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These two maxima actually correspond to values of ϕ slightly less than 3 π rad and 5 π rad. Since the total length of the arc of the phasor diagram is always N Δ E 0 , the radius of the arc decreases as ϕ increases. As a result, E 1 and E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π rad and 5 π rad, respectively. The exact values of ϕ for the maxima are investigated in [link] . In solving that problem, you will find that they are less than, but very close to, ϕ = 3 π , 5 π , 7 π , 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 ϕ at the center of the circle,

N Δ E 0 = r ϕ

and

sin ( ϕ 2 ) = E 2 r .

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 = 2 r sin ϕ 2 = 2 N Δ E o ϕ sin ϕ 2 .

Now defining

β = ϕ 2 = π D sin θ λ

we obtain

E = N Δ E 0 sin β β

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

I = I 0 ( sin β β ) 2

where I 0 = ( N Δ E 0 ) 2 / 2 μ 0 c is the intensity at the center of the pattern.

For the central maximum, ϕ = 0 , β is also zero and we see from l’Hôpital’s rule that lim β 0 ( sin β / β ) = 1 , so that lim ϕ 0 I = I 0 . For the next maximum, ϕ = 3 π rad, we have β = 3 π / 2 rad and when substituted into [link] , it yields

I 1 = I 0 ( sin 3 π / 2 3 π / 2 ) 2 = 0.045 I 0 ,

in agreement with what we found earlier in this section using the diameters and circumferences of phasor diagrams. Substituting ϕ = 5 π 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 sin β = 0 , which occurs when β = m π rad. This corresponds to

π D sin θ λ = m π ,

or

D sin θ = m λ ,

which is [link] .

Figure a shows a graph of I by I0 versus beta. There is a crest at the center of the graph at beta equal to 0. The y-value of this is 1. The graph has ripples on both sides of this which grow smaller as you go outwards. The graph has zeroes at minus 3 pi, minus 2 pi, minus pi, pi, 2 pi, 3 pi. Figure b shows a strip with alternating light and dark regions. The central portion is brightest.
(a) The calculated intensity distribution of a single-slit diffraction pattern. (b) The actual diffraction pattern.

Intensity in single-slit diffraction

Light of wavelength 550 nm passes through a slit of width 2.00 μ 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 sin θ = m λ . 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 θ gives us θ m = sin −1 ( m λ / D ) , so that
    θ 1 = sin −1 ( ( + 1 ) ( 550 × 10 −9 m ) 2.00 × 10 −6 m ) = + 16.0 °

    and
    θ 2 = sin −1 ( ( + 2 ) ( 550 × 10 −9 m ) 2.00 × 10 −6 m ) = + 33.4 ° .
  2. The halfway point between θ 1 and θ 2 is
    θ = ( θ 1 + θ 2 ) / 2 = ( 16.0 ° + 33.4 ° ) / 2 = 24.7 ° .

[link] gives

β = π D sin θ λ = π ( 2.00 × 10 −6 m ) sin ( 24.7 ° ) ( 550 × 10 −9 m ) = 1.52 π or 4.77 rad .

From [link] , we can calculate

I I o = ( sin β β ) 2 = ( sin ( 4.77 ) 4.77 ) 2 = ( −0.9985 4.77 ) 2 = 0.044 .

Significance

This position, halfway between two minima, is very close to the location of the maximum, expected near β = 3 π / 2 , or 1.5 π .

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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 ° , 0.0083 I 0

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If the slit width D is varied, the intensity distribution changes, as illustrated in [link] . The central peak is distributed over the region from sin θ = λ / D to sin θ = + λ / D . For small θ , this corresponds to an angular width Δ θ 2 λ / D . Hence, an increase in the slit width results in a decrease in the width of the central peak    . For a slit with D λ , the central peak is very sharp, whereas if D λ , it becomes quite broad.

Figures a through c show graphs of I by I0 versus theta in degrees. Each has a wave crest with y value 1 at x=0. Figure a, labeled D equal to lambda has a broad arc. Figure b, labeled D equal to 5 lambda has a narrower crest. It has zeroes roughly between 10 and 15 and between minus 10 and minus 15. Figure c, labeled D equal to 10 lambda has a narrow crest. It has zeroes at plus and minus 5, roughly between 10 and 15 and between minus 10 and minus 15.
Single-slit diffraction patterns for various slit widths. As the slit width D increases from D = λ to 5 λ and then to 10λ , the width of the central peak decreases as the angles for the first minima decrease as predicted by [link] .

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 ( sin β β ) 2 ,

    where β = ϕ 2 = π D sin θ λ , D is the slit width, λ is the wavelength, and θ is the angle from the central peak.

Conceptual questions

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

The parameter β = ϕ / 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 β and is related to the curvature of the arc that forms the resultant phasor that determines the light intensity.

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Problems

A single slit of width 3.0 μ m is illuminated by a sodium yellow light of wavelength 589 nm. Find the intensity at a 15 ° angle to the axis in terms of the intensity of the central maximum.

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

I / I 0 = 2.2 × 10 −5

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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.

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Consider the single-slit diffraction pattern for λ = 600 nm , D = 0.025 mm , and x = 2.0 m . Find the intensity in terms of I o at θ = 0.5 ° , 1.0 ° , 1.5 ° , 3.0 ° , and 10.0 ° .

0.63 I 0 , 0.11 I 0 , 0.0067 I 0 , 0.0062 I 0 , 0.00088 I 0

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Questions & Answers

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
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
why we have physics
Anil Reply
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
Physics Reply
explain l-s coupling
Depk Reply
how can we say dirac equation is also called a relativistic equation in one word
preeti Reply
what is the electronic configration of Al
usman Reply
what's the signeficance of dirac equetion.?
Sibghat Reply
what is the effect of heat on refractive index
Nepal Reply
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
Priyojit Reply
please tell
Swati
yes
Chemist
1 millimeter is How many metres
Darling Reply
1millimeter =0.001metre
Gitanjali
The photoelectric effect is the emission of electrons when light shines on a material. 
Chris Reply
Practice Key Terms 1

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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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