Wave Optics Class 12 Notes CBSE Physics Chapter 10 (Free PDF Download)

Section–A (1 Mark Questions)

1. State the reason why two independent sources of light cannot be considered as coherent sources. 

Ans. Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

2. Draw a diagram to show refraction of a plane wave front incident in a convex lens and hence draw the refracted wave front. 

Ans.

3. Differentiate between a ray and a wave front. 

Ans. Ray defines the path of light. Wave front is the locus of points in the light wave having the same phase of oscillation at any instant.

4. In Young's experiment the monochromatic light is used to illuminate two slits A and B as shown in figure. Interference fringes are observed on a screen placed in front of the slits. Now a thin glass plate is placed normally in the path of beam coming from the slit A, then what changes will be observed in the fringe width?

Ans.     

When we placed a glass plate in the path of one of the beams, there would be a variation in the position of fringes on the screen. All the fringes appearing on the screen will be shifted. But it will not cause a change in the width of the fringe. 

5. Find the angular reAnsution of a 10 cm diameter telescope at a wavelength of $5000A^{\circ}$ .

Ans. Angle reAnsution of telescope is given by;

$\Rightarrow d\theta=\dfrac{1\cdot 22\lambda }{d}$

Where, 

$d\theta$ is the angular reAnsution of telescope, d is the diameter of a telescope.

$\Rightarrow d\theta \dfrac{1\cdot 22\times 5000\times 10^{-10}}{0\cdot 1}=6\cdot 1\times 10^{-6}rad$ .

Section–B (2 Marks Questions)

6. In a single slit diffraction pattern, the distance between the first minima on the left and the first minima on the right is 5 mm. The screen on which the diffraction pattern is displayed is at a distance of 80 cm from the slit. The wavelength is 6000 Å. Then find the slit width.

Ans. Slit width, $d=\dfrac{2D\lambda }{\beta _{centre}}$

Given, $D=80cm=80\times 10^{-2}m$ ,

$\lambda =6000A^{\circ}=6000\times 10^{-10}m$

and $\beta =5mm=5\times 10^{-3}m$

$\therefore d=\dfrac{80\times 10^{-2}\times 6000\times 10^{-10}\times 2}{5\times 10^{-3}}$

=0.192mm

. 

7. In Young’s double slit experiment, the distance between two slits is halved and the distance between the screen and slit is made three times. Then, find the new width of the fringe.

Ans. Given that, the distance between two slits ${d}'=\dfrac{d}{2}$ and the distance between screen and slit D’ = 3D

We know that, the fringe width is 

$\beta =\dfrac{D\lambda }{d}$ … (i)

According to question

$\Rightarrow \beta=\dfrac{{D}'\lambda }{{d}'}\Rightarrow {\beta }'=\dfrac{3D\cdot \lambda }{d/2}$

$\Rightarrow \beta =\dfrac{6D\lambda }{d}$ … (ii)

On dividing Eq. (ii) by Eq. (i), we get

$\dfrac{{\beta }'}{\beta }=\dfrac{\dfrac{6D\lambda }{d}}{\dfrac{D\lambda }{d}}\Rightarrow \dfrac{{\beta }'}{\beta }=6\Rightarrow {\beta }'=6\beta$

8. Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. Then find the ratio of intensities of the sources.

$\dfrac{I_{max}}{I_{min}}=\dfrac{(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}}{(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}}=\dfrac{25}{I}$

Ans. 

$\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}=\dfrac{5}{1}\Rightarrow \dfrac{I_{1}}{I_{2}}=\dfrac{9}{4}$

9. In Young’s double slit experiment, the aperture screen distance is 2 m. The slit width is 1 mm and light of 600 nm is used. If a thin plate of glass ( $\mu$ = 1.5) of thickness 0.06 mm is placed over one of the slits, then find the lateral displacement of the fringes that will take place there.

Ans. When a thin glass plate of thickness t is placed over one of the slits, then lateral displacement is given by

Given,

$X=\dfrac{(\mu -1)tD}{d}$

Given, $\mu$ =1.5, t=0.06mm= $6\times 10^{-5}m$

and D=2m, d=1mm= $1\times 10^{-3}m$

Putting the values in the above relation, we get

$X=\dfrac{(1\cdot 5-1)\times 6\times 10^{-5}\times 2}{1\times 10^{-3}}=0\cdot 5\times 12\times 10^{-12}=6cm$

10. Two Nicols are oriented with these principal planes making an angle of 60°. Find the percentage of incident unpolarised light which passes through the system.

