CBSE Class 12 Physics Diffraction and Polarisation of light Theory

CBSE Class 12 Physics Diffraction and Polarisation of light Theory. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Class 12 Physics: Diffraction and Polarisation of Light (Theory and Examples)

1. Diffraction

1.1 Meaning of diffraction

It is the spreading of waves round the corners of an obstacle whose size is of the order of the wavelength.

1.2 Definition of diffraction

The phenomenon of bending of light waves around the sharp edges of opaque obstacles or apertures, and their encroachment into the geometrical shadow of the obstacle or aperture, is defined as diffraction of light.

1.3 Necessary conditions for diffraction of waves

The size of the obstacle (\(a\)) must be of the order of the wavelength of the waves (\(\lambda\)), i.e. \( a\approx\lambda \). The condition is also written as \( \dfrac{a}{\lambda}>1 \) (in practice \(a\approx\lambda\), as in the table below).

Note: The greater the wavelength of a wave, the higher is its degree of diffraction. This is the reason that diffraction of sound and radio waves is easily observed, but for the diffraction of light, additional arrangements have to be made.

\[ \lambda_{sound}>\lambda_{light} \]

The wavelength of sound is nearly equal to the size of ordinary obstacles. If the size of the obstacle is \(a\) and the wavelength is \(\lambda\), then:

S.No.	\(a\) versus \(\lambda\)	Diffraction
1	\(a\ll\lambda\)	Not possible
2	\(a\gg\lambda\)	Not possible
3	\(a\approx\lambda\)	Possible

1.4 Interpretation of diffraction

As a result of diffraction, maxima and minima of light intensity are found, which have unequal intensities. Diffraction is the result of the superposition of waves from an infinite number of coherent sources on the same wavefront after the wavefront has been distorted by the obstacle.

1.5 Examples of diffraction

1. When an intense source of light is viewed with a partially opened eye, colours are observed in the light.
2. Sound produced in one room can be heard in the nearby room.
3. The appearance of a shining circle around the Sun just before sunrise.
4. A coloured spectrum is observed if a distant light source is seen through a thin cloth.

1.6 Two types of diffraction

Fresnel diffraction: Fresnel diffraction involves non-plane (spherical) wavefronts, so that the source S and the point P (where the diffraction effect is to be observed) are at a finite distance from the diffracting obstacle.Fraunhofer diffraction: Fraunhofer diffraction deals with wavefronts that are plane on arrival and an effective viewing distance of infinity. It follows that Fraunhofer diffraction is an important special case of Fresnel diffraction. In Young's double slit experiment, we assume the screen to be relatively distant, so that we have Fraunhofer conditions.

1.7 Difference between interference and diffraction of light

S.No.	Interference	Diffraction
1	Two coherent sources are necessary	One coherent source is necessary
2	All fringes have the same width	Fringes have unequal width
3	The width of a bright fringe is equal to that of the other fringes	The central bright fringe has just double the width of the other fringes
4	The intensity of a dark fringe is zero	The intensity of a dark fringe is not zero
5	All bright fringes have equal intensity	As the order of the bright fringes increases, the intensity goes down
6	For a bright fringe: (a) path difference \(\Delta=n\lambda\); (b) phase difference \(\delta=2n\pi\)	For a bright fringe: (a) path difference \(\Delta=(2n-1)\lambda/2\); (b) phase difference \(\delta=(2n-1)\pi\)
7	For a dark fringe: (a) path difference \(\Delta=(2n-1)\lambda/2\); (b) phase difference \(\delta=(2n-1)\pi\)	For a dark fringe: (a) path difference \(\Delta=2n\lambda/2=n\lambda\); (b) phase difference \(\delta=2n\pi\)

1.8 Resolving power (R.P.)

A large number of images are formed as a consequence of light diffraction from a source. If two sources are separated such that their central maxima do not overlap, their images can be distinguished and are said to be resolved. The R.P. of an optical instrument is its ability to distinguish two neighbouring points.

\[ \text{Linear R.P.}=\frac{d}{\lambda D},\qquad \text{Angular R.P.}=\frac{d}{\lambda} \]

where \(D\) = observed distance and \(d\) = distance between the two points. (Formulas as printed in the source.)

