# Young's Double Slit Experiment

Interference occurs when two coherent waves of same frequency but different phase superpose. \begin{align} &E_1=E_{01}\sin(kx-\omega t)\\ &E_2=E_{02}\sin(kx-\omega t+\delta)\nonumber\\ &E=E_1+E_2=E_0\sin(kx-\omega t+\epsilon) \nonumber \\ & E_0=\sqrt{{E_{01}}^2 + {E_{02}}^2 + 2 E_{01} E_{02}\cos\delta} \nonumber\\ &\tan\epsilon=\frac{E_{02\sin\delta}}{E_{01}+E_{02}\cos\delta}\nonumber \end{align}

The path difference between interferening waves in a Young's double slit experiment is given by \begin{align} \Delta x=\frac{dy}{D} \end{align}

The phase difference is given by \begin{align} \delta=\frac{2\pi}{\lambda}\Delta x \end{align}

The conditions for constructive and destructive interference are: for integer $n$, \begin{align} \delta=\left\{\begin{array}{ll} 2n\pi, & \text{constructive};\\ (2n+1)\pi, & \text{destructive}, \end{array}\right.\nonumber \end{align} \begin{align} \Delta x=\left\{\begin{array}{ll} n\lambda, & \text{constructive}; \\ \left(n+\tfrac{1}{2}\right)\lambda, & \text{destructive} \end{array}\right.\nonumber \end{align}

The inetensity is given by \begin{align} I&=I_1+I_2+2\sqrt{I_1I_2}\cos\delta,\nonumber\\ I_\text{max}&=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2,\\ I_\text{min}&=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \end{align} If $I_1=I_2=I_0$ then \begin{align} I&=4I_0\cos^2\tfrac{\delta}{2},\\ I_\text{max}&=4I_0, \\ I_\text{min}&=0 \end{align}

The fringe width is given by \begin{align} w=\frac{\lambda D}{d} \end{align}

The ptical path is given by \begin{align} \Delta x^\prime=\mu\Delta x \end{align}

## Exercise

1. In the Young’s double slit experiment the size of the apertures forming the two coherent sources is given as 0.02 mm. Plot the intensity pattern seen on the screen if the distance between the apertures is kept (a) 0.09 mm (b) 0.03 mm.

## Problems from IIT JEE

Problem (IIT JEE 2012): Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe width recorded are $\beta_G,$ $\beta_R$, and $\beta_B,$ respectively. Then,

1. $\beta_G > \beta_B > \beta_R$
2. $\beta_B > \beta_G > \beta_R$
3. $\beta_R > \beta_B > \beta_G$
4. $\beta_R > \beta_G > \beta_B$

Solution: The fringe width $\beta$ is related to the wavelength $\lambda$ by, $\beta={\lambda D}/{d}$. In the visible spectrum $\lambda_B<\lambda_G <\lambda_R$ (VIBGYOR), and hence $\beta_R>\beta_G>\beta_B$.

Problem (IIT JEE 2000): In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern,

1. the intensities of both maxima and the minima increase.
2. the intensity of the maxima increases and the minima has zero intensity.
3. the intensity of maxima decreases and that of minima increases.
4. the intensity of maxima decreases and the minima has zero intensity.

Solution: Let the intensity of light passing through the two slits be $I_1$ and $I_2$. The intensity at a point having a phase difference $\delta$ is given by, \begin{align} I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta. \end{align} The above equation gives the intensity at maxima ($\delta=0$) and minima ($\delta=\pi$) as, \begin{align} &I_\text{max}=I_1+I_2+2\sqrt{2I_1I_2},\\ &I_\text{min}=I_1+I_2-2\sqrt{I_1I_2}. \end{align} When the slits are of same width, $I_1=I_0$, $I_2=I_0$, $I_\text{max}=4I_0$ and $I_\text{min}=0$. When the width of one of the slit is doubled, \begin{align} I_1&=I_0,\\ I_2&=2I_0,\\ I_\text{max}&=(3+2\sqrt{2})I_0=5.83I_0\\ I_\text{min}&=(3-2\sqrt{2})I_0=0.17I_0. \end{align} The readers are encouraged to show that both $I_\text{max}$ and $I_\text{min}$ increase if $I_2=xI_0$ for some $x>1$.