Question

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

• 6533 Å

• 6531 Å

• 6535 Å

• 6351 Å

Answer 1

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 6531 Å.

Explanation:

In a biprism experiment, wavelength of the light is given as

`lambda = ("d"beta)/"D"`    ...(i)

where, β is the fringe width.

d is 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 = 0.84 × 10^-3 m and

`beta = (2.799 xx 10^-2)/30 = 9.33 xx 10^-4`

Substituting these values in Eq. (i), we get

`lambda = (0.84 xx 10^-3 xx 9.33 xx 10^-4)/1.2`

= 6.531 × 10^-7 m

= 6531 Å