Question

The intensity at the maximum in a Young's double slit experiment is $I_0$. Distance between two slits is d = $5\lambda$,where $\lambda$ is the wavelength of light used in the experiment . What will be the intensity in front of one the slits on the screen place at a distance D = 2 m

Answer 1

Given that,

maximum intensity $=I_0$ silt width (d) = $5\lambda$ .

at front of a slid, $y=\frac{d}{2}=\frac{5\lambda }{2}$ at A

then path difference, $△x=\frac{dy}{D}=\frac{5\lambda }{2}\left(\frac{5\lambda }{2}\right)$

$△x=\frac{25{\lambda }^2}{4}$

thus, total phare difference, $8=\frac{2\pi }{\lambda }△x=\frac{2\pi }{x}.\frac{25{\lambda }^2}{4}=\frac{25\lambda }{2}\pi$

so intensity at point A,

$I=I_{0}co{s}^{2}8/2=I_{0}co{s}^2\left(\frac{25\lambda }{2}\pi \right)$

If $\lambda$ is even, then $co{s}^2\left(\frac{25\lambda \pi }{2}\right)=1,I=I_0\Rightarrow$ maximum

but $\lambda$ is odd means $co{s}^2\left(\frac{25\lambda \pi }{2}\right)=0I=0\Rightarrow$ maximum