Question

A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance $d$ and the screen is placed parallel to the plane of the slits. If the incident beam makes an angle $\theta =si{n}^{-1}\left(\frac{\lambda }{2d}\right)$ with the normal to the plane of the slits, there will be......at the centre $p_0$ of the pattern

• A. $\frac{1}{4}$the maximum intensity
• B. half the maximum intensity
• C. bright
• D. dark

Answer 1

The correct option is D dark

Given,

$\theta ={sin}^{-1}\left(\frac{\lambda }{2d}\right)sin\theta =\frac{\lambda }{2d}$

We know that path difference

$△x=(2n-1)\frac{\lambda }{2}$

For $n=1$

$△x=(2-1)\frac{\lambda }{2}=\frac{\lambda }{2}$

In young's double slit experiment

$sin\theta =\frac{y}{d}$

By comparing

$y=\frac{\lambda }{2}$

We can see as path difference is odd ,multiple therefore, fringe at point $P$ will be dark.

Correct option is D.