Question

Interference effects are produced at point $P$ on a screen as a result of direct rays from a $500\text{nm}$ source and reflected rays from a mirror, as shown in the figure. If the source is $100\text{m}$ to the left of the screen and $1.00\text{cm}$ above the mirror, find the distance $y$ to the first dark band above the mirror.

• A. $y=1.5\text{mm}$
• B. $y=2.5\text{mm}$
• C. $y=3.5\text{mm}$
• D. $y=0.5\text{mm}$

Answer 1

The correct option is B $y=2.5\text{mm}$


Here, the actual source and image of the source acts as two coherent sources to produce an interference pattern.

The interference occurs by actual ray and reflected ray.

Unlike, the proper $YDSE$ set up, here reflected ray will have an additional path difference of $\frac{\lambda }{2}$ due to reflection.

So, central fringe $O$ will be dark fringe.

The first minima shall be formed at a distance of one fringe width from $O$.

Fringe width,

$\beta =\frac{\lambda D}{d}=\frac{500\times {10}^{-9}\times 100}{2\times {10}^{-2}}=2.5\times {10}^{-3}\text{m}=2.5\text{mm}$

So, the first dark band will be formed at a distance of $2.5\text{mm}$ above the mirror.

Hence, option $\left(B\right)$ is correct.