Ans. Intensity of polarized light from first polarizer $=\dfrac{100}{2}=5$ 

$I=I_{0}\;cos^{2}\;\theta =50\;cos^{2}\;60^{\circ}$

$\dfrac{50}{4}=12\cdot 5$

$\dfrac{12\cdot 5}{100}\times 100=12\cdot 5%$ . 

PDF Summary - Class 12 Physics Wave Optics Notes (Chapter 10)

1. Wave Front:

A light source is a point which emits disturbance in all directions. In a homogeneous medium, the disturbance reaches all those particles of the medium in phase, which are located at the same distance from the source of light and hence at all the time, every particle must be vibrating in phase with each other. The locus of all the particles of medium, which at any instant are vibrating in the same phase, is called the wave front. 

Depending upon the shape of the source of light, wave front can be the following types:

1. Spherical wavefront

2. Cylindrical wavefront

1.1. Spherical Wave Front: 

A point source of light produces a spherical wave front. This is because the locus of every points, which are equidistant from the point source, is a sphere figure (a).

1.2. Cylindrical Wave Front: 

If the light source is linear (such as a slit), it produces a cylindrical wave front. Here, every points, which are equidistant from the linear source, lie on the surface of a cylinder figure (b). 

1.3. Plane Wave Front: 

A wave front will appear plane if it is a small part of a spherical or a cylindrical wave front I originating from a distant source. So it is called a plane wave front figure (c). 

1.4. Ray of Light: 

The path along which light travels is known as a ray of light. If we draw an arrow normal to the wave front and which points in the direction of propagation of disturbance represents a ray of light. In a ray diagram, thick arrows represent the rays of light. 

It is also called as the wave normal because the ray of light is normal to the wave front. 

Key Points: 

• If we take any two points on a wave front, the phase difference between them will be zero. 

2. Huygens’s Principle: 

Huygens’s principle is a geometrical construction, which can be used to obtain new position of a wave front at a later time from its given position at any instant. Or we can quote that this principle gives a method gives an idea about how light spreads out in the medium. 

It is developed on the following assumptions: 

1. All the points on a given or primary wave front acts as a source of secondary wavelets, which sends out disturbance in all directions in a similar manner as the primary light source. 

2. The new position of the wave front at any instant (called secondary wave front) is the envelope of the secondary wavelets at that instant. 

These two assumptions are known as Huygens principle or Huygens’ construction.

Key Point: 

• Huygens principle is simply a geometrical construction to find the position of wave front at a later time. 

3. Principle of Superposition:

If two or more than two waves superimpose each other at a common particle of the medium then the resultant displacement \[\left( \text{y} \right)\] of the particle is equal to the vector sum of the displacements (\[{{\text{y}}_{\text{1}}}\]  and \[~{{\text{y}}_{\text{2}}}\] ) produced by individual waves  .i.e  $\overrightarrow{\text{y}}\text{=}\overrightarrow{{{\text{y}}_{\text{1}}}}\text{+}\overrightarrow{{{\text{y}}_{\text{2}}}}$

3.1. Graphical View: 

3.2. Phase/Phase difference/Path difference/Time difference

i. Phase: Phase is defined as the argument of sine or cosine in the expression for displacement of a wave. For displacement  $\text{y = asin }\!\!\omega\!\!\text{ t}$ ; term $\text{ }\!\!\omega\!\!\text{ t =}$ phase or instantaneous phase.

ii. Phase Difference $\left( \phi  \right)$: Phase difference is the difference between the phases of two waves at a point. i.e. if ${{\text{y}}_{\text{1}}}\text{=}{{\text{a}}_{\text{1}}}\text{sin }\!\!\omega\!\!\text{ t}$ and ${{\text{y}}_{\text{2}}}\text{=}{{\text{a}}_{\text{2}}}\text{sin}\left( \text{ }\!\!\omega\!\!\text{ t+}\phi  \right)$ so phase difference $=\phi $ 

iii. Path Difference  $\left( \Delta  \right)$: Path difference between the waves at that point is the difference in path length’s of two waves meeting at a point. Also $\text{ }\!\!\Delta\!\!\text{ =}\frac{\text{ }\!\!\lambda\!\!\text{ }}{\text{2 }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\phi $.

