In a Young's experiment, two coherent sources are placed `0.90mm` apart and the fringes are observed one metre away. If is produces the second dark fringe at a distance of `1mm` from the central fringe, the wavelength of monochromatic light used would be

Options

A. `6xx10^(-4)cm`
B. `8xx10^(-4)cm`
C. `8xx10^(-5)cm`
D. `6xx10^(-5)cm`

Answer

D

Explanation

The distance of the second dark fringe from the centre
`=(3X)/(2) and X=(lamdaD)/(d)`
`therefore 1xx10^(-3)=(3lamdaD)/(2d)`
`therefore lamda=(2dxx10^(-3))/(3D)=(2xx0.9xx10^(-3)xx10^(-3))/(3)`
`=6xx10^(-7)m=6xx10^(-3)cm`