This interference film is used to measure the thickness of slides, paper, etc. The arrangement is as shown in fig. For the sake of clarity, the two strips are shown thick. Consider the wedge formed in between strips 1 and 2. If the interference pattern because of the two waves reflected from wedge surface is observed, then from the observed data we can compute thickness of paper, refractive index of the medium filled in wedge, number of bonds formed, etc.
Considre the strips to be thick as compared to wavelength of ligth and light is incident normally.
Neglect the effect due to reflection from top surface of strip 1 and bottom surface of strip 2. Take `L = 5 cm` and `lambda_(air) = 40 nm`.

Consider an air wedge formed by two glass plates. having refractive index 1.5 by placing a piece of paper of thickness 20 mm. Determine the number of dark bands formed.

Options

A. 1000
B. 500
C. 5000
D. 500

Answer

b

Explanation

For constructive interference,`2 t = (2 n - 1) (lambda_("air"))/(2)`
For destructive interference, `2 t = n lambda_("air")` As due to reflection at the top surface of bottom layer, an additional path difference of `lambda//2` occurs, contact point would be dark. Let at distance x, nth dark band forms.
`:. 2 t = n lambda_("air")`
`(t)/(x) = (h)/(L)`
`(2 h x)/(L) = n lambda_("air")`
`implies x = n [(lambda_("air") xx L)/(2 h)]`
`implies x = n [(400 xx 10^(-9) xx 5 xx 10^(-2))/(2 xx 20 xx 10^(-6))]`
`= (5 xx 10^(-4) n)`
Maximum value of n would be for `x = L`.