Q: The intensity at the maximum in a Young’s double slit experiment is I_{0} . Distant between two slits id d = 5 λ , where λ is wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?

(a) I_{0}/4

(b) 3/4 I_{0}

(c) I_{0}/2

(d) I_{0}

Ans: (c)

Solution: Here, d = 5 λ , I = ? D = 10 d , x = d/2

Path difference , $\displaystyle \Delta x = \frac{x d}{D} $

$\displaystyle \Delta x = \frac{d}{2}.\frac{5 \lambda}{10 d} $

$\displaystyle \Delta x = \frac{\lambda}{4} $

Phase difference , $\displaystyle \phi = \frac{2\pi}{\lambda}.\frac{\lambda}{4} $

$\displaystyle \phi = \frac{\pi}{2} $

$\displaystyle I = I_0 cos^2\frac{\phi}{2} $

$\displaystyle I = I_0 cos^2\frac{\pi}{4} $

$\displaystyle I = I_0 \frac{1}{2} = \frac{I_0}{2} $