Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

A square piece of thin film with RI = 1.4 has a wedge shaped section so that its thickness at two opposite sides are $t_1$&$ t_2$. If with$\lambda = 5800 \mathring{A}$, the number of fringes observed is 10, find$t_2$ – $t_1$.

μ =1.4, λ = 5800 Å, number of fringe is 10

Condition for darkness is 2μtcos(r+θ) =nλ

For normal incidence and small wedge angle cos(r+θ)=1

Let n$^{th}$ fringe is obtained at thickness t$_1$.

2μt$_1$ = nλ ………………………………..(1)

Thickness t$_2$ is where 10$^{th}$ fringe is observed

Therefore,

2μt$_2$ = (n+10) λ………………………………..(2)

$ t_2 - t_1 = \frac{10 λ}{2μ} = 2.07 \times 10^{-4} \ cm $