What is the interference of light?

Word interference means to interfere. It is a process in which at least two waves interfere to produce a resultant wave that may or may not have the same amplitude. Interference of light can be defined as the process in which at least two light waves are superposed in such a way that the resultant wave may have the same or entirely different amplitude. 

How does interference of light happen?

The process of interference of light was explained through Young’s Double-slit experiment. In this experiment, Thomas Young took a coherent light source (in which both the waves were in phase) consisting of monochromatic light (word monochromatic light signifies light having a single wavelength) and passed it through 2 slits. Two coherent light sources were formed because of the 2 slits. Hence, in conclusion, the light waves interfered constructively and destructively. In interference, we can draw the position of the dark and bright fringes easily.

Due to these slits, diffraction (which is not abductive) occurred and light waves produced semi-circular waves. During which, the light fell on the screen and got scattered into the observer’s eyes, and hence, in conclusion, the pattern could be seen clearly.

Types of interference

Constructive interference

The word "constructive" signifies addition in interference. Constructive interference occurs when the path of the waves that start from the slit differs by 1 entire wavelength or in the intervals of the same. Hence, the waves that are presented on the screen are in phase (crest to crest/trough to trough). The waves interfering constructively forms bright fringes on the screen.

Destructive interference

The word "destructive" signifies subtraction in interference. Destructive interference occurs when the path of the waves that start from the slit differs by half-integral numbers. Hence, the waves that are presented on screen aren’t in phase (crest to trough and vice-versa). Waves interfering destructively form dark fringes on the screen.

Derivation

Interference of light is shown below which will, in turn, draw the interference pattern, and hence, as an entailment, we'll obtain the dark and bright fringes.

  Position of Bright and Dark Fringes can be given as:

Path difference for a fringe having the maximum intensity is:

$∆z=n\lambda \left(n=0,\pm 1,\pm 2,.....\right)$

Therefore,

$\begin{array}{rcl}\frac{xd}{D}& =& n\lambda \\ x& =& \frac{n\lambda D}{d}\end{array}$

Distance of  fringe having maximum intensity from the center/base is

$x=\frac{n\lambda D}{d}$

Likewise, the distance of  fringe having maximum intensity from the center/base is

$x_{n-1}=\frac{\left(n-1\right)\lambda D}{d}$

Therefore, a width of fringe can be given as:

$\begin{array}{rcl}\beta & =& x_n-x_{n-1}\\ & =& \frac{n\lambda D}{d}-\frac{\left(n-1\right)\lambda D}{d}\\ & =& \frac{\lambda D}{d}\end{array}$

Similarly, the path difference for a fringe having the maximum intensity is:

$∆z=\left(2n+1\right)·\frac{\lambda }{2}\left(n=0,\pm 1.\pm 2,......\right)$

Therefore,

$\begin{array}{rcl}\frac{xd}{D}& =& \left(2n+1\right)\frac{\lambda D}{2d}\\ x& =& \frac{\left(2n+1\right)\lambda D}{2d}\end{array}$

Distance of  fringe having minimum intensity from the center/base is

$x=\frac{\left(2n+1\right)\lambda D}{2d}$

Likewise, the distance of  fringe having minimum intensity from the center/base is

$x_{n-1}=\frac{\left(2\left(n-1\right)+1\right)\lambda D}{2d}$

Therefore, a width of fringe can be given as:

$\begin{array}{rcl}\beta & =& x_n-x_{n-1}\\ & =& \frac{\left(2n+1\right)\lambda D}{2d}-\frac{\left(2\left(n-1\right)+1\right)\lambda D}{2d} \left(n=0,\pm 1,\pm 2,....\right)\\ & =& \frac{\lambda D}{d}\end{array}$

Fringe’s angular width can be given as

Let be the angular position fringe having a maximum intensity.

