# 3.3 Multiple-slit interference

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By the end of this section, you will be able to:
• Describe the locations and intensities of secondary maxima for multiple-slit interference

Analyzing the interference of light passing through two slits lays out the theoretical framework of interference and gives us a historical insight into Thomas Young’s experiments. However, much of the modern-day application of slit interference uses not just two slits but many, approaching infinity for practical purposes. The key optical element is called a diffraction grating, an important tool in optical analysis, which we discuss in detail in Diffraction . Here, we start the analysis of multiple-slit interference by taking the results from our analysis of the double slit ( $N=2$ ) and extending it to configurations with three, four, and much larger numbers of slits.

[link] shows the simplest case of multiple-slit interference, with three slits, or $N=3$ . The spacing between slits is d , and the path length difference between adjacent slits is $d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ , same as the case for the double slit. What is new is that the path length difference for the first and the third slits is $2d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ . The condition for constructive interference is the same as for the double slit, that is

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

When this condition is met, $2d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ is automatically a multiple of $\lambda$ , so all three rays combine constructively, and the bright fringes that occur here are called principal maxima . But what happens when the path length difference between adjacent slits is only $\lambda \text{/}2$ ? We can think of the first and second rays as interfering destructively, but the third ray remains unaltered. Instead of obtaining a dark fringe, or a minimum, as we did for the double slit, we see a secondary maximum    with intensity lower than the principal maxima.

In general, for N slits, these secondary maxima occur whenever an unpaired ray is present that does not go away due to destructive interference. This occurs at $\left(N-2\right)$ evenly spaced positions between the principal maxima. The amplitude of the electromagnetic wave is correspondingly diminished to $1\text{/}N$ of the wave at the principal maxima, and the light intensity, being proportional to the square of the wave amplitude, is diminished to $1\text{/}{N}^{2}$ of the intensity compared to the principal maxima. As [link] shows, a dark fringe is located between every maximum (principal or secondary). As N grows larger and the number of bright and dark fringes increase, the widths of the maxima become narrower due to the closely located neighboring dark fringes. Because the total amount of light energy remains unaltered, narrower maxima require that each maximum reaches a correspondingly higher intensity.

## Summary

• Interference from multiple slits ( $N>2$ ) produces principal as well as secondary maxima.
• As the number of slits is increased, the intensity of the principal maxima increases and the width decreases.

## Problems

Ten narrow slits are equally spaced 0.25 mm apart and illuminated with yellow light of wavelength 580 nm. (a) What are the angular positions of the third and fourth principal maxima? (b) What is the separation of these maxima on a screen 2.0 m from the slits?

a. $0.40\text{°},0.53\text{°};$ b. $4.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{m}$

The width of bright fringes can be calculated as the separation between the two adjacent dark fringes on either side. Find the angular widths of the third- and fourth-order bright fringes from the preceding problem.

For a three-slit interference pattern, find the ratio of the peak intensities of a secondary maximum to a principal maximum.

1:9

What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by $d=2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\text{mm}$ ? (b) 50 slits with the same separation? Assume that $\lambda =600\phantom{\rule{0.2em}{0ex}}\text{nm}$ .

#### Questions & Answers

as a free falling object increases speed what is happening to the acceleration
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
explain l-s coupling
how can we say dirac equation is also called a relativistic equation in one word
what is the electronic configration of Al
what's the signeficance of dirac equetion.?
what is the effect of heat on refractive index
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Swati
yes
Chemist
1 millimeter is How many metres
1millimeter =0.001metre
Gitanjali
The photoelectric effect is the emission of electrons when light shines on a material.