In a YDSE experiment,if a slab whose refracti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a $YDSE$ experiment,the intensity at any point on the screen is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference

Vedclass · Accepted Answer

Vedclass · Answer

(C) In a $YDSE$ experiment,the intensity at any point on the screen is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference between the two waves.At the mid-point of the screen,the path difference due to the geometry is zero.When a slab of thickness $t$ and refractive index $\mu$ is placed in front of one slit,an additional path difference $\Delta x = (\mu - 1)t$ is introduced.The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} (\mu - 1)t$.The intensity at the mid-point is $I = 4I_0 \cos^2(\frac{\pi t}{\lambda}(\mu - 1))$.As $\mu$ increases from $1$,the term $(\mu - 1)$ increases,causing the phase difference $\phi$ to increase.The intensity $I$ follows a $\cos^2$ variation,starting from a maximum value $(4I_0)$ at $\mu = 1$ (where $\phi = 0$) and decreasing as $\mu$ increases,following the shape of a squared cosine function.

Similar Questions

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{57}{2}\right) \lambda$. The band obtained at that point is

In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

In Young's double slit interference experiment,the slit separation is made $3$ folds. The fringe width becomes ......... times.

If the Young's double slit experiment is performed with white light,then which of the following is not true?

If yellow light in the Young's double slit experiment is replaced by red light,the fringe width will

Vedclass Test Series

Exam Paper Generator

Online Exam Module