iv. Time Difference (T.D): Time difference between the waves meeting at a point is given by T.D $\text{= }\frac{\text{T}}{\text{2 }\!\!\pi\!\!\text{ }}\times \phi $ 

3.3. Resultant Amplitude and Intensity

If we have two waves   ${{\text{y}}_{\text{1}}}\text{= }{{\text{a}}_{\text{1}}}\text{sin }\!\!\omega\!\!\text{ t}$  and  ${{\text{y}}_{\text{2}}}\text{= }{{\text{a}}_{\text{2}}}\text{sin}\left( \text{ }\!\!\omega\!\!\text{ t+}\phi  \right)$ where  ${{\text{a}}_{\text{1}}}\text{,}{{\text{a}}_{\text{2}}}\text{=}$ Individual amplitudes,  $\phi =$   Phase   difference between   the   waves at an instant when they are meeting a point.     ${{\text{I}}_{\text{1}}}\text{,}{{\text{I}}_{\text{2}}}\text{=}$Intensities of Individual waves.

Resultant Amplitude: 

After superimposition of the given waves resultant amplitude (or the amplitude of resultant wave) is given by  $\text{A=}\sqrt{{{\text{a}}_{\text{1}}}^{\text{2}}\text{+}{{\text{a}}_{\text{2}}}^{\text{2}}\text{+2}{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{cos}}\phi $ 

For the interfering waves ${{\text{y}}_{\text{1}}}\text{=}{{\text{a}}_{\text{1}}}\text{sin }\!\!\omega\!\!\text{ t}$and ${{\text{y}}_{\text{2}}}\text{=}{{\text{a}}_{\text{2}}}\text{sin}\left( \text{ }\!\!\omega\!\!\text{ t+}\phi  \right)$, Phase difference between them is ${{90}^{o}}$ . So resultant amplitude $\text{A=}\sqrt{{{\text{a}}_{\text{1}}}^{\text{2}}\text{+}{{\text{a}}_{\text{2}}}^{\text{2}}}$

Resultant Intensity:  

As we know intensity $\text{ }\!\!\alpha\!\!\text{ }{{\left( \text{Amplitude} \right)}^{\text{2}}}$ $\Rightarrow {{\text{I}}_{\text{1}}}\text{-k}{{\text{a}}_{\text{1}}}^{\text{2}}\text{,}{{\text{I}}_{\text{2}}}\text{-k}{{\text{a}}_{\text{2}}}^{\text{2}}\text{ and I=k}{{\text{A}}^{2}}(k$  is a proportionality constant) . Hence from the formula of resultant amplitude, we get the following formula of resultant intensity

$\text{I-}{{\text{I}}_{\text{1}}}\text{+}{{\text{I}}_{\text{2}}}\text{+2}\sqrt{{{\text{I}}_{\text{1}}}{{\text{I}}_{\text{2}}}}\text{cos}\phi $ 

The term $\text{2}\sqrt{{{\text{I}}_{\text{1}}}{{\text{I}}_{\text{2}}}}\text{cos}\phi $  is called interference term. For incoherent interference this term is zero so resultant intensity  $\text{I=}{{\text{I}}_{\text{1}}}\text{+}{{\text{I}}_{\text{2}}}$ 

3.4. Coherent Sources:

Coherent sources are the sources of light which emits continuous light waves with same wavelength, frequency and in phase or having a constant phase difference.

4. Interference of Light:

If intensity of light at some points is maximum while at some other point intensity is minimum due to the simultaneous superposition of two waves of exactly same frequency (coming from two coherent sources) travels in a medium and in the same direction, this phenomenon is called Interference of light.

4.1. Types of Interference:

4.2. Resultant Intensity Due to Two Identical Waves:

The resultant intensity for two coherent sources is given by

$\text{I=}{{\text{I}}_{\text{1}}}\text{+}{{\text{I}}_{\text{2}}}\text{+2}{{\sqrt{{{\text{I}}_{\text{1}}}\text{I}}}_{\text{2}}}\cos \phi $ 

For identical source ${{\text{I}}_{\text{1}}}\text{=}{{\text{I}}_{\text{2}}}\text{=}{{\text{I}}_{\text{0}}}$   

