In a Young's double slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths 650 nm and 520 nm is used tc obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is.....

Options

A. `9.75 mm`
B. `15.6` mm
C. `1.56` mm
D. `7.8` mm

Answer

D

Explanation

`n_(1)barX_(1)=n_(2)barX_(2)`
`(n_(1)lambda_(1)D)/(d)=(n_(2)lambda_(2)D)/(d)`
`:. n_(1) lambda_(1)=n_(2)lambda_(2)`
`:.(n_(1))/(n_(2))=(lambda_(2))/(lambda_(1))=(520)/(650)=(4)/(5)`
`:.n_(1)=4 and n_(2)=5`
Now `x=(n_(1)lambda_(1)D)/(d) or x=(n_(2) lambda_(2)D)/(d)`
`=(4xx65xx10^(-6)xx150)/(6.65)`
`=780000xx10^(-6) cm`
`=7.8 mm`