Question

Light from a laser with wavelength 400 nm passed through two small openings and produces an interference pattern on a screen 1 m away. If the first point of destructive interference occurs at an angle of 0.1 radians, what is the distance between the two openings?

Answer 1

To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,

$\text{Path difference}= dsin\theta = (2n-1)\frac{\lambda}{2}$

Here,

$\lambda$ = Wavelength

$\theta$ = Angle when occurs the interference point of destructive interference

Our values are given as,

$\text{Wavelength} = \lambda = 400nm = 4*10^{-7}m$

$\text{Distance of Screen} = D = 1m$

Using the previous expression we have,

$d \times \theta = \frac{\lambda}{2}$

$d \times (0.1) = \frac{4*10^{-7}}{2}$

$d = 2*10^{-6} m$

$d = 2\mu m$$d = 2\mu m$

Therefore the distance between the two openings is $2\mu m$