Telescope: limit of resolution

\[ \theta=\sin^{-1}\frac{1.22\lambda}{a};\quad \text{for small angles }\theta=\frac{1.22\lambda}{a},\qquad \text{resolving power}=\frac{1}{\text{limit of resolution}} \]

(\(a\) is the aperture of the objective.)

Microscope: limit of resolution (the smallest distance between two objects)

\[ x_{min}=\frac{1.22\lambda}{2\mu\sin\theta} \]

Prism: \( \text{R.P.}=t\,\dfrac{d\mu}{d\lambda}=\dfrac{\lambda}{d\lambda} \)Eye: the limit of resolution of the human eye is 1′ of arc (one minute of arc).

1.9 Fraunhofer diffraction for a single slit

In this diffraction pattern the central maximum is bright, and on both sides of it maxima and minima occur symmetrically.

The diffraction condition for a slit of width \(a\) (\(n=1,2,3,\dots\)):

1. For diffraction maxima (secondary): \( a\sin\theta=(2n+1)\dfrac{\lambda}{2} \). (The source prints \((2n-1)\lambda/2\); with \(n=1\) that would put the first maximum at \(\lambda/2\), which is not correct.)
2. For diffraction minima: \( a\sin\theta=n\lambda \).
3. The maxima or minima are observed due to the superposition of waves emerging from an infinite number of secondary sources between the points A and B of the slit.
4. Fringe width: the distance between the two secondary minima formed on the two sides of the central maximum is known as the width of the central maximum.

\[ W=\frac{2f\lambda}{a} \]

where \(f\) = focal length of the convex lens and \(a\) = width of the slit. Angular width \( W_\theta=\dfrac{2\lambda}{a} \).

Examples based on Fraunhofer diffraction for a single slit

When a wave of wavelength 0.2 cm is made incident normally on a slit of width 0.004 m, then the semi-angular width of the central maximum of the diffraction pattern will be

1. \(60^\circ\)
2. \(30^\circ\)
3. \(90^\circ\)
4. \(0^\circ\)

Solution:

\[ \theta=\sin^{-1}\left(\frac{\lambda}{a}\right) \]

Here \( \lambda=2\times10^{-3} \) m and \( a=4\times10^{-3} \) m, so

\[ \theta=\sin^{-1}\left(\frac12\right)=30^\circ \]

Answer: (B) \(30^\circ\)

A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1 mm. When the diffraction pattern is seen on a screen placed at a distance of 2 m, the width of the principal maximum is found to be 2.5 mm. The wavelength of light is

1. \(6250\ \text{Å}\)
2. \(6200\ \text{Å}\)
3. \(5890\ \text{Å}\)
4. \(6000\ \text{Å}\)

Solution: The width of the principal maximum is \( \beta=2.5 \) mm, so its half-width is \( \beta/2=1.25\times10^{-3} \) m. The diffraction angle is

\[ \theta=\frac{\beta/2}{D}=\frac{1.25\times10^{-3}}{2}=6.25\times10^{-4}\ \text{rad} \]

Since \( a\theta=\lambda \):

\[ \lambda=a\theta=10^{-3}\times6.25\times10^{-4}=6.25\times10^{-7}\ \text{m}=6250\ \text{Å} \]

Answer: (A) \(6250\ \text{Å}\). (The source question says 1.5 mm, but its solution and the options correspond to 2.5 mm.)