$\Rightarrow \text{I=}{{\text{I}}_{\text{0}}}\text{+}{{\text{I}}_{\text{0}}}\text{+2}\sqrt{{{\text{I}}_{\text{0}}}{{\text{I}}_{\text{0}}}}\cos \phi =\text{4}{{\text{I}}_{\text{0}}}{{\cos }^{2}}\frac{\phi }{2}$

$\left[ 1+\cos \theta =2{{\cos }^{2}}\frac{\theta }{2} \right]$

Note:

• Redistribution of energy takes place in the form of maxima and minima in interference

• Average Intensity: ${{\text{I}}_{\text{av}}}\text{=}\frac{{{\text{I}}_{\text{max}}}\text{+}{{\text{I}}_{\text{min}}}}{\text{2}}\text{=}{{\text{I}}_{\text{1}}}\text{+}{{\text{I}}_{\text{2}}}\text{=}{{\text{a}}_{\text{1}}}^{\text{2}}\text{+}{{\text{a}}_{\text{2}}}^{\text{2}}$ 

• Ratio of Maximum and Minimum Intensities:

$\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}{{\left( \frac{\sqrt{{{I}_{1}}/{{I}_{2}}}+1}{\sqrt{{{I}_{1}}/{{I}_{2}}}-1} \right)}^{2}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}={{\left( \frac{{{a}_{1}}/{{a}_{2}}+1}{{{a}_{1}}/{{a}_{2}}-1} \right)}^{2}}$ 

• Also $\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}=\frac{{{a}_{1}}}{{{a}_{2}}}=\left( \frac{\sqrt{\frac{{{I}_{\max }}}{{{I}_{\min }}}}+1}{\sqrt{\frac{{{I}_{\max }}}{{{I}_{\min }}}}-1} \right)$ 

• If two waves having equal intensity $\left( {{\text{I}}_{\text{1}}}\text{=}{{\text{I}}_{\text{2}}}\text{=}{{\text{I}}_{\text{0}}} \right)$  meets at two locations $\text{P}$  and $\text{Q}$ with path difference ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$  respectively then the ratio of resultant intensity at point

• $\text{P}$ and $\text{Q}$  will be $\frac{{{I}_{p}}}{{{I}_{Q}}}=\frac{{{\cos }^{2}}\frac{{{\phi }_{1}}}{2}}{{{\cos }^{2}}\frac{{{\phi }_{2}}}{2}}=\frac{{{\cos }^{2}}\left( \frac{\pi {{\Delta }_{1}}}{\lambda } \right)}{{{\cos }^{2}}\left( \frac{\pi {{\Delta }_{2}}}{\lambda } \right)}$ 

5. Young’s Double Slit Experiment (YDSE):

An interference pattern is obtained on the screen when monochromatic light (single wavelength) falls on two narrow slits ${{\text{S}}_{1}}\text{ and }{{\text{S}}_{2}}$ which are very close together acts as two coherent sources, and when waves coming from these two sources superimposes on each other. Alternate bright and dark bands obtained on the screen in this experiment. These bands are called Fringes.

$d=$ Distance between slits.

$D=$ Distance between slits and screen

$\lambda =$ Wavelength of monochromatic light emitted from source.

1. At the central position $\phi ={{0}^{o}}\text{ or }\Delta \text{=0}$. So, the Central fringe will always be bright.

2. The fringe pattern formed by a slit will be brighter than that due to a point.

3. The minima will not be completely dark if the slit widths are unequal. So, uniform illumination occurs for a very large width.

4. No interference pattern is observed on the screen if one slit is illuminated with red light and the other is illuminated with blue light.

5. The central fringe will be dark instead of bright if the two coherent sources consist of object and its reflected image.

5.1. Path Difference:

Path difference between the interfering waves meeting at a point $P$ on the screen is given by $\text{x=}\frac{\text{yd}}{\text{D}}\text{=d}\sin \theta $  where $x$ is the position of point $P$ from central maxima.

For maxima at$P$: $\text{x = n }\!\!\lambda\!\!\text{ }$ 

Where $n=0,\pm 1,\pm 2,...$ 

And for minima at$P$: $x=\frac{\left( 2n-1 \right)\lambda }{2}$ 

Where $n=0,\pm 1,\pm 2,...$

Note: 

If the slits are horizontal path difference is $d\cos \theta $,so as $\theta $ increases,$x$ decreases. But if the slits are vertical, the path difference $\left( x \right)$is $d\sin \theta $, so as $\theta $ increases,$\Delta $ also increases.

