NCERT Class 12 Physics Chapter 10 Notes Wave Optics - Download PDF

NCERT Notes for Class 12 Physics Chapter 10: Download PDF

Download the NCERT Wave Optics Notes in PDF format for quick and offline access. It is the best for learning at your own convenience.

Download PDF

NCERT Notes for Class 12 Physics Chapter 10 Wave Optics

Wave Optics Notes are essential to understand the behaviour of light not explained by its ray nature. These notes are concise and explain all topics in simple language, making it easier for students to review and comprehend the material.

Wavefront

A light source is a spot that radiates noise in all directions. In a homogeneous medium, the disturbance hits all of the medium's particles in phase, which is all placed at the same distance from the light source, and so every particle must vibrate in phase with each other at all times. A wavefront represents the surface of constant phase in the propagation of a wave.

• The direction of propagation of light is perpendicular to the wavefront.
• The time taken by the light to travel from one wavefront to another is the same along any ray.
• The phase difference between various particles on the wavefront is zero.

Wavefronts can be one of the following sorts, depending on the shape of the light source:

1. Spherical WaveFront:
   A spherical wavefront is created by a point source of light. This is because the locus of all equidistant points from a point source is a sphere.
2. Cylindrical WaveFront:
   A cylindrical wavefront is produced when the light source is linear (such as a slit). Every equidistant point from the linear source is located on the surface of a cylindrical
3. Plane WaveFront:
   If a wavefront is a small component of a spherical or cylindrical wavefront originating from a distant source, it will appear as a plane.

Huygens' Principle

According to the Huygens principle, every point on the given wavefront acts as a source of a new disturbance called secondary wavelets. And a common tangent to these secondary wavelets in the forward direction at any instant gives the new wavefront at that instant as shown in the below figure. This is called secondary wavefront.

Properties of Light explained using Huygens' Principle

Light waves follow the Laws of Reflection and Refraction. These can be explained using Huygens' principle.

Refraction of Plane Wave

Consider a plane wavefronts travels towards a plane $A C$ as shown in the above figure.
Let $A B$ and $C D$ as the incident and refracted wavefronts respectively.
Let at $\mathrm{t}=0$ wave is at A and at $t=\tau$ wave is at C .
if $\mathrm{v}_1$ is the velocity of the wave in medium 1 then $B C=v_1 \tau$
similarly $\mathrm{v}_2$ is the velocity of the wave in the medium 2 then $A D=v_2 \tau$
For $\triangle A B C \rightarrow \sin (i)=\frac{B C}{A C}=\frac{v_1 \tau}{A C}$
similarly
For $\triangle A C D \rightarrow \sin (r)=\frac{A D}{A C}=\frac{v_2 \tau}{A C}$
So we get $\frac{\sin (i)}{\sin (r)}=\frac{v_1 \tau}{v_2 \tau}=\frac{v_1}{v_2}$
And we know $\mu \propto v$
So we get

$
\frac{\sin (i)}{\sin (r)}=\frac{v_1}{v_2}=\frac{\mu_2}{\mu_1}=\mu_{21}=\text { constant }
$
This verifies the first law of refraction.

Similarly from the figure, we can say that the incident wavefront, the refracted wavefront and normal lie in the same plane.

This again verifies the second law of Refraction.

Reflection of Plane Wave

Consider a plane wavefronts travels towards a plane reflecting surface as shown in the figure.

Let $A B$ and $C D$ as the incident and reflected wavefronts respectively.
Let at $\mathrm{t}=0$ wave is at A and at $t=\tau$ wave is at C.
if $\mathrm{v}_1$ is the velocity of the wave $B C=A D=v_1 \tau$
And as $\triangle A B C \cong \triangle A D C$ So we get $i=r$
This verifies the first law of reflection which states that the angle of incidence i and angle of reflection $r$ are always equal.
Similarly from the figure, we can say that the incident wavefront, the reflected wavefront and normal lie in the same plane.
This again verifies the second law of reflection.
Therefore, the two laws of Reflection are verified using Huygens's Principle.

Addition of Coherent Waves

Coherent sources of light are light sources that emit continuous light waves of the same wavelength, frequency, and phase (or with a constant phase difference).

Principle of Superposition

Suppose there are two sources of waves $S_1$ and $S_2$.

Now, the two waves from $S_1$ and $S_2$ meet at some point (say $P$ ). Then, according to principle of superposition net displacement at $P$ (from its mean position) at any time is given by

$y=y_1+y_2$

Consider the superposition of two sinusoidal waves of same frequency (means sources are coherent) at some point. Let us assume that the two waves are travelling in the same direction with same velocity. The equation of the two waves reaching at a point can be written as

$y_1=A_1 \sin (k x-\omega t)$

$y_2=A_2 \sin (k x-\omega t+\phi)$

$A =\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi} $

The same relation can be achieved by vector addition of the two amplitudes with the angle between them equal to the phase difference.

