In biprism experiment ,the distance between source and eyepiece is 1.2 m,the distance between two virtual sources is 0.84 mm. Then the wavelength of light used if eyepiece is to be moved transversely through a distance of 2. 799 cm to shift 30 fringes is

Options

A. 6535 Å
B. 6527 Å
C. 6535 Å
D. 6351 Å

Explanation

In a biprism experiment, wavelength of the light is given as
`lamda=(dbeta)/D` …(i)
where, `beta` is the fringe width,
dis the distance between the two sources and D is the distance between the source and eyepeice.
Given, D =1.2 m.
d = 0.84 mm = `0.84xx10^(-3)` m and
`beta=(2.799xx10^(-2))/30=9.33xx10^(-4)`
Substituting these values in Eq. (i), we get
`lamda=(0.84xx10^(-3)xx9.33xx10^(-4))/(1.2)`
`=6.531xx10^(-7)` m =6531 A