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

Answer: 1

Phase difference, $\varphi =\frac{2\pi }{\lambda }(\Delta x)$
For path difference $\lambda ,$ phase difference ${\varphi }_1=2\pi$
For path difference $\lambda /4,$ phase difference ${\varphi }_2=\pi /2$
Using, $I=4I_{0}co{s}^2\frac{\varphi }{2}$
$\therefore \frac{I_1}{I_2}=\frac{co{s}^2\left({\varphi }_1/2\right)}{co{s}^2\left({\varphi }_2/2\right)}$
$\therefore \frac{K}{I_2}=\frac{co{s}^2\left(2\pi /2\right)}{co{s}^2\left(\frac{\pi /2}{2}\right)}=\frac{1}{1/2}$
$\Rightarrow I_2=\frac{K}{2}=0.5K=\frac{1}{2}K$