Wave Optics

Wave optics describes the behaviour of light as waves. According to wave optics, Light consists of oscillating electric and magnetic field components, forming electromagnetic waves propagating through space. This wave nature explains phenomena such as interference, diffraction, and polarization.Wave optics provides a comprehensive framework for understanding the behaviour of light as waves, explaining a wide range of optical phenomena observed in nature and utilized in technology and science.

1.0Huygens Wave Theory of Light

• Huygen showed that a wave theory of light could also explain reflection and refraction. The Locus of all particles vibrating in the same phase is called a wavefront. Light travels in a medium in the form of a wavefront.
• When light travels in a medium, the particles of the medium start vibrating, and consequently, a disturbance is created in the medium.
• Each point on the wavefront acts as a source of secondary wavelets, which propagate at the speed of light and emit further wavelets in all directions.
• The tangent plane to these secondary wavelets represents the new position of the wavefront.

2.0Wavefront and Types of Wavefront

The Locus of all particles vibrating in the same phase is called the wavefront; its shape depends upon the shape of the light source from which it originates.

1. Spherical Wavefront-Originates from a point source: $\mathrm{A}\propto \frac{1}{r}$

2. Cylindrical Wavefront- Originates from linear source: $A\propto \frac{1}{\sqrt{r}}$

3. Plane Wavefront-Originates from a source situated at a very large distance: A = Constant

3.0Coherent and Incoherent Sources

• Coherent Sources- If they emit light waves of same frequency and have a constant phase difference.
• Incoherent Sources-Phase Difference changes with time.

4.0Methods of Obtaining Coherent Sources

1. Division of Wavefront
2. Division of Amplitude

5.0Interference of Light

• When waves emitting from coherent sources propagating in the same direction are superposed over one another, energy redistribution occurs; this phenomenon is called interference, which is based on energy conservation.
• Resultant Intensity 

${\mathrm{I}}_{\mathrm{R}}={\mathrm{I}}_1+{\mathrm{I}}_2+2\sqrt{I_1}\sqrt{I_2}\mathrm{Cos}\Delta \varphi$

$2\sqrt{I_1}\sqrt{I_2}\cos \Delta \varphi =InterferenceFactor$

Constructive Interference

• $\Delta x=n\lambda$ (Path difference)
• $\Delta \varphi =2n\pi$ (Phase difference)
• $A_{\max }=A_1+A_2$
• $I_{\max }={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$

Destructive Interference

• $\Delta x=\frac{2(n-1)\lambda }{2}$
• $\Delta \varphi =(2n-1)\pi =(2n-1)$
• A_min = A₁ - A₂
• $I_{\min }={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2$

6.0Conditions for Sustained Interference

• Two sources must be coherent.
• Separation between two coherent sources must be small.
• Distance between source and screen must be large.
• The amplitude of two interfering waves must be as nearly equal as possible for good contrast between maxima and minima.

7.0Young's Double Slit Experiment (YDSE)

Light wave nature is proved experimentally by YDSE, in this experiment division of wavefront takes place.

If bright fringes is formed

• $\Delta x=n\lambda$
• $\frac{dy}{D}=\mathrm{n}\lambda$
• $\mathrm{y}=\frac{n\lambda D}{d}$
• $y_{\text{max }nth}=\frac{n\lambda D}{d}$

If dark fringes is formed

• $\Delta x=\frac{(2n-1)\lambda }{2}$
• $\frac{dy}{D}=\frac{(2n-1)\lambda }{2}$
• $y=\frac{(2n-1)\lambda D}{2d}$
• $y_{\min nth}=\frac{(2n-1)\lambda D}{2d}$

8.0Fringe width($\beta$)

• Distance between two successive dark or bright fringes.
• $\beta =\frac{\lambda D}{d}$

Angular Fringe Width ($\theta$)

• It is the angle subtended by linear fringe width at the centre of the plane of the slit.
• The width remains unchanged regardless of the distance between the slit plane and the screen.

