Question 247: Waves Optics

Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is ' $I$ '. The intensity at a point where the path difference is $\lambda / 6$ is $\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right] [\lambda=$ wavelength of light $][\cos \pi=-1]$

Accepted Answer

YoungÃ¢ÂÂs Double Slit Experiment (YDSE):
We know the
intensity at a point
on the screen in YDSE is given by:
$ I(\Delta \phi) = I_0 \left[ 1 + \cos (\Delta \phi) ight] $
where $I_0$ is the intensity due to each slit, and $\Delta \phi$ is the phase difference.
Equivalently, one can write:
$ I(\Delta \phi) = 4I_1 \cos^2\!\left(\frac{\Delta \phi}{2}ight) $
where $I_1$ = intensity from each slit.
Step 1: Relation between path difference and phase difference
Phase difference:
$ \Delta \phi = \frac{2\pi}{\lambda} \cdot (	ext{path difference}) $
Step 2: Given condition
At path difference = $\lambda$:
$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi $
So intensity at this point is maximum:
$ I_{\max} = 4I_1 $
They have denoted this maximum intensity value as $I$.
So, $I = 4I_1$.
Step 3: Required condition
At path difference = $\lambda/6$:
$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} $
Now intensity is:
$ I' = 4I_1 \cos^2\!\left(\frac{\Delta \phi}{2}ight) = 4I_1 \cos^2\!\left(\frac{\pi}{6}ight) $
$ = 4I_1 \left(\frac{\sqrt{3}}{2}ight)^2 = 4I_1 \cdot \frac{3}{4} = 3I_1. $
Step 4: Express in terms of given $I$
Since $I = 4I_1$,
$ I' = \frac{3}{4}I $
Ã¢ÂÂ Final Answer:
The intensity at path difference $\lambda/6$ is:
$ \boxed{	frac{3I}{4}} $
Correct Option: B

Answer

I

Answer

$\frac{3 \mathrm{I}}{4}$

Answer

$\frac{1}{2}$

Answer

$\frac{\mathrm{I}}{4}$

Waves Optics

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \lambda is ' I '. The intensity at a point where the path difference is \lambda / 6 is \left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right] [\lambda= wavelength of light ][\cos \pi=-1]

Choose the correct answer: