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In Young's double slit experiment, the s...

In Young's double slit experiment, the slits are `0.5mm` apart and interference pattern is observed on a screen placed at a distance of `1.0m` from the plane containg the slits. If wavelength of the incident light is `6000Å`, then the separation between the third bright fringe and the central maxima is

A

(a) `4.0mm`

B

(b) `3.5mm`

C

(c) `3.0mm`

D

(d) `2.5mm`

Text Solution

Verified by Experts

The correct Answer is:
B
Separation of `n^(th)` bright fringe and central maxima is `x_n=(nlambdaD)/(d)`
So, `x_3=(3xx6000xx10^(-10)xx1)/(0.5xx10^-3)=3.5mm`
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