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The intensity at the maximum in a Young'...

The intensity at the maximum in a Young's double slit experiment is `I_(0)`. Separation between two slits is d = `5lambda`, 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?

A

`I_(0)/(4)`

B

`(3)/(4) I_(0)`

C

`I_(0)/(2)`

D

`I_(0)`

Text Solution

Verified by Experts

Path difference = `sqrt((50lambda)^(2)+(5lambda)^(2))-50lambda=(lambda)/(4)`
Path difference,
`phi=(2pi)/(lambda)xx` path difference = `(2pi)/(lambda)xx(lambda)/(4)=(pi)/(2)`
Now I =`I_(0) "cos"^(2)(phi)/(2)=I_(0)cos^(2)"(pi)/(4)=(I_(0))/(2)`
The option C is correct.
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