Question

A broad source of light of wavelength 680nm illuminates normally two glass plates 120mm long that meet at one end and are separated by a wire 0.048 mm in diameter at the other end. Find the number of bright fringes formed over the 120 mm distance.

Answer 1

141

Answer 2

The interference pattern is formed due to the wedge-shaped air film between the glass plates. The thickness of the air film varies linearly. Due to reflection from the top and bottom surfaces of the air film, interference occurs. The condition for a bright fringe is

$2t = (n - \frac{1}{2})\lambda$,

where $t$ is the thickness of the air film and $n = 1, 2, 3, \dots$. The thickness ranges from 0 to $d$. We need to find the number of integers $n \ge 1$ such that the corresponding thickness $t$ is less than or equal to $d$. Substituting the condition for a bright fringe, we get

$(n - \frac{1}{2})\frac{\lambda}{2} \le d$.

Solving for $n$, we get

$n \le \frac{2d}{\lambda} + \frac{1}{2}$.

Given $\lambda = 680 \text{ nm}$ and $d = 0.048 \text{ mm}$, we calculate

$\frac{2d}{\lambda} = \frac{2 \times 0.048 \times 10^{-3}}{680 \times 10^{-9}} \approx 141.176$.

So, $n \le 141.176 + 0.5 = 141.676$. Since $n$ is an integer and $n \ge 1$, the possible values of $n$ are $1, 2, \dots, 141$. The number of bright fringes is 141.