Question

Infrared light of wavelength $2.5\mu \mathrm{m}$illuminates a $0.20-mm$diameter hole. What is the angle of the first dark fringe in radians? In degrees?

Short answer

Angle of the first dark fringe in radians and in degree is $0.01525\mathrm{rad}\text{or}{0.87}^{\circ }$

Step-by-step solution

Introduction

Dark Fringe:

Destructive interference causes the dark fringes to appear.

If the resultant amplitude and hence the resultant intensity are both zero, interference is said to be destructive.

Find angle of dark fringe 

The angle at which the first dark fringe appears in the circular aperture experiment is given by

${\theta }_1=\frac{1.22\lambda }{D}$

As a result, in our scenario, we'll have

localid="1650215776407" ${\theta }_1=\frac{1.22\times 2.5\times {10}^{-6}m}{2\times {10}^{-4}m}$

$=0.01525\mathrm{rad}$

Dark fringe 

It's easy to convert to degrees if we remember that the complete angle $\left(360°\right)$equals $2\mathrm{\pi }$. As a result, our perspective will be

localid="1650213775027" ${\theta }_1=0.01525rad\times \frac{360}{2\pi }$

$=0.87°$