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

We know, `y = (n_1 lambda_1 D)/(d) = (n_2 lambda_2 D)/(d)`
`rArr n_(1) lambda_(1) = n_(2) lambda_(2) `
`rArr n_(1) ( 500 nm ) = n_2 ( 600 nm ) `
`rArr 5n_1 = 6 n_2`
`therefore n_1 = 6, n_2 =5 ( therefore n_1 ne o , n_2 ne 0)`
Consider
`y = (n_1 lambda_1 D)/(d)`
`rArr 2.5 mm = (6xx 500 nm)/(3mm) xx D`
`rArr D = (2.5 xx 3 xx 10^(-6))/(6xx 500 xx 10%^(-9))`
`therefore D = 2 . 5 m`