# 4.2 Intensity in single-slit diffraction  (Page 2/3)

 Page 2 / 3

These two maxima actually correspond to values of $\varphi$ slightly less than $3\pi$ rad and $5\pi$ rad. Since the total length of the arc of the phasor diagram is always $N\text{Δ}{E}_{0},$ the radius of the arc decreases as $\varphi$ increases. As a result, ${E}_{1}$ and ${E}_{2}$ turn out to be slightly larger for arcs that have not quite curled through $3\pi$ rad and $5\pi$ rad, respectively. The exact values of $\varphi$ for the maxima are investigated in [link] . In solving that problem, you will find that they are less than, but very close to, $\varphi =3\pi ,5\pi ,7\pi ,\text{…}\phantom{\rule{0.2em}{0ex}}\text{rad}.$

To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of [link] . Since the arc subtends an angle $\varphi$ at the center of the circle,

$N\text{Δ}{E}_{0}=r\varphi$

and

$\text{sin}\phantom{\rule{0.2em}{0ex}}\left(\frac{\varphi }{2}\right)=\frac{E}{2r}.$

where E is the amplitude of the resultant field. Solving the second equation for E and then substituting r from the first equation, we find

$E=2r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\varphi }{2}=2\frac{N\text{Δ}{E}_{o}}{\varphi }\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\varphi }{2}.$

Now defining

$\beta =\frac{\varphi }{2}=\frac{\pi D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{\lambda }$

we obtain

$E=N\text{Δ}{E}_{0}\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\beta }{\beta }$

This equation relates the amplitude of the resultant field at any point in the diffraction pattern to the amplitude $N\text{Δ}{E}_{0}$ at the central maximum. The intensity is proportional to the square of the amplitude, so

$I={I}_{0}{\left(\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\beta }{\beta }\right)}^{2}$

where ${I}_{0}={\left(N\text{Δ}{E}_{0}\right)}^{2}\text{/}2{\mu }_{0}c$ is the intensity at the center of the pattern.

For the central maximum, $\varphi =0$ , $\beta$ is also zero and we see from l’Hôpital’s rule that ${\text{lim}}_{\beta \to 0}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\beta \text{/}\beta \right)=1,$ so that ${\text{lim}}_{\varphi \to 0}I={I}_{0}.$ For the next maximum, $\varphi =3\pi$ rad, we have $\beta =3\pi \text{/}2$ rad and when substituted into [link] , it yields

${I}_{1}={I}_{0}{\left(\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}3\pi \text{/}2}{3\pi \text{/}2}\right)}^{2}=0.045{I}_{0},$

in agreement with what we found earlier in this section using the diameters and circumferences of phasor diagrams. Substituting $\varphi =5\pi$ rad into [link] yields a similar result for ${I}_{2}$ .

A plot of [link] is shown in [link] and directly below it is a photograph of an actual diffraction pattern. Notice that the central peak is much brighter than the others, and that the zeros of the pattern are located at those points where $\text{sin}\phantom{\rule{0.2em}{0ex}}\beta =0,$ which occurs when $\beta =m\pi$ rad. This corresponds to

$\frac{\pi D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{\lambda }=m\pi ,$

or

$D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda ,$

which is [link] .

## Intensity in single-slit diffraction

Light of wavelength 550 nm passes through a slit of width $2.00\phantom{\rule{0.2em}{0ex}}\mu \text{m}$ and produces a diffraction pattern similar to that shown in [link] . (a) Find the locations of the first two minima in terms of the angle from the central maximum and (b) determine the intensity relative to the central maximum at a point halfway between these two minima.

## Strategy

The minima are given by [link] , $D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda$ . The first two minima are for $m=1$ and $m=2.$ [link] and [link] can be used to determine the intensity once the angle has been worked out.

