In Young's double-slit experiment, the maximum intensity is I₀. If λ is the wavelength of monochromatic light used in the experiment then the distance between the slits is d = 5λ. Find the intensity of light in front of one of the slits on a screen at a distance D = 10 d.

Both the sources have equal intensity. So, resultant intensity at $A=I_o{\cos }^2\left(\frac{\varphi }{2}\right)$ where, ϕ is the phase difference between two sources at A.
From the diagram, path difference is given by:

$\begin{array}{l}\mathrm{\Delta x}=\mathrm{ytan}\left(\mathrm{\theta }\right)\\ \mathrm{\Delta x}=\frac{\mathrm{yd}}{\mathrm{D}}\left(\mathrm{y}=\frac{\mathrm{d}}{2}\right)\\ \mathrm{\Delta x}=\frac{\left(\frac{5\lambda }{2}\right)5\mathrm{\lambda }}{10\times 5\mathrm{\lambda }}=\frac{\mathrm{\lambda }}{4}\end{array}$

Phase difference at A, $\mathrm{\Delta }\varphi$

$\begin{array}{l}=\mathrm{k\Delta x}\\ \mathrm{\Delta \varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times \frac{\mathrm{\lambda }}{4}\\ \mathrm{\Delta \varphi }=\frac{\mathrm{\pi }}{2}\end{array}$

Total intensity at A by interference of equal intensity waves is given by :

$\begin{array}{l}\mathrm{I}={\mathrm{I}}_{\mathrm{o}}{\cos }^2\frac{\mathrm{\varphi }}{2}\\ \mathrm{I}={\mathrm{I}}_{\mathrm{o}}{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right)\\ \mathrm{I}=\frac{{\mathrm{I}}_0}{2}\end{array}$