$\theta =\frac{\lambda }{d}$

Variation of Intensity with phase difference

• If slits are of equal width then intensity is also equal 

$I=4I_0{\mathrm{Cos}}^2\left(\frac{\Delta \varphi }{2}\right)$

Variations in YDSE

• A Wave traveling in a medium of refractive index ,then time taken by the  wave to cover x distance is calculated: $\mathrm{t}=\frac{\mu x}{c}$
• Optical Path-Distance covered by the wave in the same time t in vacuum: $\mathrm{d}=\mu x$
• Effect of white light in YDSE
• The fringe closest to either side of the central white fringe is red, and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Thin Film in Front of a slit in YDSE

• Path difference$(\Delta x)=\lambda =\frac{dy}{D}-(\mu -1)t$
• Maxima, $\mathrm{n}\lambda =\frac{dy}{D}-(\mu -1)t$
• Minima, $\frac{(2n-1)\lambda }{2}=\frac{dy}{D}-(\mu -1)t$
• Central Maxima is not at the center of the screen: $\mathrm{y}=\frac{(\mu -1)tD}{D}=y_{cm}$

Thin Film is placed in front of both the slit

• Central maxima is shifted to that side where the value of (-1)t is more or path difference due to slab is more.
• For CBF shift upwards

$\left({\mu }_1-1\right)t_1>\left({\mu }_2-1\right)t_2$

• For CBF shift downwards

$\left({\mu }_1-1\right)t_1<\left({\mu }_2-1\right)t_2$

• For no shift in CBF 

$\left({\mu }_1-1\right)t_1=\left({\mu }_2-1\right)t_2$

$y_{cm}=\frac{\left(\left({\mu }_2-1\right)t_2-\left({\mu }_1-1\right)t_1\right)D}{d}$

$n=\frac{\left(\left({\mu }_2-1\right)t_2-\left({\mu }_1-1\right)t_1\right)D}{\lambda }$

9.0Path Difference in Thin Film Interference

• For Reflected System

1. Maxima: $2\mu t\cos r-\frac{\lambda }{2}=n\lambda$
2. Minima: $2\mu t\cos r-\frac{\lambda }{2}=\frac{(2n-1)\lambda }{2}$                     

• For Transmitted System

1. Maxima: $2\mu t\cos r=n\lambda$
2. Minima: $2\mu t\cos r=\frac{(2n-1)\lambda }{2}$

• For thin films to be observable thickness of film must be in order of wavelength of light $\approx 10^{-7}\mathrm{m}$.
• For Reflected System, thickness tends to zero

$\Delta x=2\mu t\cos r-\frac{\lambda }{2}$

$|\Delta x|=\frac{\lambda }{2}$, this condition represents destructive interference therefore dark fringes are formed.

10.0Diffraction

The bending of light rays from the sharp edges of an opaque obstacle or aperture and its spreading in the geometrical shadow region is called diffraction.

Diffraction depends on  

1. Size of obstacle/Aperture
2. Wavelength of wave

Condition of Diffraction

1. Size of the obstacle/aperture must be equal to the wavelength.
2. If the obstacle/aperture size is much larger than the wavelength, diffraction is not observable.

Types of Diffraction

• Fresnel Diffraction-Source and screen are near  the diffraction device.
• Fraunhofer Diffraction-Source and screen are effectively at an  infinite distance from the diffracting device.

• Path Difference : $\Delta x=a\sin \theta$
• n^th Secondary Minima

$\Delta \varphi =2n\pi$

$a\sin \theta =n\lambda$

If angle is small

$\mathrm{a}\mathrm{Tan}\theta =\mathrm{n}\lambda$

$\mathrm{a}\frac{y}{D}=\mathrm{n}\lambda$

$\mathrm{y}=\frac{n\lambda D}{A}$

nth secondary Maxima

• $\Delta \varphi =(2n+1)\pi$
• $\Delta x=\frac{(2n+1)\lambda }{2}$
• $a\mathrm{Sin}\theta =\frac{(2n+1)\lambda }{2}$

If angle is small 

$a\tan \theta =\frac{(2n+1)\lambda }{2}$

$a\frac{y}{D}=\frac{(2n+1)\lambda }{2}\Rightarrow y=\frac{(2n+1)\lambda D}{2a}$

Central Bright Fringe(CBF)

• It is the region between the first minima on both sides of the centre of the screen.
• Linear Width of CBF = $\frac{2\lambda D}{a}$
• Angular Width of CBF $=2{\mathrm{Sin}}^{-1}\left(\frac{\lambda }{a}\right)$
• For Small angle, angular width $=\frac{2\lambda }{a}$

Fresnel’s Distance (Zf)

• Distance upto which ray optics is a good approximation: $Z_f=\frac{a^2}{\lambda }$
• Upto Fresnel’s Distance diffraction is not observable.
• After Fresnel’s Distance spreading due to diffraction dominates.

11.0Polarisation of Light Wave

• It is the phenomenon of constraining the vibration of light (electric vector) to a specific direction perpendicular to the wave's direction of propagation.
• Polarisation confirms the transverse nature of waves.

Unpolarised light

• Light waves exhibit electric field oscillations in every possible direction perpendicular to their propagation direction.

Polarised light

• Light waves in which the electric field oscillates predominantly in a specific direction, typically along a single plane perpendicular to the direction of propagation.

