Question 234: Waves Optics

Question

In Young's double slit experiment, the intensity on screen at a point, where path difference is $\frac{\lambda}{4}$ is $\frac{K}{4}$. The intensity at a point when path difference is ' $\lambda$ ' will be $\left[\cos \frac{\pi}{2}=0, \cos 2 \pi=1\right]$

Accepted Answer

The intensity is given by
$$ I=4 I_0 \cos ^2\left(\frac{\Delta \phi}{2}ight) $$
When path difference is $\frac{\lambda}{4}$, i.e., $\phi=\frac{\pi}{2}$, intensity will be
$$ \frac{K}{4}=4 I_0 \cos ^2\left(\frac{\pi}{4}ight)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $$
When path difference is $\lambda$,i.e., $\phi=2 \pi$, intensity will be
$$ 	herefore \quad \mathrm{I}=4 \mathrm{I}_0 \cos ^2(\pi) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
$	herefore \quad$ From (i) and (ii), we get
$$ \begin{aligned} & \frac{K}{4 I}=\frac{4 I_0 \cos ^2\left(\frac{\pi}{4}ight)}{4 I_0 \cos ^2(\pi)}=\frac{\cos ^2\left(\frac{\pi}{4}ight)}{\cos ^2(\pi)}=\frac{1}{2} \ & I=\frac{K}{2} \end{aligned} $$

Answer

4 K

Answer

2 K

Answer

K

Answer

$\frac{\mathrm{K}}{2}$

Waves Optics

In Young's double slit experiment, the intensity on screen at a point, where path difference is \frac{\lambda}{4} is \frac{K}{4}. The intensity at a point when path difference is ' \lambda ' will be \left[\cos \frac{\pi}{2}=0, \cos 2 \pi=1\right]

Choose the correct answer: