In Young's double-slit experiment, the separation between the slits is d, distance between the slit and screen is `D (D gt gt d)`, In the interference pattern, there is a maxima exactly in front of each slit. Then the possilbe wavelength(s) used in the experiment are

Options

A. `(d^(2))/D,(d^(2))/(2D),(d^(2))/(3D)`
B. `(d^(2))/D,(d^(2))/(3D),(d^(2))/(5D)`
C. `(d^(2))/(2D),(d^(2))/(4D),(d^(2))/(6D)`
D. none of these

Explanation

Fringe width `=n (lambdaD)/d`
From given situation
`F.W. =d/2`
`rArr n (lambdaD)/d=d/2 rArr lambda=(d^(2))/(2nD)`
Hence (C) is possible .