5.2. More About Fringes:

(i) Every fringes will have equal width. Width of one fringe is $\text{ }\!\!\beta\!\!\text{  = }\frac{\text{ }\!\!\lambda\!\!\text{ D}}{\text{d}}$ and angular fringe width $\text{ }\!\!\theta\!\!\text{  = }\frac{\text{ }\!\!\lambda\!\!\text{ }}{\text{d}}$ 

(ii) If the YDSE setup is taken in one medium then changes into another, so $\beta $ changes. E.g. in water ${{\lambda }_{w}}=\frac{{{\lambda }_{a}}}{{{\mu }_{w}}}\Rightarrow {{\beta }_{w}}=\frac{{{\beta }_{a}}}{{{\mu }_{w}}}=\frac{3}{4}{{\beta }_{a}}$ 

(iii) Fringe width $\text{ }\!\!\beta\!\!\text{ }\propto \frac{\text{1}}{\text{d}}$ i.e if separation between the sources increases, $\beta $ decreases.

(iv) Position of ${{n}^{th}}$ bright fringe from central maxima ${{x}_{n}}=\frac{n\lambda D}{d}=n\beta ;n=0,1,2,..$ 

(v) Position of  ${{n}^{th}}$dark fringe from central maxima ${{x}_{n}}=\frac{\left( 2n-1 \right)\lambda D}{2d}=\frac{\left( 2n-1 \right)\beta }{2};n=1,2,3,..$ 

(vi) In YDSE, if ${{n}_{1}}$ fringes are visible in a field of view with light of wavelength ${{\lambda }_{1}}$ , while ${{n}_{2}}$ with light of wavelength ${{\lambda }_{2}}$ in the same field, then ${{\text{n}}_{\text{1}}}{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}}\text{=}{{\text{n}}_{\text{2}}}{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}$ 

5.3. Shifting of Fringe Pattern in YDSE:

The fringe pattern will get shifted if a transparent thin film of mica or glass is placed in the path of one of the waves. If this film is placed in the path of upper wave, the pattern shifts upward and if the film is placed in the path of lower wave, the pattern will shift downward.

Fringe shift $\text{= }\frac{\text{D}}{\text{d}}\left( \text{ }\!\!\mu\!\!\text{ -1} \right)\text{t = }\frac{\text{ }\!\!\beta\!\!\text{ }}{\text{ }\!\!\lambda\!\!\text{ }}\left( \text{ }\!\!\mu\!\!\text{ -1} \right)\text{t}$ 

$\Rightarrow $ Additional path difference $=\left( \mu -1 \right)t$ 

$\Rightarrow $If the shift is equivalent to $\text{n}$fringes, then $\text{n=}\frac{\left( \text{ }\!\!\mu\!\!\text{ -1} \right)\text{t}}{\text{ }\!\!\lambda\!\!\text{ }}$  or $\text{t=}\frac{\text{n }\!\!\lambda\!\!\text{ }}{\left( \text{ }\!\!\mu\!\!\text{ -1} \right)}$ 

$\Rightarrow $Fringe shift is independent of the order of fringe (i.e shift of zero order maxima = shift of ${{\text{n}}^{\text{th}}}$ order maxima)

$\Rightarrow $Also, the shift is independent of wavelength.

6. Illustrations of Interference

Interference effects are commonly observed in thin films when their thickness is comparable to wavelength of incident light (If it is too thin as compared to wavelength of light it appears dark and if it’s too thick, this will return in uniform illumination of film). Thin layer of oil on water surface and soap bubbles shows various colours in white light due to interference of waves reflected from the two surfaces of the film.

6.1 Thin Films:

In case of thin films, interference occur between the waves reflected from its two surfaces and waves refracted through it.

Note:

For interference in visible light, the thickness of the film must be in the order of $\text{10,000}{{\text{A}}^{\text{o}}}$ 

7. Doppler’s Effect in Light:

The phenomenon due to relative motion between the source of light and the observer which causes apparent change in frequency (or wavelength) of the light is called Doppler’s effect.

According to special theory of relativity,

$\frac{\text{v }\!\!'\!\!\text{ }}{\text{v}}\text{=}\frac{\text{1 }\!\!\pm\!\!\text{ v/c}}{\sqrt{\text{1-}{{\text{v}}^{\text{2}}}\text{/}{{\text{c}}^{\text{2}}}}}$ 

If $v=$actual frequency, $v'=$ apparent frequency,$v=$ speed of source with respect to stationary observer, $c=$ speed of light.

