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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

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Let us firstly, the parameters given in the question, `d=0.28 mm,D=1.4m`
Also, separation between first dark fringe and fourth bringe fringe is 0.6 cm.
Now position of 1st dark fringe from central maximum be `y_1=(lamda D)/(2d)`
Position of 4th bright fringe will be `y_2=(4 lamda D)/d`
Separation between them is equal to `(4lamda D)/d-(lamda D)/(2d)=0.6 cm`
`implies (T lamda D)/(2d)=0.6 cm`
`implies lamda =(0.6 times 10^-2 times 2 times 0.28 times 10^-3)/(7 times 1.4)m=3.428 times 10^-7 m`
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