Light of wavelength \(6328\ \text{Å}\) is incident normally on a slit having a width of 0.2 mm. The angular width of the central maximum, measured from minimum to minimum of the diffraction pattern on a screen 9.0 m away, will be about

1. \(0.360^\circ\)
2. \(0.180^\circ\)
3. \(0.720^\circ\)
4. \(0.090^\circ\)

Solution: Slit width \(a=0.2\) mm. The angular width is

\[ W_\theta=\frac{2\lambda}{a}=\frac{2\times6328\times10^{-10}}{0.2\times10^{-3}}=6.33\times10^{-3}\ \text{rad}=0.36^\circ \]

Answer: (A) \(0.360^\circ\). (The options are angles in degrees; the distance of the screen is not needed for the angular width. The linear width on the screen would be \(2\lambda D/a\approx5.7\) cm.)

1.10 Difference between Fresnel and Fraunhofer diffraction

S.No.	Fresnel	Fraunhofer
1	The source is near and on one side of the obstacle, and the screen is also near the obstacle and on the other side.	The source and the screen are both effectively at infinite distances from the obstacle.
2	Spherical or cylindrical wavefronts are used.	Plane wavefronts are used.
3	No sophisticated equipment is required.	A spectrometer is required.
4	No lens is required.	A convex lens is required.
5	The diffraction pattern may have both a bright and a dark central fringe.	Only a bright central fringe is possible.
6	Examples: circular aperture, disc, ring, etc.	Examples: single slit, double slit, grating, etc.

1.11 Comparative study of diffraction of light and sound

1. Sound travels in the form of waves, so it is also diffracted. Generally the diffraction of sound waves is more easily observed than that of light, because the wavelength of sound waves is of the order of the size of obstacles, but the wavelength of light is very small in comparison to the obstacles.
2. (a) Ordinary audible sound has a wavelength of the order of 1 m and the size of ordinary obstacles is of the same order, so diffraction is easily observed. (b) Ordinary light has a wavelength of about \(10^{-7}\) m and ordinary obstacles are much greater in size than the wavelength, so a diffraction pattern is not observed.
3. Generally the diffraction of ultrasonic waves is not observed because their wavelength is of the order of 1 cm or less.

1.12 Rectilinear motion of light

1. The rectilinear motion of light can be explained by the diffraction of light.
2. If the size of the obstacle is of the order of the wavelength of light, then diffraction of light takes place and its rectilinear motion is not possible.
3. If the size of the obstacle is much greater than the wavelength of light, then rectilinear motion of light is observed.

1.13 Points to remember

1. Grimaldi discovered diffraction.
2. The intensity of the diffraction pattern decreases if the size of the obstacle is increased.
3. Superposition of waves causes both diffraction and interference. The superposition of secondary waves originating from two coherent sources generates interference. The superposition of secondary wavelets generated from the same wavefront is called diffraction.
4. Interference fringes have equal width but diffraction fringes have unequal width.
5. In a diffraction pattern, the intensity of bright fringes is different, but for interference it is equal.
6. The intensity of dark fringes of interference is zero but for diffraction it is not zero.
7. The effect of diffraction can be observed only near the edges of the geometrical shadow.
8. In diffraction, the bright central fringe has double the width in comparison to the others.
9. In Fresnel diffraction neither is the wavefront plane, nor are the rays parallel.

2. Polarisation

2.1 Unpolarised light

1. An ordinary beam of light consists of a large number of waves emitted by the atoms or molecules of the light source. Each atom produces a wave with its own orientation of the electric vector \( \vec E \). However, because all directions are equally probable, the resulting electromagnetic wave is a superposition of waves produced by the individual atomic sources. This wave is called unpolarised light.
2. All the vibrations of an unpolarised light at a given instant can be resolved in two mutually perpendicular directions, and hence unpolarised light is equivalent to the superposition of two mutually perpendicular identical plane polarised lights.

