In a Young's double slit experiment on interference distance between two vertical slits was 0.5 mm and distance of the screen from the plane of slits was 100 cm . If was observed that the 4 th bright band was 2.945 mm away from the second dark band. Find the wavelength of light used.

Explanation

Form Fig. 6.18 it is observed that the distance between second dark band to 4th bright band `=2.5 xx` band width. If band width of fringe width is y then ,
`2.5 xx y =2.945`
`:. Y (2.945)/(2.5) = 1.178` mm
=0.1178
As , `y=(D lambda)/(2d)`
So `lambda = (2d.y)/(D) = (0.05 xx 0.1178 )/(100)`
`=5890 xx 10^(-8)` cm =5890 Å
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