# Interference Limited to Young's Double Slit Experiment

## 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} \label{yta:eqn:1} 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, $I_1=I_0$, $I_2=2I_0$, $I_\text{max}=(3+2\sqrt{2})I_0=5.83I_0$ and $I_\text{min}=(3-2\sqrt{2})I_0=0.17I_0$. The readers are encouraged to show that both $I_\text{max}$ and $I_\text{min}$ increase if $I_2=xI_0$ for some $x>1$.