In Young.s double slit experiment, slits are separated by 2 mm and the screen is placed at a distance of 1.2 m from the slits. Light consisting of two wavelengths 6500Å and 5200Å are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

Options

A. 0.312 mm
B. 0.123 mm
C. 0.213 mm
D. 0.412 mm

Answer

A

Explanation

`x=(nlamdaD)/d`
`x_(1)-x_(2)=(4D)/d(lamda_(1)-lamda_(2))`
= `(4xx1.2)/(2xx10^(-3))(6500-5200)xx10^(-10)`
= `0.312xx10^(-3)m`
= 0.312mm