Polariser - Device that converts unpolarised light into polarised light. e.g Nicol Prism, Tourmaline Crystal

Analyser - Device is used to determine whether or not the light is plane polarised.

Note: The polariser and analyzer are made up of the same material; the placement position is different.

12.0Methods of obtaining Plane Polarised Light

• By Selective Absorption
• By Reflection
• By Refraction
• By Scattering
• By Double Refraction

Malus law

This law states that the intensity of polarized light passing through a polarizer varies proportionally with the square of the cosine of the angle θ, which represents the alignment between the polarizer's transmission axis and the light's polarization direction.

$\mathbf{I}=I_0{\cos }^2\theta$

Brewster’s Law( Polarisation by Reflection)

• When unpolarised light is incident on the interface of two mediums such that,Reflected and refracted rays are oriented perpendicularly to each other. the reflected ray is perfectly polarised normal to the plane of incidence, and then the refracted ray in the same plane is partially polarised.

$\mathrm{Tan}{\theta }_p=\frac{{\mu }_2}{{\mu }_1}$

Polarisation By Refraction

• According to Brewster's Law, the reflected light will be plane-polarised with vibrations perpendicular to the plane of incidence, and the transmitted light will be partially polarised. Since in one reflection, about 15% of the light with vibration perpendicular to the plane of paper is reflected, after passing through a number of plates, emerging light will become plane-polarised with vibrations in the plane of paper.

Polarisation by Scattering

• When light scatters on small particles such as dust or air molecules, it is absorbed by electrons in the atoms and re-radiated in all directions.
• Light scattered in a plane at a right angle to the incident light is always plane-polarised.

13.0Solved Examples

Question 1. If the amplitude of light at 10 m from a small light bulb is A₀ then find amplitude of light at a distance 50 m from the same light bulb?

Solution:

$\mathrm{A}\propto \frac{1}{r}\Rightarrow \frac{A_1}{A_2}=\frac{r_2}{r_1}\Rightarrow \frac{A_0}{A_2}\Rightarrow A_2=\frac{A_0}{5}$

Question 2. Find resultant amplitude of two superimposing waves given by y₁ = 6 sin($\omega$t), y₂ = 8 cos($\omega$t)

Solution:

$y_1=6\sin (\omega t)\text{ and }{y}_2=8\cos (\omega t)=(8\mathrm{Sin}\left(\omega t+90^{\circ }\right)$              

$\Delta \varphi =(\omega t+90)-(\omega t)=90^{\circ }$

$A_{\text{Resultant }}=\sqrt{6^2+8^2+2(6)(8)\mathrm{Cos}90^0}=\sqrt{100}=10$

Question 3. In YDSE two narrow slits are 1 mm apart are illuminated by a source of light of wavelength 500 nm. How far apart adjacent bright bands in the interference pattern observed on a screen 2 m away and also find distance of the third bright fringe from center of screen?

Solution:

$\beta =\frac{\lambda D}{d}=\frac{5\times 10^{-7}\times 2}{10^{-3}}=10^{-3}\mathrm{m}=1\mathrm{mm}\Rightarrow y_{\max 3}=\frac{3\lambda D}{d}=3\beta =3\mathrm{mm}$

Question 4. The distance between the coherent source is 0.3 mm and the screen is 90 cm from the sources. The second dark band is 0.3 cm away from the central bright fringe. Find the wavelength and the distance of the fourth bright fringe from the central bright fringe?

Solution: 

$y_{\min 2}=0.3\times 10^{-2}\mathrm{m}$

$\frac{3\lambda D}{d}=0.3\times 10^{-2}\mathrm{m}$

$\frac{3\times \lambda \times 90\times 10^{-2}}{2\times 3\times 10^{-4}}=3\times 10^{-3}$

$\lambda =\frac{6\times 10^{-7}}{90\times 10^{-2}}=666\mathrm{nm}$

Question 5. A YDSE produces interference fringe for sodium light of wavelength 5890 A° that are 0.40° apart. What is the angular fringe separation if the entire arrangement is immersed in water (μ = 4/3).

Solution:

${\alpha }_{\text{new }}=\frac{\alpha }{\mu }=\frac{0.40^0}{\frac{4}{3}}=0.30^0$

Question 6. In Y.D.S.E, $\lambda$ = 400 nm is used when a thin film of refractive index 1.5 and thickness 4 $\mu$m is placed in front of the upper slit, find the number of maxima that crossed the centre line?

Solution:

$\mathrm{n}=\frac{(\mu -1)t}{\lambda }=\frac{(1.5-1)}{4\times 10^{-7}}\times 4\times 10^{-6}=5$