The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d=5\lambda$, where $\lambda$ is the 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 $D=10d?$

$\text{ Here, }d=5\lambda ,D=10d,y=\frac{d}{2}$

$\text{ Resultant Intensity at }y=\frac{d}{2},I_y=?$

$\text{ The path difference between two waves at }y=\frac{d}{2} is$

$\Delta x=d\tan \theta =d\times \frac{y}{D}=\frac{d\times \frac{d}{2}}{10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}$

$\text{ Corresponding phase difference, }\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{\pi }{2}\text{ . }$

$\text{ Now, maximum intensity in Young's double slit experiment, }$

$\begin{array}{lll} & I_{\max }=I_1+I_2+2I_1{I}_2& \\ & I_0=4I& \left(\because I_1=I_2=I\right)\\ \therefore & I=\frac{I_0}{4}& \end{array}$

intensity at y is  $I_y=I_1+I_2+2I_1{I}_2\cos \frac{\pi }{2}=2I=\frac{I_0}{2}$