1.

The least distance from the central maxima where the fringes due to both wavelength coincide is

Answer»

`3.02 mm`
`2.12 mm`
`1.56 mm`
`1.22 mm`

Solution :The bright fringes of both wavelengths will coincide, when NTH bright fringe of ONE COINCIDES with (n + 1) th bright fringe due to other wavelength.
Now fringe width `beta = (n lambda D)/(d)`
`therefore (n lambda_(1)D)/(d) = (n + 1)lambda_(2).D/d`
then `nlambda_(1) = (n + 1) lambda_(2)`
` nxx 6500 XX 10^(-8) = (n + 1) 5200 xx 10^(-8)`
Then `n = 4` for `lambda_(1)`
`x_(4) = (4lambdaD)/(d) = 1.56 mm`


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