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In YDSE, the slits are seperated by 0.28...

In YDSE, the slits are seperated by `0.28` mm and the screen is placed 1.4 m away. The distance between the first dark fringe and fourth bright fringe is obtained to be `0.6` cm Determine the wavelength of the light used in the experiment.

Text Solution

Verified by Experts

Let us first identify the parameters given in the question
d=0.28 mm,
D=1.4 m
Also, separation between first dark fringe and fourth bright fringe is 0.6 cm.
Now position of `1^(st)` dark fringe central maximum be
`y_(1)=(lambdaD)/(2d)`
Position of `4^(th)` bright fringe will be
`y_(2)=(4lambdaD)/d`
Separation between the is equal to
`(4lambdaD)/d-(lambdaD)/(2d)=0.6 cm`
`rArr(7lambdaD)/(2d)=0.6 cm`
`rArr lambda=(0.6xx10^(-2)xx2xx0.28xx10^(-3))/(7xx1.4)m`
`=3.428xx10^(-07)m`
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