Resultant Intensity

We know,

$I=\frac{1}{2} \rho \omega^2 A^2 v \quad \text { or } \quad I \propto A^2$

So, if $\rho, \omega$ and $v$ are same for the both interfering waves then Eq. (i) can also be written as

$I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$

For identical sources-

$I_1=I_2=I_0 \implies I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi=4 I_0 \cos ^2 \frac{\phi}{2}$

Important Terms for Superposition of Waves

Phase difference ($\phi$): The difference between the phases of two waves at a given place is known as phase difference.

Path difference ($\Delta x$): The difference in path lengths of two waves meeting at a place is referred to as path difference.

$\Delta x =\frac{\lambda}{2 \pi} \times \phi$

Time Difference (T.D.): It is the time difference between two waves meeting at a place. = $\frac{T}{2 \pi} \times \phi$

Constructive Interference

In this condition, the two waves are completely in phase. For this, their path difference must be an integral multiple of the wavelength.

$\Delta x=n \lambda$
$\therefore \phi = 2 n \pi$

The intensity is,
$I_{max} = 4 I_0 \cos ^2 \frac{ 2 n \pi}{2} = 4I_0$

Destructive Interference

In this condition, the two waves are completely out of phase. For this, their path difference must be an odd multiple of half-wavelength.

$\Delta x=(2n+1) \frac{\lambda}{2}$
$\therefore \phi = (2 n +1)\pi$

The intensity is,
$I_{min} = 4 I_0 \cos ^2 \frac{(2 n +1)\pi}{2} = 0$

Young’s Double Slit Experiment (YDSE)

When monochromatic light (single wavelength) falls on two narrow slits S₁ and S₂ that are very close together, they act as two coherent sources, and the waves from these two sources superimpose on each other, an interference pattern appears on the screen. In this experiment, bright and dark bands alternated on the screen. Fringes are the name given to these bands.

d is the distance between the slits.
D be the distance between the slits and the screen.
$\lambda$ is the wavelength of the monochromatic light source.

The path difference between interfering waves that collide at point P on the screen is given by

$x=d \sin \theta=\frac{y d}{D}$

where y be the position of point P from central maxima

For maxima at P:

$x=n \lambda \quad n=0, \pm 1, \pm 2 \ldots$.
$\therefore y = \frac{D}{d} n \lambda$

And for minima at P:

$x=\frac{(2 n+1) \lambda}{2} \quad n=0, \pm 1, \pm 2 \ldots$
$\therefore y = \frac{D}{d} \frac{(2 n+1) \lambda}{2}$

Bandwidth

The distance between any two consecutive bright or dark bands is called bandwidth. For YDSE, this comes out the be a constant value.

Taking the consecutive dark or bright fringe,

$\begin{aligned} x_{n+1}-x_n & =\frac{(n+1) \lambda D}{d}-\frac{(n) \lambda D}{d} \\ x_{n+1}-x_n & =\frac{\lambda D}{d} \\ \beta & =\frac{\lambda D}{d}\end{aligned}$

Properties of Young's Double Slit Experiment:

1. At the central position Φ=0^o or Δ=0 the Central fringe will always be bright

2. A slit's fringe pattern will be brighter than a point's fringe pattern.

3. If the slit widths are mismatched, the minima will not be fully dark. As a result, uniform illumination occurs over a wide area.

4. When one slit is lit with red light and the other with blue light, no interference pattern appears on the screen.

Diffraction of Light

The phenomenon of light bending around the corners of an obstacle/aperture with a size comparable to the wavelength of light is known as Diffraction Of Light

Types of Diffraction

• Fresnel Diffraction
  • Source and screen are at a finite distance.
  • No lenses are used.
  • Wavefronts are spherical or cylindrical.
  • The diffraction pattern changes with the distance between the screen and the obstacle.
  • Example: Light bending around the edge of a razor blade.

• Fraunhofer Diffraction

  • Source and screen are at infinite distance (or made so with lenses).
  • Lenses are used to make wavefronts plane.
  • Diffraction pattern is sharp and fixed.
  • Commonly used for studying diffraction in labs.
  • Example: Diffraction pattern through a single slit using a laser and lenses.

Single Slit Diffraction

The central maxima in Fraunhofer diffraction by a single slit is the brightest and widest part of the diffraction pattern observed on a screen. When light passes through a narrow slit, it spreads out and forms a series of bright and dark fringes. The central maxima is located at the midpoint directly opposite the slit and are significantly more intense than the subsequent fringes.

(i) The Angular width d central maxima $=2 \theta=\frac{2 \lambda}{b}$
(ii) Linear width of central maxima $=2 x=2 D \theta=2 f \theta=\frac{2 \lambda f}{b}$

The secondary maxima are smaller peaks of intensity located on either side of the central maximum.