## Solution

1. Solving [link] for $\theta$ gives us ${\theta }_{m}={\text{sin}}^{-1}\left(m\lambda \text{/}D\right),$ so that
${\theta }_{1}={\text{sin}}^{-1}\left(\frac{\left(+1\right)\left(550\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}{2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}}\right)=+16.0\text{°}$

and
${\theta }_{2}={\text{sin}}^{-1}\left(\frac{\left(+2\right)\left(550\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}{2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}}\right)=+33.4\text{°}.$
2. The halfway point between ${\theta }_{1}$ and ${\theta }_{2}$ is
$\theta =\left({\theta }_{1}+{\theta }_{2}\right)\text{/}2=\left(16.0\text{°}+33.4\text{°}\right)\text{/}2=24.7\text{°}.$

[link] gives

$\beta =\frac{\pi D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{\lambda }=\frac{\pi \left(2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\left(24.7\text{°}\right)}{\left(550\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}=1.52\pi \phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}4.77\phantom{\rule{0.2em}{0ex}}\text{rad}.$

From [link] , we can calculate

$\frac{I}{{I}_{o}}={\left(\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\beta }{\beta }\right)}^{2}={\left(\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(4.77\right)}{4.77}\right)}^{2}={\left(\frac{-0.9985}{4.77}\right)}^{2}=0.044.$

## Significance

This position, halfway between two minima, is very close to the location of the maximum, expected near $\beta =3\pi \text{/}2,\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}1.5\pi$ .

Check Your Understanding For the experiment in [link] , at what angle from the center is the third maximum and what is its intensity relative to the central maximum?

$74.3\text{°}$ , $0.0083{I}_{0}$

If the slit width D is varied, the intensity distribution changes, as illustrated in [link] . The central peak is distributed over the region from $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =\text{−}\lambda \text{/}D$ to $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =+\lambda \text{/}D$ . For small $\theta$ , this corresponds to an angular width $\text{Δ}\theta \approx 2\lambda \text{/}D.$ Hence, an increase in the slit width results in a decrease in the width of the central peak    . For a slit with $D\gg \lambda ,$ the central peak is very sharp, whereas if $D\approx \lambda$ , it becomes quite broad.

A diffraction experiment in optics can require a lot of preparation but this simulation by Andrew Duffy offers not only a quick set up but also the ability to change the slit width instantly. Run the simulation and select “Single slit.” You can adjust the slit width and see the effect on the diffraction pattern on a screen and as a graph.

## Summary

• The intensity pattern for diffraction due to a single slit can be calculated using phasors as
$I={I}_{0}{\left(\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\beta }{\beta }\right)}^{2},$

where $\beta =\frac{\varphi }{2}=\frac{\pi D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{\lambda }$ , D is the slit width, $\lambda$ is the wavelength, and $\theta$ is the angle from the central peak.

## Conceptual questions

In [link] , the parameter $\beta$ looks like an angle but is not an angle that you can measure with a protractor in the physical world. Explain what $\beta$ represents.

The parameter $\beta =\varphi \text{/}2$ is the arc angle shown in the phasor diagram in [link] . The phase difference between the first and last Huygens wavelet across the single slit is $2\beta$ and is related to the curvature of the arc that forms the resultant phasor that determines the light intensity.

## Problems

A single slit of width $3.0\phantom{\rule{0.2em}{0ex}}\mu \text{m}$ is illuminated by a sodium yellow light of wavelength 589 nm. Find the intensity at a $15\text{°}$ angle to the axis in terms of the intensity of the central maximum.

A single slit of width 0.1 mm is illuminated by a mercury light of wavelength 576 nm. Find the intensity at a $10\text{°}$ angle to the axis in terms of the intensity of the central maximum.

$I\text{/}{I}_{0}=2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

The width of the central peak in a single-slit diffraction pattern is 5.0 mm. The wavelength of the light is 600 nm, and the screen is 2.0 m from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at 4.5 mm from the center of the pattern to the intensity at the center.

Consider the single-slit diffraction pattern for $\text{λ}=600\phantom{\rule{0.2em}{0ex}}\text{nm}$ , $D=0.025\phantom{\rule{0.2em}{0ex}}\text{mm}$ , and $x=2.0\phantom{\rule{0.2em}{0ex}}\text{m}$ . Find the intensity in terms of ${I}_{o}$ at $\theta =0.5\text{°}$ , $1.0\text{°}$ , $1.5\text{°}$ , $3.0\text{°}$ , and $10.0\text{°}$ .

$0.63{I}_{0},\phantom{\rule{0.2em}{0ex}}0.11{I}_{0},\phantom{\rule{0.2em}{0ex}}0.0067{I}_{0},\phantom{\rule{0.2em}{0ex}}0.0062{I}_{0},\phantom{\rule{0.2em}{0ex}}0.00088{I}_{0}$

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