Interference of light

When two light waves from two sources meet at a point, they produce a resultant wave according to the principle of superposition. As a result, there is redistribution of light energy and bright and dark bands are produced. This phenomenon is called interference of light.

During superposition, if crest of one wave falls on the crest of the other or the trough of one fall on the trough of the other, the resultant amplitude becomes maximum. This type of interference is called constructive interference and intensity of light is maximum in this case. On the other hand, if the crest of one wave falls on the trough of the other or vice versa, the resultant amplitude becomes minimum. This type of interference is called destructive interference and the intensity of light becomes minimum.

The condition for constructive interference is, path difference (x) = [latex]n\lambda[/latex]

The condition for destructive interference is, path difference (x) = [latex](2n-1)\frac{\lambda}{2}[/latex]

Coherent sources

Coherent sources are sources which emit waves of the same frequency and possess constant phase difference (i.e., phase difference remains constant with time ‘t’). Since, two independent sources can never be coherent because, they may emit light of same wavelength or frequency, they are not in same phase or they may not have the constant phase difference. So, coherent sources are obtained from a source by division of wavefront. Eg. Laser light.

Intensity in Double Slit Experiment

Let us consider, two light waves from two coherent sources S₁ and S₂ meet at a point P on the screen. The displacement of the waves is given by:

[latex]y_1 = a_1sin\omega t[/latex] and

[latex]y_2 = a_2sin(\omega t + \phi )[/latex], where, [latex]a_1[/latex] and [latex]a_2[/latex] be their amplitudes and [latex]\phi[/latex] be the constant phase difference between them.

According to principle of superposition, the displacement of the resultant wave is

[latex]y = y_1 + y_2[/latex]

= [latex]a_1sin\omega t + a_2sin(\omega t + \phi)[/latex]

= [latex]a_1sin\omega t + a_2sin\omega t.cos\phi + a_2cos\omega t.sin\phi[/latex]

= [latex](a_1 + a_2cos\phi)sin\omega t + a_2cos\omega t.sin\phi[/latex] ………………….. (1)

Suppose, [latex]a_1 + a_2cos\phi = Acos\theta[/latex] …………………. (2)

And, [latex]a_2sin\phi = Asin\theta[/latex] ………………………… (3)

Then, equation (1) becomes,

[latex]y = Asin\omega t.cos\theta + Acos\omega t.sin\theta[/latex]

= [latex]Asin(\omega t + \theta)[/latex]

This is the equation for resultant wave on interference of two waves.

To find its amplitude (A), squaring and adding equations (2) and (3), we get,

[latex](a_1 + a_2 cos\phi)^2 + (a_2^2sin^2\phi) = A^2cos^2\theta + A^2sin^2\theta[/latex]

Or, [latex]a_1^2 + 2a_1a_2cos\phi + a_2^2cos^2\phi + a_2^2sin^2\phi = A^2(cos^2\theta + sin^2\theta)[/latex]

Or, [latex]a_1^2 + 2a_1a_2cos\phi + a_2^2 = A^2[/latex] ………………………. (4)

This is the expression for the amplitude of resultant wave in interference.

As we know, Intensity [latex]\propto (amplitude)^2[/latex]

So, for first wave, [latex]I_1 \propto a_1^2[/latex]

Or, [latex]I_1 = ka_1^2[/latex]

For second wave, [latex]I_2 = ka_2^2[/latex]

And, for resultant wave, [latex]I = kA^2[/latex]

So, equation (4) becomes,

[latex]\frac{I_1}{k} + 2\sqrt{\frac{I_1}{k}}.\sqrt{\frac{I_2}{k}}cos\phi + I_2 = I[/latex]

So, [latex]I = I_1 + 2\sqrt{I_1 I_2}cos\phi + I_2 = I[/latex]

This is the required expression for the intensity of resultant wave during interference.

Constructive Interference

Intensity of light will be maximum if: [latex]cos\phi = 1[/latex]

Or, [latex]cos\phi = cos0, cos2\pi, cos4\pi, cos6\pi, ……..[/latex]

Or, [latex]\phi = 0, 2\pi, 4\pi, 6\pi, ……[/latex]

Or, [latex]\frac{2\pi}{\lambda}x = 0, 2\pi, 4\pi, 6\pi, ….[/latex]

Or, [latex]x = 0, \lambda, 2\lambda, 3\lambda, …….. n\lambda[/latex]

Or, path difference (x) = [latex]n\lambda[/latex], where, n = 0, 1, 2, 3, ….

Destructive Interference

Intensity of light will be minimum if: [latex]cos\phi = – 1[/latex]

Or, [latex]cos\phi = cos\pi, cos3\pi, cos5\pi, …….[/latex]

Or, [latex]\phi = \pi, 3\pi, 5\pi, …….[/latex]

Or, [latex]\frac{2\pi}{\lambda}x = \pi, 3\pi, 5\pi, …….[/latex]

Or, [latex]x = \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, …….. \frac{(2n-1)\lambda}{2}[/latex]

Or, path difference (x) = [latex]\frac{(2n-1)\lambda}{2}[/latex], where, n = 1, 2, 3, 4, ……..

