Young's Double Slit Experiment.

We begin our discussion of interference with a situation shown in

Figure 10.1: Young's double slit experiment I

Figure 10.1. Light from a distant point source is incident on a screen with two thin slits. The separation between the two slits is $d$. We are interested in the image of the two slits on a screen which is at a large distance from the slits. Note that the point source is aligned with the center of the slits as shown in Figure 10.2. Let us calculate the intensity at a point $P$ located at an angle $\theta$ on the screen.

Figure 10.2: Young's double slit experiment II

The radiation from the point source is well described by a plane wave by the time the radiation reaches the slits. The two slits lie on the same wavefront of this plane wave, thus the electric field oscillates with the same phase at both the slits. If $\tilde{E}_1 (t)$ and $\tilde{E}_2(t)$ be the contributions from slits 1 and 2 to the radiation at the point P on the screen, the total electric field will be

$$\tilde{E}(t) = \tilde{E}_1 (t) + \tilde{E}_2(t)$$ (10.1)

Both waves originate from the same source and they have the same frequency. We can thus express the electric fields as $\tilde{E}_1(t) = \tilde{E}_1 e^{i \omega t},~\tilde{E}_2(t) = \tilde{E}_2 e^{i \omega t }$ and $\tilde{E}(t)= \tilde{E}e^{i \omega t}$. We then have a relations between the amplitudes

$$\tilde{E}= \tilde{E}_1 + \tilde{E}_2$$

. It is often convenient to represent this addition of complex amplitudes graphically as shown in Figure 10.3. Each complex amplitude can be represented by a vector in the complex plane, such a vector is called a phasor. The sum is now a vector sum of the phasors.

Figure 10.3: Summation of two phasors

The intensity of the wave is

$$I = \langle E(t) E(t) \rangle = \frac{1}{2} \tilde{E}\tilde{E}^*$$ (10.2)

where we have dropped the constant of proportionatily in this relation. It is clear that the square of the length of the resultant phasor gives the intensity. Geometrically, the resultant intensity $I$ is the square of the vector sum of two vectors of length $\sqrt{I_1}$ and $\sqrt{I_2}$ with angle $\phi_2-\phi_1$ between them as shown in Figure 10.3. Consequently, the resulting intensity is

$$I=I_1+I_2 + 2 \sqrt{I_1 I_2} \cos(\phi_2-\phi_1) \,.$$ (10.3)

Calculating the intensity algebraically, we see that it is

$$I = \frac{1}{2} [ \tilde{E}_1 \tilde{E}_1^*+ \tilde{E}_2 \tilde{E}_2^* + \tilde{E}_1 \tilde{E}_2^* + \tilde{E}^*_1 \tilde{E}_2]$$ (10.4)

$$= I_1 + I_2 + \frac{1}{2} E_1 E_2 \left[e^{i ( \phi_1 - \phi_2 )} + e^{i ( \phi_2 - \phi_1 ) } \right]$$ (10.5)

$$I = I_1 + I_2 + 2 \sqrt{ I_1 I_2 } \cos (\phi_2 - \phi_1)$$ (10.6)

The intensity is maximum when the two waves have the same phase

(10.7)

and it is minimum when $\phi_2 - \phi_1 = \pi$ i.e the two waves are exactly out of phase

$$I=I_1 + I_2 - 2 \sqrt{I_1 I_2 } \,.$$ (10.8)

The intensity is the sum of the two intensities when the two waves are $\pi/2$ out of phase.

Figure 10.4: Young double slit interference fringes with intensity profile

In the Young's double slit experiment the waves from the two slits arrive at P with a time delay because the two waves have to traverse different paths. The resulting phase difference is

$$\phi_1 - \phi_2 = 2 \pi \frac{d \sin \theta}{\lambda}.$$ (10.9)

If the two slits are of the same size and are equidistant from the the original source, then $I_1=I_2$ and the resultant intensity,

$$I (\theta) = 2 I_1 \left[ 1 + \cos \left( \frac{2 \pi d \sin \theta } { \lambda} \right) \right]$$ (10.10)

$$= 4 I_1 \cos^2 \left( \frac{ \pi d \sin \theta } { \lambda} \right)$$ (10.11)

For small $\theta$ we have

$$I (\theta) = 2 I_1 \left[ 1 + \cos \left( \frac{2 \pi d \theta } { \lambda} \right) \right]$$ (10.12)

There will be a pattern of bright and dark lines, referred to as fringes, that will be seen on the screen as in Figure 10.4. The fringes are straight lines parallel to the slits, and the spacing between two successive bright fringes is $\lambda/d$ radians.