Question

The maximum intensity in Young's double-slit experiment is $I_{0}$ . Distance between the slits is $d=5\lambda $ , where $\lambda $ is the wavelength of monochromatic light used in the experiment. The intensity of the light in front of one of the slits on a screen at a distance $D=10d$ is

• A. $\frac{I_{0}}{2}$
• B. $\frac{3}{4}I_{0}$
• C. $I_{0}$
• D. $\frac{I_{0}}{4}$

Answer 1

Answer: (A)

Path difference at point $P$ is
$\Delta x=\frac{d x}{D}=\frac{d^{2}}{2 D}$
$\Delta x=\frac{(5 \lambda)^{2}}{2 \times 10 d}$
Therefore, phase difference is
$\Delta \phi=\frac{2 \pi }{\lambda }\times \Delta x=\frac{\pi }{2}$
If $I$ is the intensity due to each slit, then the maximum intensity $I_{0}$ is,
$I_{0}=4I$
$\Rightarrow \, I=\frac{I_{0}}{4}$
Therefore, intensity at point $P$ is
$I_{n e t}=I+I+2\sqrt{I}\sqrt{I}cos \frac{\pi }{2}$
$I_{n e t}=2I=\frac{I_{0}}{2}$