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. What will be the intensity of light in front of one of the slits on a screen at a distance $D=10d$?

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

Answer 1

Answer: (D)

$I_0/2$

Path difference $=\Delta x=\frac{xd}{D}$-------(i)

In front of one of the slits,

$x=\frac{d}{2}$ but $d=5\lambda$

$x=\frac{5\lambda }{2}$ and $D=10d=50\lambda$

Now, from equation (i), we have:

$\Delta x=\frac{5\lambda }{2}\times \frac{5\lambda }{50\lambda }=\frac{\lambda }{4}$

So, corresponding phas defference:

$\varphi =\frac{2\pi }{\lambda }.\left(\Delta x\right)=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

As, $I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

So, $I=I_0{\cos }^2\left(\frac{\pi }{4}\right)=\frac{I_0}{2}$