Question

If 490 nm light falls on a slit 0.0430 mm wide, what is the full angular width of the central diffraction peak?

Answer 1

Concepts and reason

The required concept to solve the problem is the single slit diffraction experiment.

First, from the condition for diffraction maxima, find the angle between the normal line between the screen and slit and the wave ray to a point on the screen.

Then, using the angle between the normal line between the screen, slit, and the wave ray to a point on the screen and the angular width, find the angular width of the central diffraction peak.

Fundamentals

A light can act both as a particle and as a wave. It can bend around the objects like a wave. This is the concept behind the single slit diffraction. Due to interference, the waves are added together constructively, and then, a maxima is formed. If the waves are added together destructively, then a minima is formed.

The equation for diffraction maxima is,

$d\sin \theta = m\lambda$

Here, $d$ is the width of the slit, $\theta$ is the angle between the normal line between the screenand slit, and the wave ray to a point on the screen, $m$ is the order, and $\lambda$ is the wavelength.

The angular width is,

$\Phi = 2\theta$

The condition for diffraction maxima is,

$d\sin \theta = m\lambda$

Rearrange the above equation in terms of $\theta$ .

$\begin{array}{c}\\\sin \theta = \frac{{m\lambda }}{d}\\\\\theta = {\sin ^{ - 1}}\left( {\frac{{m\lambda }}{d}} \right)\\\end{array}$

Substitute $490\,{\rm{nm}}$ for $\lambda$ , $1$ for $m$ and $0.0430\,{\rm{mm}}$ for $d$ to find $\theta$ .

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left[ {\frac{{\left( 1 \right)\left( {\left( {490\,{\rm{nm}}} \right)\left( {\frac{{{{10}^{ - 9}}\,{\rm{m}}}}{{1\,{\rm{nm}}}}} \right)} \right)}}{{\left( {0.0430\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,{\rm{mm}}}}} \right)}}} \right]\\\\ = {\sin ^{ - 1}}\left[ {\frac{{\left( {490 \times {{10}^{ - 9}}\,{\rm{m}}} \right)}}{{\left( {0.0430 \times {{10}^{ - 3}}\,{\rm{m}}} \right)}}} \right]\\\\ = {\sin ^{ - 1}}\left( {0.01139} \right)\\\\ = 0.653^\circ \\\end{array}$

The equation for the angular widthis,

$\Phi = 2\theta$

Substitute $0.653^\circ$ for $\theta$ .

$\begin{array}{c}\\\Phi = 2\left( {0.653^\circ } \right)\\\\ = 1.306^\circ \\\end{array}$

Ans:

Thus, the full angular width of the central diffraction peak is $1.306^\circ$ .