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 to obtain interference fringes on the screen. The least distance from the commom central maximum to the point where the bright fringes fue 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

For common maxima, `n_(1)lamda_(1)=n_(2)lamda_(2)`
`rArr" "n_(1)/n_(2)=lamda_(2)/lamda_(1)=(520xx10^(-9))/(650xx10^(-9))=4/5`
For `lamda_(1)`
`y=(n_(1)lamda_(1)D)/d,lamda_(1)=650nm`
`y=(4xx650xx10^(-9)xx1.5)/(0.5xx10^(-3))" or, "y=7.8mn`