Question

The intensity at the maximum in a Young's double slit experiment is $1_0$. Distance between two slits is $d=5\lambda$, where $\lambda$ is the wavelength if 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 = 10 d?

Answer 1

it is given that maximum intensity is $I_0$

Distance between slits $d=5\lambda$

Distance between screen and slits $D=10d$

Path difference is given by

$\Delta x=\frac{yd}{D}$

$\Delta x=\left(\frac{5\lambda }{2}\right)\left(\frac{5\lambda }{10\lambda }\right)(\because y=\frac{d}{2}=\frac{5\lambda }{2})$

$\Delta x=\frac{\lambda }{4}$

Phase difference is

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

$\varphi =\frac{\pi }{2}$

Thus, intensity is

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

$I=\frac{I_0}{2}$