Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths 500 nm and 600 nm. Bright fringes due to both wavelengths coincide at 2.5 mm from the central maximum. If the separation between the slits is 3 mm, then the distance between the screen and plane of the slits is 

Options

A. 1.2 m
B. 2.8 m
C. 2.5 m
D. 3.2 m

Answer

C

Explanation

Position of bright fringe on screen is given by
`x=(n lambda D)/(d)[x=2.5" mm"]`
For coinciding case, `n_(1)lambda_(1)=n_(2)lambda_(2)`
`:.` n = 6 for wavelength of 500 nm and n = 5 for wavelength of 600 nm. Using any one of those,
`25xx10^(-3)=(6xx500xx10^(-9)xxD)/(3xx10^(-3))`
`implies" D=2.5" m"`