8. Diffraction of Light:

The phenomenon of light bending around the corners of an obstacle/aperture whose size is comparable to the size of the wavelength of light.

8.1. Types of Diffraction:

The diffraction phenomenon of light is divided into two types

Types of diffraction Phenomenon:

Fresnel Diffraction	Fraunhofer Diffraction
In Fresnel’s diffraction, either source or screen or both are at finite distance	In this case both source and screen are effectively at infinite distance from the diffracting device.
In Fresnel’s diffraction, either source or screen or both are at finite distance from the diffracting device (obstacle or aperture).	Common examples:     Diffraction at single slit, double slit and diffraction grating.

8.2. Diffraction of Light at a Single Slit:

In case of diffraction at a single slit, we get a central bright band with alternate bright (maxima) and dark (minima) bands of decreasing intensity as shown

(i) Width of central maxima ${{\text{ }\!\!\beta\!\!\text{ }}_{\text{o}}}\text{=}\frac{\text{2 }\!\!\lambda\!\!\text{ D}}{\text{d}}$  and angular width $\text{=}\frac{\text{2 }\!\!\lambda\!\!\text{ }}{\text{d}}$ 

(ii) The path difference between the waves from the two ends of the aperture is given by $\text{ }\!\!\Delta\!\!\text{  = n }\!\!\lambda\!\!\text{ }$ ; where $\text{n = 1,2,3,}...\text{i}\text{.e}\text{. dsin }\!\!\theta\!\!\text{  = n }\!\!\lambda\!\!\text{ }$ as the minima occurs at a point on either side of the central maxima.

$\Rightarrow \text{sin }\!\!\theta\!\!\text{  = }\frac{\text{n }\!\!\lambda\!\!\text{ }}{\text{d}}$ 

(iii) The secondary maxima occurs, where the path difference between the waves from the two ends of the aperture is given by $\text{ }\!\!\Delta\!\!\text{ =}\left( \text{2n+1} \right)\frac{\text{ }\!\!\lambda\!\!\text{ }}{\text{2}}$ ; 

where,

$\text{n=1,2,3,}...\text{i}\text{.e}\text{. dsin }\!\!\theta\!\!\text{ =}\left( \text{2n+1} \right)\frac{\text{ }\!\!\lambda\!\!\text{ }}{\text{2}}\Rightarrow \text{sin }\!\!\theta\!\!\text{  = }\frac{\left( \text{2n+1} \right)\text{ }\!\!\lambda\!\!\text{ }}{\text{2d}}$

8.3. Comparison Between Interference and Diffraction: 

8.4. Diffraction and Optical Instruments:

Objective lens of instrument like telescope or microscope etc. acts like a circular aperture. By diffraction of light at the circular aperture, a converging lens doesn’t form a point image of an object rather it produces a brighter disc surrounded by alternate dark and bright concentric rings known as Airy disc.

The angular half width of Airy disc$\text{= }\!\!\theta\!\!\text{ =}\frac{\text{1}\text{.22 }\!\!\lambda\!\!\text{ }}{\text{D}}$ (where $D=$ aperture of lens)

The lateral width of the image $=f\theta $ (where $f=$ focal length of the lens)

Note:

Diffraction of light limits the ability of optical instruments to form clear images of objects when they are close to each other.

9. Polarization of Light:       

Light travel as transverse EM waves. While comparing to magnitude of magnetic field, the magnitude of electric field is much larger. We generally describe light as electric field oscillations. 

9.1. Unpolarized Light: 

Light with electric field oscillations in every directions in the plane perpendicular to the propagation of it is called Unpolarised light. The oscillation of light is divided into horizontal and vertical components. 

9.2. Polarized Light: 

Polarized or plane polarized light is the light with oscillations only in one plane is. 

• Plane of oscillation is the plane in which oscillation occurs in the polarized light. 

• Plane of polarization is the plane perpendicular to the plane of oscillation. 

• By transmitting through certain crystals such as tourmaline or Polaroid light can be polarized. 

9.4. Polarization of Light by Reflection: 

If unpolarized light is reflected using a surface, the reflected light can be obtained as completely polarised, partially polarized or unpolarized. The nature of reflected light depends on the angle of incidence. 

