Light reflected from a soap film of RI 1.33 shows two consecutive overlapping dark bands corresponding to $\lambda s$ 6100 & 6000$ \mathring{A}$. Calculate the film thickness if the incidence angle is $Sin^{-1}(4/5)$.

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

μ =1.33,

$ μ= \frac{sin \ i}{sin \ r} \\[2ex] sin \ r = \frac{sin \ i}{μ} $

sin r = 0.60

Therefore r=36.97

cosr = 0.8

Consecutive means if one dark band is ‘n’ then other will be ‘n+1’corresponding to wavelengths $λ_1$ and $λ_2$ respectively.

For dark bands 2μ t cos r = nλ

Therefore,

2μ t cos r = nλ$_1$………………………………………(1)

2μ t cos r = (n+1)λ$_2$ …………………………………(2)

From equation 1 and 2

nλ$_1$ = (n+1)λ$_2$

nλ$_1$ = nλ$_2$ + λ$_2$

n( λ$_1$ - λ$_2$ ) = λ$_2$

Therefore, n= $\frac{λ_2}{λ_1 - λ_2}$

n= $\frac{6000}{6100-6000}$

n =60

substituting value of n in equation 1

2x1.33x t x0.8 = 60x6100x10$^{-5}$

Thickness t = 0.0017 cm