen must be far enough to produce wide fringes.
6. The sources must emit light in the same state of polarization.
7. The sources must be monochromatic.

For Constructive Interference

Phase difference, φ = 2nπ

Path difference, Δx = nλ

where, n = 0, 1, 2, 3,…

For Destructive Interference

Phase difference, φ = (2n – 1)π

Path difference, Δx = (2n – 1)π / 2

where, n = 1, 2, 3, …

If two waves of exactly same frequency and of amplitude a and b interfere, then amplitude of resultant wave is given by

R = √a2 + b2 + 2ab cos φ

where φ is the phase difference between two waves.

Rmax = (a + b)

Rmin = (a – b)

Intensity of wave

∴ I = a2 + b2 + 2ab cos φ

= I1 + I2 + 2 √I1 I2 cos φ

where I1 and I2 are intensities of two waves.

∴ I1 / I2 = a2 / b2 = ω1 / ω2

Where ω1 and ω2 are width of slits.

Energy remains conserved during interference.

Interference fringe width

β = Dλ / d

where, D = distance of screen from slits, λ = wavelength of light and d = distance between two slits.

Distance of nth bright fringe from central fringe xn = nDλ / d

#### Law of conservation of energy in constructive & Destructive interference :

Energy distributes ununiformly by interference. Neither energy produces nor distroy. It changes from constructive to destructive & from destructive to constructive. Energy redistributes. It follows law of conservation of energy. comparative study of constructive & destructive interference.

Distance of nth dark fringe from central fringe x’n = (2n – 1) Dλ / 2d

#### Coherent Sources of Light

The sources of light emitting light of same wavelength, same frequency having a zero or constant phase difference are called coherent sources of light.

When a transparent sheet of refractive index μ and of thickness t is introduced in one of the path of interfering waves, then fringe pattern shifts in that direction by a distance Y

Y = D / d (μ – 1) t = β / λ (μ – 1) t

where, β = fringe width.

#### Fresnel’s Biprism

It is a combination of two prisms of very small refracting angles placed base to base. It is used to obtain two coherent sources from a single light source.

#### Lyod’s Mirror

The shape of interference fringes are usually hyperbolic.

When screen is held at 900 to the line joining foci of the hyperbola, the fringes are circular.

When distance of screen (D) is very large compare to the distance between the slits (d), the Cringes are straight.

#### Diffraction

The bending of light waves around the corners of an obstacle or aperture is called diffraction of light.

#### diffraction Type

(a) Fresnel class
(b) Fraunhofer class

#### Fraunhofer Diffraction at a Single Slit

Linear Width 0f central maximum 2Dλ / a = 2fλ / a

Angular width of central maximum = 2λ / a

where, λ = wavelength of light, a = width of single slit, D = distance of screen from the slit and f = focal length of convex lens.

#### For Secondary Minima

(a) Path difference = nλ

(b) Linear distance = nDλ / a = nfλ / a

(c) Angular spread = nλ / a

where, n = 1, 2, 3,.,.

#### For Secondary Maxima

(a) Path difference = (2n + 1 ) &lamda; / 2

(b) LInear distance = (2n + 1 ) D&lamda; / 2a = (2n + 1 ) f&lamda; / 2a

(c) Angular spread = (2n + 1 ) &lamda; / 2

Important Points

• A soap bubble or oil film on water appears coloured in white light due to interference of light reflected from upper and lower surfaces of soap bubble or oil film.
• In interference fringe pattern all bright and dark fringes are of same width,
• In diffraction fringe pattern central bright fringe is brightest and widest. and I remaining secondary maximas are of gradually decreasing intensities.
• The difference between interference and diffraction is that the interference is the superposition between the wavelets coming from two coherent sources while the diffraction is the superposition between the wavelets coming from the single wavefront

#### Polarisation

The phenomena of restructuring of electric vectors of light into a single direction is called polarisation.

Ordinary light has electric vectors in all possible directions in a plane perpendicular to the direction of propagation of light.

#### Nicol Prism

A nicol prism is an optical device which is used for producing   plane polarised light and analysing light the same.

The nicol prism consists of two calcite crystal cut at 68° with its principal axis joined by a glue called Canada balsam.

#### Law of Malus

When a beam of completely plane polarised light is incident on an analyser, the intensity of transmitted light from analyser is directly proportional to the square of the cosine of the angle between plane of transmission of analyser and polariser, i.e.,

I ∝ cos2 θ

When ordinary light is incident on a polariser the intensity of transmitted light is half of the intensity of incident light.

When a polariser and analyser are perpendicular to each other, then intensity of transmitted light from analyser becomes O.

#### Brewster’s Law

When unpolarised light is incident at an angle of polarisation (ip) on the interface separating air from a medium of refractive index μ, then reflected light becomes fully polarised, provided

μ = tan ip

If angle of polarisation is ip and angle of refraction is μ then

ip + r = 90°

Refractive index μ = tan ip = 1 / sin C

where, C = critical angle

#### Double Refraction

When unpolarised light is incident on a calcite or quartz crystal it splits up into two refracted rays. one of which follows laws of refraction. called ordinary ray (O-ray) and other do not follow laws of refraction. called extraordinary ray (E-ray). This phenomena is called double refraction.

#### Dichroism

Few double refracting crystals have a property of absorbing one of the two refracted rays and allowing the other to emerge out. This property of crystal is called dichroism.

#### Polaroid

It is a polarising film mounted between two glass plates. It is used to produce polarised light.

#### Application of Polaroid

(i) Polaroids are used in sun glasses. They protect the eyes from glare.
(ii) The polaroids are used in window panes of a train and especially of an aeroplane. They help to control the light entering through the window.
(iii) The pictures taken by a stereoscopic camera. When seen with the help of polarized spectacles, create three dimensional effect.
(iv) The windshield of an automobile is made of polaroid. Such a mind shield protects the eyes of the driver of the automobile from the dazzling light of the approaching vehicles

### Nootan Solutions Interference of Light

ISC Class-12 Physics Ch-20 Nageen Prakashan Numericals of Kumar and Mittal

(Page-815 , 816)

Question 1:

Two waves of equal frequency have their amplitudes in the ratio of 3 : 5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

Question 2:

Two coherent light source whose  intensity ratio is 81 : 1 produce interference ………………………………. minima in the fringe system.

Question 3:

………………………

…………………….

……………………..

Question 8:

A double slit is ………………………………. two adjacent minima.

Question 9:

In young’s double-slit ………………………………………. wavelength of light used.

Question 10:

………………………

………………………

………………………

Question 17:

In Young’s double slit experiment the distance ……………………………. distance between the fringes ?

Question 18:

In Young’s double-slit experiment the ………………………… and the same light used ?

Question 19:

In an experiment with sodium light ……………………………… light form the second source?

Question 20:

A double-slit is illuminated by sodium …………………………….. fringe  on the other side.

Question 21:

In young’s double-slit experiment, the slits …………………………  side of the central bright fringe.

Question 22:

In young’s double-slit experiment, the separation ……………………………… the refractive index of water is 4/3.

Question 23:

The fringe-width in young’s experiment, when the separation between the slit is 0.5 mm and the distance of the screen from the slits in 1.5 m, is 1.8 mm. Find the path difference  between the interfering waves at the position of first-order maximum.

—: End of Nootan Solutions Interference of Light ISC Class-12 Physics Ch-20 Numericals :–

CONTACT FOR LIVE CLASSES -9335725646

Thanks

You might also like