Question

Monochromatic light with lambda equals 630 space n m is incident on a single slit. The slit width is 0.8 mm. If the distance between the screen and slit is 1.8 m, what is the width of the central bright fringe on the screen

Answer 1

Answer:

The angular width is "$1.575\times 10^{-3} \ rad$" and the linear width is "$2.835\times 10^{-3} \ m$".

Explanation:

Given:

Wavelength,

$\lambda = 630 \ nm$

or,

  $=630\times 10^{-9} \ m$

Slit width,

$a = 0.8 \ mm$

or,

  $=0.8\times 10^{-3} \ m$

Distance between slit and screen,

$D = 1.8 \ m$

As we know,

The central bright fringe's angular height,

⇒ $\theta = \frac{2 \lambda}{a}$

       $= \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}$

       $=1.575\times 10^{-3} \ rad$

and,

The linear width will be:

⇒ $x_o=\frac{2D \lambda}{a}$

By substituting the values, we get

        $=\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}$

        $=2.835\times 10^{-3} \ m$