2.2 Plane polarised light

1. If somehow we confine the vibrations of the electric vector in one direction perpendicular to the direction of wave motion, the light is said to be plane polarised, and the plane containing the direction of vibration and wave motion is called the plane of polarisation.
2. If an unpolarised light is converted into plane polarised light, its intensity reduces to half.
3. Polarisation is a convincing proof of the transverse wave nature of light. (The source says 'wave nature'.)

2.3 Partially polarised light

If, in the case of unpolarised light, the electric vector in some plane is either more or less than in its perpendicular plane, the light is said to be partially polarised.

2.4 Methods of polarisation

(a) By reflection: Brewster discovered that when light is incident at a particular angle on a transparent substance, the reflected light is completely plane polarised with vibrations in a plane perpendicular to the plane of incidence. This specific angle of incidence is called the polarising angle \( \theta_p \) and is related to the refractive index \( \mu \) of the material through the relation

\[ \tan\theta_p=\mu \qquad (1) \]

This is known as Brewster's law. In the case of polarisation by reflection:

1. For \( i=\theta_p \), the reflected light is plane polarised.
2. For \( i=\theta_p \), the reflected and refracted rays are perpendicular to each other.
3. For \( i<\theta_p \) or \( i>\theta_p \), both the reflected and refracted light become partially polarised.

(The source says 'refracted light is plane polarised' in the first point; the reflected light is the plane polarised one, as in Brewster's law above.)

(b) By refraction: In this method, a pile of glass plates is formed by taking 20 to 30 microscope slides and the light is made to be incident at the polarising angle. Since, in one reflection, about 15% of the light with vibration perpendicular to the plane of the paper is reflected, after passing through a number of plates, as shown in the figure, the emerging light becomes plane polarised with vibrations in the plane of the paper.(c) By dichroism: Some crystals such as tourmaline and sheets of iodosulphate of quinine have the property of strongly absorbing the light with vibrations perpendicular to a specific direction (called the transmission axis) and transmitting the light with vibrations parallel to it. This selective absorption of light is called dichroism. So if unpolarised light passes through a proper thickness of these, the transmitted light will be plane polarised with vibrations parallel to the transmission axis. Polaroids work on this principle.

2.5 Intensity of light emerging from a polaroid

If plane polarised light of intensity \( I_0 (=KA^2) \) is incident on a polaroid and its vibrations of amplitude \(A\) make an angle \(\theta\) with the transmission axis, the polaroid will pass only those vibrations which are parallel to its transmission axis, i.e. \(A\cos\theta\), so the intensity of the emergent light will be

\[ I=K(A\cos\theta)^2=KA^2\cos^2\theta \]

\[ \text{or}\quad I=I_0\cos^2\theta\quad[\text{as } I_0=KA^2] \qquad (2) \]

This law is called Malus' law. From this it is clear that:

1. If the incident light is unpolarised, then as vibrations are equally probable in all directions (in a plane perpendicular to the direction of wave motion), \(\theta\) can have any value from 0 to \(2\pi\) and \( (\cos^2\theta)_{av}=\tfrac12 \), so \[ I=\tfrac12 I_0 \] i.e. if an unpolarised light is converted into plane polarised light, its intensity becomes half.
2. If light of intensity \(I_1\) emerging from one polaroid, called the polariser, is incident on a second polaroid (usually called the analyser), the intensity of the light emerging from the second polaroid in accordance with Malus' law will be given by \[ I_2=I_1\cos^2\theta' \] where \(\theta'\) is the angle between the transmission axes of the two polaroids.

So if the two polaroids have their transmission axes parallel to each other, i.e. \(\theta'=0^\circ\), \( I_2=I_1\cos^2 0^\circ=I_1 \), and if the two polaroids are crossed, i.e. have their transmission axes perpendicular to each other, \(\theta'=90^\circ\), \( I_2=I_1\cos^2 90^\circ=0 \). So, if an analyser is rotated from \(0^\circ\) to \(90^\circ\) with respect to the polariser, the intensity of the emergent light changes from the maximum value \(I_1\) to the minimum value zero.