Polarizing angle or Brewster’s angle $\left( {{{i}}_{{p}}} \right)$ is the angle of incidence when the reflected light is completely plane polarized.  (Not in the updated syllabus)

9.5. Polaroid: 

The device used to produce the plane polarised light is known as a Polaroid. It is based on the principle of selective absorption. Also, it is more effective than the tourmaline crystal.

It can also be described as a thin film of ultramicroscopic crystals of quinine iodo sulphate which has its optic axis parallel to each other. 

(i) A Polaroid only allows light oscillations which are parallel to the transmission axis to pass through them. 

(ii) Polarizer is the crystal or Polaroid on which unpolarised light is incident. Crystal or polaroid on which polarised light is incident is called analyzer.

Note: If an unpolarized light is passed through a polarizer, the intensity of the transmitted polarized light will become half of the intensity of unpolarised light.

(i) Polaroids are used in making wind shields of automobiles, sun glasses etc. They helps to reduce head light glare of cares and improve colour contrast in old paintings. Polaroids are also used in 3-D motion pictures are in optical stress analysis.

9.6. Malus Law: 

The intensity of a polarised light passed through an analyser will change as the square of the cosine of the angle between the plane of transmission of the analyser and the plane of the polariser. This is known as Malus law.

$\text{I=}{{\text{I}}_{\text{o}}}\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{  and }{{\text{A}}^{\text{2}}}\text{=}{{\text{A}}_{\text{o}}}^{\text{2}}\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }\Rightarrow \text{A=}{{\text{A}}_{\text{o}}}\text{cos }\!\!\theta\!\!\text{ }$ 

If $\text{ }\!\!\theta\!\!\text{ =}{{\text{0}}^{\text{o}}}\text{,I=}{{\text{I}}_{\text{o}}}\text{,A=}{{\text{A}}_{\text{o}}}$ 

If $\text{ }\!\!\theta\!\!\text{ =4}{{\text{5}}^{\text{o}}}\text{,I=}\frac{{{\text{I}}_{\text{o}}}}{2}\text{,A=}\frac{{{\text{A}}_{\text{o}}}}{\sqrt{\text{2}}}$ 

If $\text{ }\!\!\theta\!\!\text{ =9}{{\text{0}}^{\text{o}}}\text{,I=0,A=0}$ 

(ii) If ${{I}_{i}}=$ Intensity of unpolarised light.

So, ${{I}_{o}}=\frac{{{I}_{i}}}{2}$ i.e. if an unpolarised light is converted into plane polarized light ( say by passing it through a Polaroid or a Nicole-prism), its intensity becomes half and $\text{I=}\frac{{{\text{I}}_{\text{i}}}}{\text{2}}\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }$ 

Note:

Percentage of polarisation$=\frac{\left( {{I}_{\max }}-{{I}_{\min }} \right)}{\left( {{I}_{\max }}+{{I}_{\min }} \right)}\times 100$ 

9.7. Brewster’s Law: 

When a beam of unpolarised light is reflected from a transparent medium (having refractive index$=\mu $ ), the reflected light will be completely plane polarised at a certain angle of incidence (called the angle of polarisation ${{\theta }_{p}}$ ). This is known as Brewster’s law.

Also $\mu =\tan {{\theta }_{p}}$    --- Brewster’s law

i. For $\text{i}{{\text{ }\!\!\theta\!\!\text{ }}_{\text{p}}}\text{ or i}{{\text{ }\!\!\theta\!\!\text{ }}_{\text{p}}}$

Both reflected and refracted rays becomes partially polarized

ii. For glass ${{\text{ }\!\!\theta\!\!\text{ }}_{\text{p}}}\approx \text{5}{{\text{7}}^{\text{o}}}$ , for water ${{\theta }_{p}}\approx {{53}^{o}}$ 

10. Validity of Ray Optics:

By diffraction of light travels, a parallel beam of light travels up to distances as large as few meters can be broadened.  (Not in the updated syllabus)

10.1 Fresnel Distance: 

The minimum distance a beam of light can travel before its deviation from straight line path becomes significant/ noticeable is known as Fresnel distance.

${{\text{Z}}_{\text{F}}}\text{ = }\frac{{{\text{a}}^{\text{2}}}}{\text{ }\!\!\lambda\!\!\text{ }}$ 

As the wavelength of light is very small, the deviation will be also very small and light can be assumed as travelling in a straight line. 