$\Delta x=b \sin \theta=(2 n+1) \frac{\lambda}{2} ;$ where $\mathrm{n}=1,2,3 \ldots \ldots$
$x_n=\frac{D}{b}\frac{(2 n+1) \lambda }{2}$

The secondary minima are the points where the intensity of the diffracted light falls to zero between the central maximum and subsequent maxima.

$\Delta x=b \sin \theta=n \lambda$
$x_n=\frac{n \lambda D}{b}$

Comparison Between Interference and Diffraction

Polarisation

• Unpolarized light:

Unpolarized light is light that has electric field oscillations in all directions in a plane perpendicular to its propagation. The horizontal and vertical components of light oscillation are separated.

• Polarized light:

Plane or polarized light is light that has just one plane of oscillations.
The plane of oscillation is the plane in which polarized light oscillates.
The plane perpendicular to the plane of oscillation is the plane of polarization.
Light can be polarized by passing through particular crystals like tourmaline or Polaroid..

• Polaroid:

A Polaroid is the name of the device that produces plane-polarized light. It is based on the selective absorption principle. It's also more powerful than the tourmaline crystal. It is a thin layer of ultramicroscopic quinine iodide sulphate crystals with optic axes parallel to one another.

Only light oscillations that are parallel to the transmission axis can travel through a Polaroid.

Malus Law

The square of the cosine of the angle between the analyzer’s plane of transmission and the plane of the polarizer will vary the intensity of polarized light passing through an analyzer. This is referred to as the Malus's Law.

$\begin{aligned} & I=I_0 \cos ^2 \theta \\ & A^2=A_0{ }^2 \cos ^2 \theta \Rightarrow A=A \cos \theta \\ \\ & \theta=0^{\circ}, I=I_0, A=A_0 \\ & \theta=45^{\circ}, I=\frac{I_0}{2}, A=\frac{A_0}{\sqrt{2}} \\ & \theta=90^{\circ}, I=0, A=0\end{aligned}$

Brewster’s law

When unpolarized light is reflected from a clear medium (with refractive index=μ), the reflected light will be totally plane-polarized at a specific angle of incidence (known as the angle of polarization θ_p). Brewster's law is the name for this rule.

$\mu=\tan \theta_p$

Resolvong Power:

If two-point objects are near together, their image diffraction patterns will likewise be close together and overlap.

The minimum distance between two objects that can be seen independently by the object instrument is known as the instrument's limit of resolution.

• Resolving power of Microscope:

R.P. of microscope = $\frac{2 \mu \sin \theta}{\lambda}$

• Resolving power of Telescope:

R.P of telescope = $\frac{1}{d \theta}=\frac{D}{1.22 \lambda}$

Where D is the aperture of the telescope

Wave Optics Previous Year Question and Answer

Q.1 Two coherent light waves of intensity $5 \times 10^{-2} \mathrm{Wm}^{-2}$ each super-impose and produce the interference pattern on a screen. At a point where the path difference between the waves is $\frac{\lambda}{6}, \lambda$ being wavelength of the wave, find the
(a) phase difference between the waves.
(b) resultant intensity at the point.
(c) resultant intensity in terms of the intensity at the maximum

Answer:

a) Phase difference,
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}$

b) Resultant Intensity,
$\begin{aligned}
I&=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \\
& =3 \times I_1=15 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2
\end{aligned}$

c)
$\begin{aligned}
& I_{\max }=4 I_1 \\
& I_{\max }=4 I / 3 \\
& I=\frac{3}{4} I_{\max }
\end{aligned}$

Q.2 Write the conditions of path difference under which (i) constructive (ii) destructive interference occurs in Young’s double-slit experiment.

Answer:

(i) For constructive interference
Path difference should be even multiple of half wavelength.
$\Delta x=2n \frac{\lambda}{2} \quad n = 0,1,2\ldots$

(ii) For destructive interference
Path difference should be odd multiple of half wavelength.
$\Delta x=(2 n+1) \frac{\lambda}{2}\quad n = 0,1,2\ldots$

Q.3 A beam of plane polarised light is passed through a polaroid. Show graphically, the variation of the intensity of the transmitted light with angle of rotation of the polaroid.

Answer:

The intensity follows Malus' Law,

$I=I_0 \cos ^2 \theta $

The graph shows the variation of the intensity (I) of light with the angle of rotation $(\theta)$ of the polaroid.

NCERT Class 12 Physics Notes Chapterwise

Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Solutions
• NCERT Exemplar Class 12 Maths
• NCERT Exemplar Class 12 Physics
• NCERT Exemplar Class 12 Chemistry
• NCERT Exemplar Class 12 Biology

Subject Wise NCERT Solutions

• NCERT Solutions for Class 12 Mathematics
• NCERT Solutions for Class 12 Chemistry
• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

NCERT Books and Syllabus