The maximum intensity in Young's double slit experiment is I₀. Distance between the slits is d = 5 λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10 d

$\frac{I_0}{2}$

Suppose P is a point infront of one slit at which intensity is to be calculated from figure it is clear that $x = \frac{d}{2}$. Path difference between the waves reaching at P

$\Delta=\frac{x d}{D}=\frac{\left(\frac{d}{2}\right) d}{10 d}=\frac{d}{20}=\frac{5 \lambda}{20}=\frac{\lambda}{4}$

Hence corresponding phase difference $\varphi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$

Resultant intensity at P

$I=I_{\max } \cos ^2 \frac{\varphi}{2}=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2}$