So, we can neglect broadening of beam due to diffraction up to distances as large as a few meters, i.e., we can assume that light travels along straight lines and ray optics can be taken as a limiting case of wave optics. 

Therefore, Ray optics can be considered as a limiting case of wave optics. 

11. Resolving Power:

If two point objects are close to each other, images diffraction patterns of those objects will also be close and overlap each other. 

Limit of resolution of the instrument is the minimum distance between two objects which can be seen separately by the object instrument.

Resolving power (R.P)$=\dfrac{1}{{Limit of Revolution}}$  (Not in the updated syllabus)

11.1. Resolving power of Microscope:

R.P. of microscope$=\dfrac{2\mu \sin \theta }{\lambda }$  (Not in the updated syllabus)

11.2. Resolving power of Telescope:

R.P of telescope ${=}\dfrac{{1}}{{d }\!\!\theta\!\!{ }}{=}\dfrac{{D}}{{1}{.22 }\!\!\lambda\!\!{ }}$ 

Where $D$ is the aperture of the telescope.  (Not in the updated syllabus)

Class 12 Physics Chapter 10 Notes: Laws to Support Wavelength Properties

These laws were formed to support wavelength properties, some of which are mentioned below.

• The original Huygen’s principle formulation and the laws of reflection and refraction.

• Polarisation phenomenon based transverse light and electromagnetic waves.

• The diffraction phenomenon supporting HuygensFresnel principle.

• The occurrence of interfering following the principle of superposition.

Topics Related to Wave Optics

Apart from the laws in Class 12 Physics Wave Optics Revision Notes, students have to understand certain topics, the details of which are provided below.

What Does Wave Optics Mean in Revision Notes Class 12 Chapter 10?

Wave Optics in Class 12 Physics Notes explains the association between rays of light and waves. According to this theory, a transverse wave is actually a light that travels through a medium.

What is Wavefront?

Wavefront is the centre of all those particles that vibrate in the same phase at any instant. The Physics Class 12 Chapter 10 Revision Notes explain that a surface having the same phase of vibrating particles at any moment at every point on it is wavefront.

What is Hyun’s Principle?

Huygens’ principle is a geometrical structure which forms the shape of the wavefront. This takes place at the indefinite time, which helps in determining the wavefront’s figure at a later time. According to Huygens’ principle in NCERT Solutions Chapter 10 Class 12 Physics Revision Notes:

• Each point on a wavefront performs like a light source. It also emits secondary wavelets.

• The secondary wavelets stretch in multiple directions in a vacuum or space. It is related to the velocity of light.

• The cover of the wavefront which belongs to secondary wavelets gives a new position after a given time. It moves along forward direction of a wavefront.

Huygens’ wave theory can also work as a verification for the laws of reflection and refraction. This theory explains the diffraction, phenomenon of polarisation, and interference.

Here ‘V’ is a frequency which is the source in Class 12 Physics Wave Optics Revision Notes, gives v = 1/T. This shows that frequency remains identical as light moves from one medium to a new one. This gives the law that wavelength is inversely relative to the refractive index (μ) of the given medium.

What are Laws of Reflection Based On Huygens’ Wave Theory in Class 12 Notes Wave Optics?

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In the above figure, wavefront AB is an incident on the plane that reflects surface XY. Here both the reflecting surface and wavefront are perpendicular to the plane. A wavefront passes from the reflecting body at B and then at towards C.

In Huygens’ principle, at point BC, secondary wavelets start increasing with the speed c. Disturbance from A reaches point C during the time at the secondary wavelets from B.

Here, we see B has spread over a radius hemisphere that gives BD = AC = ct. Here t is the time taken by the disturbance to travel from A to C. The digression plane CD drawn from point C over this radius hemisphere ct becomes new reflected wavefront. It is where the angle of incidence is equal to the angle of reflection. This finally gives the first law of reflection.

Apart from these topics, NCERT Class 12 Revision Notes Physics Chapter 10 solution has a wide array of questions. A student desiring to gain quality answers for CBSE exams can check the official website of Vedantu. It is a trustworthy educational site offering Class 12 Physics Chapter 10 Revision Notes And Solutions. Moreover, students can check their live classes and training sessions available for a budget-friendly price. Download the official app to study from Class 12 Physics Revision Notes Solution Chapter 10 today!

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