Young’s double slit experiment

In 1801, Thomas young demonstrated first experiment on the interference of light and proved the wave nature of light. The experimental setup of this experiment consists of two narrow slits separated by a small distance. These two slits A and B are illuminated by a monochromatic (consisting of single color) light source S. A screen is placed at some distance away from the slit.

Light waves from the slit A and B are in same amplitude, same wavelength and are in same phase. Hence, two slits behave as two coherent sources. When light waves from the two slits fall on the screen, alternate bright and dark bands are produced due to interference of light waves.

Calculation of fringe width

Consider, a monochromatic source S illuminates two narrow slits S₁ and S₂ which are very close to each other and separated by a small distance ‘d’.  A screen is placed at a distance D from the plane of slit. Let, [latex]\lambda[/latex] be the wavelength of light used. Draw perpendiculars S₁Q and S₂R on the screen, the point O is equidistant from S₁ and S₂. Hence, waves coming from S₁ and S₂ reach in phase and meet constructively and bright fringe is observed at O. Consider, another point P on the screen at a distance y from the centre of the screen O. Here, the two waves from S₁ and S₂ and reaching at P, have some path difference x which is given by:

x = S₂P – S₁P

Now, from the figure [latex]\Delta S_2PR[/latex] using Pythagorus theorem,

S₂P² = S₂R² + PR²

            = D² + (y+ OR)²

            = D² + (y+d/2)²

And for [latex]\Delta S_1PR[/latex], using Pythagorus theorem,

S₁P² = S₁Q² + PQ²

            = D² + (y-d/2)²

Then, S₂P² – S₁P² = D² + (y+d/2)² – D² – (y-d/2)²

Or, (S₂P – S₁P) . (S₂P + S₁P) = y² + yd +  – y² + yd –

Or, S₂P – S₁P = [latex]\frac{2yd}{(S_2P + S_1P)}[/latex]

In practice, P is close to O, so that,

S₁P ≈ S₂P ≈ D

[latex]\therefore[/latex] S₂P – S₁P = [latex]\frac{2yd}{2D} = \frac{yd}{D}[/latex]

[latex]\therefore x = \frac{yd}{D}[/latex]

For the constructive interference, the path difference is equal to the integral multiple of the wavelength, i.e.

Path difference (x) = n[latex]\lambda[/latex]

Or, [latex]\frac{yd}{D} = n\lambda[/latex]

Or, y_n = [latex]\frac{n\lambda D}{d}[/latex]   [∵ n = 0, 1, 2, 3….]

For, n = 0, y_o = 0

For, n = 1, y₁ = [latex]\frac{\lambda D}{d}[/latex]

For, n = 2, y₂ = [latex]\frac{2\lambda D}{d}[/latex]

………………….

…………………..

For, n = n, y_n = [latex]\frac{n\lambda D}{d}[/latex]

The distance between two consecutive bright fringes is given by:

y₂ – y₁ = [latex]\frac{2\lambda D}{d} – \frac{\lambda D}{d} = \frac{\lambda D}{d}[/latex] ……….. (i)

Again, for the destructive interference, the path difference is equal to the odd integral multiple of half wavelength.

x = (2n-1)[latex]\frac{\lambda}{2}[/latex]

Or, [latex]\frac{yd}{D} = (2n-1)\frac{\lambda}{2}[/latex]

Or, y_n = [latex]\frac{(2n – 1)}{2d}\lambda D[/latex]    [∵ n = 1, 2, 3, 4……]

For, n = 1, y₁ = [latex]\frac{\lambda D}{2d}[/latex]

For, n = 2, y₂ = [latex]\frac{3\lambda D}{2d}[/latex]

…………………..

…………………..

For, n = n, y_n = [latex]\frac{(2n – 1)}{2d}\lambda D[/latex]

The distance between any two consecutive dark fringes is given by:

y₂ – y₁ = [latex]\frac{3\lambda D}{2d} – \frac{\lambda D}{2d}[/latex]

            = [latex]\frac{\lambda D}{d}[/latex] ………. (ii)

The distance between any two consecutive bright or dark fringes is called fringe width and is denoted by β.

Therefore, β = [latex]\frac{\lambda D}{d}[/latex]

Form eq^n. (i) and (ii), it is clear that the distance between any two consecutive bright fringe and dark fringe are equal i.e. bright and dark fringes are equally spaced.

Condition for sustained interference

The conditions for sustained interference are:

1. The two sources of light must be coherent.
2. The light used must be monochromatic light.
3. The two slits should be narrow.
4. The two sources must be close to each other.
5. The two sources should emit waves continuously.
6. Distance between the source and the screen should be large.