Question

In a double slit experiment, the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed exactly in front of one of the slits its order is,

• A. $\frac{d}{\lambda }$
  
• B. $\frac{{\lambda }^2}{dD}$
  
• C. $\frac{D^2}{2\lambda D}$
  
• D. $\frac{d^2}{2D\lambda }$
  

Answer 1

The correct option is D $\frac{d^2}{2D\lambda }$


When bright fringe is formed in front of one of the slits its distance$\left(y\right)$ from the center is $y=\frac{d}{2}$

The path difference at this point will be,
$\Delta x=\frac{yd}{D}$

$\therefore \Delta x=\frac{d\left(\frac{d}{2}\right)}{D}=\frac{d^2}{2D}$

Let the $n^{\text{th}}$ bright fringe is formed at this point. So the path difference sould be equal to $n\lambda$.

$\Rightarrow n\lambda =\frac{d^2}{2D}$

$\therefore n=\frac{d^2}{2D\lambda }$

Hence, $\left(D\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this question :}\\ \text{Tip : In YDSE, for}{n}^{th}\text{bright fringe, path difference is equal to n times}\lambda \text{.}\end{array}$$