Here,

$\begin{array}{rcl}\tan {\theta }_n& =& \frac{\lambda n}{d}\\ {\theta }_n& =& \frac{n\lambda }{d} \left(\tan {\theta }_n\approx {\theta }_n\right)\end{array}$

Position of  fringe having maximum intensity can be given as:

${\theta }_{n+1}=\frac{\left(n+1\right)\lambda }{d}$

Thus, Fringe’s angular width will be :

$\begin{array}{rcl}\theta & =& {\theta }_{n+1}-{\theta }_n\\ & =& \frac{\left(n+1\right)\lambda }{d}-\frac{n\lambda }{d}\\ & =& \frac{\lambda }{d}\end{array}$

The intensity of the Fringes can be given as

The net intensity of light waves due to two coherent sources will be:

$I=I_1+I_2+2\sqrt{\left(I_1·I_2\right)\cos \varphi }$

$Put I_1=I_2=I_o \left(as d<<<D\right)$

Hence,

$\begin{array}{rcl}I& =& I_o+I_o+2\sqrt{\left(I_o{I}_o\right)\cos \varphi }\\ I& =& 2I_o+2I_o\cos \varphi \\ I& =& 2I_o\left(1+\cos \varphi \right)\\ I& =& 4I_o{\cos }^2\left(\frac{\varphi }{2}\right)\end{array}$

Relevant formulas

The width of a fringe is given as

$\beta =\frac{\lambda D}{d}$

Angular fringe width is given as:

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

Fringe’s intensity in Young’s Double-slit experiment is given as

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

Where,

$$\begin{gathered} \beta =Fringe width \\ \lambda =Wavelength of monochromatic light \\ D=Dis\tan ce between the slits and the screen \\ d=Dis\tan ce in between the slits \\ \theta =Angular fringe width \\ I=Intensity of the fringe \end{gathered}$$

Common Mistakes

These are the known facts that the experiment done by Thomas Young for the interference of light is only applicable for the wave nature of light. In fact, there are many experiments done in quantum theory proving that light possesses dual nature i.e. of a wave and a particle. Also, it might become a little difficult to infer abductive reasoning, logical reasoning, inductive reasoning, and deductive reasoning. In the case of light, it can lead to infelicity in calculation or infelicitous results.

Context and Applications

The concept of interference of light is studied in courses like

• Bachelors in Science in Physics 
• Masters in Science in Physics

Practice Problems

1. Which of the following would not show any interference?
   1. Wedge-shaped film
   2. Soap bubble
   3. Thin film
   4. Thick film

Answer: c. Thin film

Explanation: If a thin film is taken to obtain an interference pattern, then the path difference would be minimal. Hence, minima will be obtained on the screen.

2. As a part of an illation, what is the phase difference of coherent light sources?
   1. It is always changing.
   2. It is always constant.
   3. It is always zero.
   4. constant or zero

Answer: d. constant

Explanation: Coherent light sources have constant or zero phases difference and the same frequency.

3. In Young’s double-slit experiment, what will be the angular separation, if the distance between slit and screen is halved to its initial value? 
   1. Twice of its initial value
   2. Thrice of its initial value
   3. Half of its initial value
   4. Remain same

Answer: a. Twice of its initial value

Explanation: As angular separation is inversely proportional to the distance between slits and the screen, it’ll get doubled if the distance between slits and the screen is halved.

4. What is the basic principle behind the interference of light waves?
   1. Superposition principle
   2. Heisenberg’s principle
   3. Principle of conservation of angular momentum
   4. Archimedes principle

Answer: a. Superposition principle

Explanation: When light waves interfere with each other, they undergo constructive interference or destructive interference. Hence, the superposition principle comes into play.

5. By what value does the intensity of waves (having equal amplitude) must change to undergo constructive interference?
   
   1. four times  
   2. Twice
   3. Thrice
   4. It will remain the same

Answer: a. four times

Explanation: The intensity of the waves is given as $\mathrm{Intensity} \propto {\left(\mathrm{Amplitude}\right)}^2$. Hence, in constructive interference of waves having an equal amplitude, the net amplitude will become 2 times and the intensity will become four times.