# 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,

- $\beta_G > \beta_B > \beta_R$
- $\beta_B > \beta_G > \beta_R$
- $\beta_R > \beta_B > \beta_G$
- $\beta_R > \beta_G > \beta_B$

**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,

- the intensities of both maxima and the minima increase.
- the intensity of the maxima increases and the minima has zero intensity.
- the intensity of maxima decreases and that of minima increases.
- 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 $x>1$.