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
$\Delta x=\frac{dx}{D}=\frac{d^2}{2D}$
$\Delta x=\frac{(5\lambda {)}^2}{2\times 10d}$
Therefore, phase difference is
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x=\frac{\pi }{2}$
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
$I_{net}=I+I+2\sqrt{I}\sqrt{I}cos\frac{\pi }{2}$
$I_{net}=2I=\frac{I_0}{2}$