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The intensity at the maximum in a Young's double slit experiment is I₀. Distance between two slits of d = 5 $\lambda$ , where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d?
Option: 1
I₀

Option: 2
$\frac{\text{I}_{0}}{4}$

Option: 3
$\frac{3}{4}\text{I}_{0}$

Option: 4
$\frac{\text{I}_{0}}{2}$

Answers (1)

best_answer

$d=5\lambda$

$D=10d=50\lambda$

$y=\left ( \Delta x \right ).\frac{D}{d}$

For position in front os slit

$y=\frac{d}{2}=\left ( \Delta x \right )\frac{D}{d}$

6r $\Delta x=\frac{d^{2}}{2D}=\frac{25\lambda ^{2}}{2.\left ( 50\lambda \right )}$

$\Delta x=\frac{\lambda }{4}$

path diff = $\frac{\lambda }{4}$

= phase difference = $\left ( \frac{2\pi }{\lambda } \right ).\frac{\lambda }{4}=\frac{\pi }{2}$

If Intensity of any one slit is I

Then $I_{R}=I+I+2\sqrt{I.I}.\cos \frac{\pi }{2}$

$I_{R}=I_{0}=\left ( \sqrt{I} \right+\sqrt{I} )^{2}-4I$

$\therefore I_{R}=\frac{I_{0}}{2}$

Posted by

Divya Prakash Singh

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