Question

In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index ‘μ’ is placed in one of the interfering paths. Now 7^th bright fringe is obtained at the same point. If ‘A’ is the wavelength of light used, the thickness of film is equal to ______.

विकल्प

• 1.5(μ − 1)λ

• `(1.5 lambda)/(mu - 1)`

• 2.5(μ − 1)λ

• `(2.5 lambda)/(mu - 1)`

Answer 1

In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index ‘μ’ is placed in one of the interfering paths. Now 7^th bright fringe is obtained at the same point. If ‘A’ is the wavelength of light used, the thickness of film is equal to `bbunderline((2.5 lambda)/(mu - 1))`.

Explanation:

For 5^th dark fringe,

`x_1^' = (2n - 1) lambda/2 D/d`

= `(9 lambda D)/(2 d)`

For 7^th bright fringe,

x₂ = `n lambda D/d`

= `(7 lambda D)/d`

∴ `x_2 - x_1^' = (mu - 1)t D/d`

∴ `(7 lambda D)/d - (9 lambda D)/d = (mu - 1)t D/d`

∴ t = `(2.5 lambda)/(mu - 1)`