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.

Explanation

Let us first identify the parameters given in the question `d=0.28 mm, D=1.4 mm` Also, separation between first dark fringe and fourth bright fringe is `0.6` cm Now, position of 1st dark fringe from central maximum be `y_(1)=(lambdaD)/(2d)`

Position of 4th bright fringe will be
`y_(2)=(4lambdaD)/d`
Separation between them is equal to `(4lambdaD)/d-(lambdaD)/(2d)=0.6 cm`
`rArr(7lambdaD)/(2d)=0.6 cm`
Wavelength of the light,
`rArr lambda=(0.6xx10^(-2)xx2xx0.28xx10^(-3))/(7xx1.4)m`
`=3.428xx10^(-7)m`