Question

The maximum intensity in Young's double-slit experiment is $I_0$. The distance between the slits is $d=5\lambda$, where $\lambda$ 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 = 10d?

• A. $\frac{I_0}{2}$
• B. $\frac{3}{4}{I}_0$
• C. $I_0$
• D. $\frac{I_0}{4}$

Answer 1

The correct option is A $\frac{I_0}{2}$


Path difference = $\Delta x=\frac{Xd}{D}...\left(1\right)$
In front of one of the slits -
X = $\frac{d}{2}$ but d = $5\lambda$
X = $\frac{5\lambda }{2}$ and D = 10d
So from equation (1)
$\Delta$x at P = $\frac{dx}{D}=\frac{d^2}{2D}=\frac{(5\lambda {)}^2}{2\times 10\times d}$
$\Delta x=\frac{(5\lambda {)}^2}{2\times 10\times 5\lambda }=\frac{\lambda }{4}$
So corresponding phase difference
$\phi =\frac{2\pi }{\lambda }\left(\Delta x\right)=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
As $I=I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$
So, $I=I_{0}co{s}^2\left(\frac{\pi }{4}\right)=\frac{I_0}{2}$