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A thin glass of thickness $\frac{2500}{3}\lambda$ ( $\lambda$ is wavelength of light used) and refractive index $\mu =1.5$ is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :
Option: 1
$2:1$

Option: 2
$1:4$

Option: 3
$4:1$

Option: 4
$4:3$

Answers (1)

best_answer

Due to the introduction of sheet in front of one slit (thickness t and refrective index $(\mu)$ the shift $=\frac{D}{d}(\mu-1) t$$
i.e., the path difference becomes $(\mu-1) t$ instead of zero at centre of screen i.e at centre $(\Delta x \neq 0)$

therefore Phase difference $=\frac{2 \pi}{\lambda} \times(\mu-1) t$$

So using $t= \frac{2500}{3}\lambda$ we get $\Delta \phi =\frac{2\pi }{\lambda }* \frac{2500}{3}\lambda*0.5= \frac{1250}{3}*2\pi$ at centre.

As Intensity at centre will same as that of point having the Phase difference as $\frac{2\pi }{3}$

i.e 120⁰

So Let new point where Phase difference is $0$ is A

So another point B which is equidistant from the slits as A will have phase difference

as $\Delta \phi =2*\frac{2\pi }{3}=\frac{4\pi }{3}$ or 240⁰

So intensity at any point is given as $I=I_ocos^2(\Delta \phi )$

where $I_o$ is the maximum intensity or Intensity of a point when phase diffence is 0.

So $I_A=I_0$

So $I_B=I_0Cos^2(\frac{4\pi }{}3)=\frac{I_0}{4}$

SO $\frac{I_A}{I_B}=4$

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