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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 sin θ , 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 2 d sin θ . The condition for constructive interference is the same as for the double slit, that is

d sin θ = m λ .

When this condition is met, 2 d sin θ is automatically a multiple of λ , 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 λ / 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.

Picture shows interference with three slits separated by distance d. Rays 1, 2, and 3 travel through the slits at the angles Theta.
Interference with three slits. Different pairs of emerging rays can combine constructively or destructively at the same time, leading to secondary 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 ( N 2 ) evenly spaced positions between the principal maxima. The amplitude of the electromagnetic wave is correspondingly diminished to 1 / 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 / 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.

Picture A shows a graph for the interference fringe patterns for two, three and four slits. As the number of slits increases, more secondary maxima appear, but the principal maxima become narrower. Picture B shows photographs of fringe patterns for two, three and four slits. As the number of slits increases, more secondary maxima appear, but the principal maxima become brighter.
Interference fringe patterns for two, three and four slits. As the number of slits increases, more secondary maxima appear, but the principal maxima become brighter and narrower. (a) Graph and (b) photographs of fringe patterns.

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 ° , 0.53 ° ; b. 4.6 × 10 −3 m

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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.

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

1:9

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What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by d = 2.0 × 10 −3 mm ? (b) 50 slits with the same separation? Assume that λ = 600 nm .

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Questions & Answers

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
Jyoti Reply
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
Rupal Reply
how does colour appear in thin films
Nwjwr Reply
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
Kimani Reply
derivation of lateral shieft
James Reply
hi
Imran
total binding energy of ionic crystal at equilibrium is
All Reply
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Bishesh Reply
Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
Sushant
What is motion
Izevbogie Reply
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
how can u say that
Chemist
do u know about black hole
Chemist
Not so more
Sushant
Radioactive substance
DHEERAJ
These substance create harmful radiation like alpha particle radiation, beta particle radiation, gamma particle radiation
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Celine Reply
 Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
Evan Reply
as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
acceleration also increases
Jay
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Michael
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
Big
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
Rafi
40 cm into change mm
Arhaan Reply
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?
sisay Reply
Practice Key Terms 2

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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