Question

In YDSE the intensities at points where path difference is $\lambda $ and $\lambda /4$ is found to be $K$ and $\frac{N}{2}K$ respectively. Find the value of $N.$

Answer 1

Phase difference, $\phi=\frac{2 \pi }{\lambda }\left(\right.\Delta x\left.\right)$
For path difference $\lambda ,$ phase difference $\phi_{1}=2\pi $
For path difference $\lambda /4,$ phase difference $\phi_{2}=\pi /2$
Using, $I=4I_{0}cos^{2}\frac{\phi}{2}$
$\therefore \frac{I_{1}}{I_{2}}=\frac{cos^{2} \left(\phi_{1} / 2\right)}{cos^{2} \left(\phi_{2} / 2\right)}$
$\therefore \frac{K}{I_{2}}=\frac{cos^{2} \left( 2 \pi / 2 \right)}{cos^{2} \left(\frac{\pi / 2}{2}\right)}=\frac{1}{1 / 2}$
$\Rightarrow I_{2}=\frac{K}{2}=0.5K=\frac{1}{2}K$