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?

`I_(0)/(4)`

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

`I_(0)/(2)`

`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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Maximum intensity in Young's double slit experiment is I_(0) . If one slit is closed the intensity would be

`I_(0)`

`I_(0)/(4)`

`I_(0)/(3)`

`I_(0)/(2)`

In Young's double slit experiment the distance between the two slits is d and wavelength of the light used is lambda . The angular width of the fringe

`(d)/(lambda)`

`(lambda)/(d)`

`(2lambda)/(d)`

`(lambda)/(2d)`

If young' s double slit experiment is performed using white light, then

the central fringe will be white

no fringe will be completely dark

the fringe adjacent to the central one will be red

the fringe adjacent to the central one will be violet