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Figure a shows horizontal rays passing from left to right through a vertical slit of length D. This is labeled theta equal to zero, bright. Figure b shows rays passing through the slit at an angle theta to the horizontal. This is labeled sine theta equal to lambda by D, dark. A dotted line perpendicular to the rays touches the top of the slit. Its perpendicular distance from the bottom of the slit is lambda and that from the center of the slit is lambda by 2. A separate view shows the dotted line being at an angle theta to the vertical. It intersects the ray starting from the bottom of the slit at a particular point. The horizontal distance of this point from the slit is delta l equal to D sine theta. Figure c shows rays passing through the slit at an angle theta to the horizontal. This is labeled sine theta equal to 3 lambda by 2 D, bright. A dotted line perpendicular to the rays touches the top of the slit. Its perpendicular distance from the bottom of the slit is 3 lambda by 2. Figure d shows rays passing through the slit at an angle theta to the horizontal. This is labeled sine theta equal to 2 lambda by D, dark. A dotted line perpendicular to the rays touches the top of the slit. Its perpendicular distance from the bottom of the slit is 2 lambda.
Light passing through a single slit is diffracted in all directions and may interfere constructively or destructively, depending on the angle. The difference in path length for rays from either side of the slit is seen to be D sin θ .

At the larger angle shown in part (c), the path lengths differ by 3 λ / 2 for rays from the top and bottom of the slit. One ray travels a distance λ different from the ray from the bottom and arrives in phase, interfering constructively. Two rays, each from slightly above those two, also add constructively. Most rays from the slit have another ray to interfere with constructively, and a maximum in intensity occurs at this angle. However, not all rays interfere constructively for this situation, so the maximum is not as intense as the central maximum. Finally, in part (d), the angle shown is large enough to produce a second minimum. As seen in the figure, the difference in path length for rays from either side of the slit is D sin θ , and we see that a destructive minimum is obtained when this distance is an integral multiple of the wavelength.

Thus, to obtain destructive interference for a single slit    ,

D sin θ = m λ , for m = ± 1 , ± 2 , ± 3 , ... ( destructive ) ,

where D is the slit width, λ is the light’s wavelength, θ is the angle relative to the original direction of the light, and m is the order of the minimum. [link] shows a graph of intensity for single-slit interference, and it is apparent that the maxima on either side of the central maximum are much less intense and not as wide. This effect is explored in Double-Slit Diffraction .

Figure shows a graph of intensity versus sine theta. The intensity is maximum at sine theta equal to 0. There are smaller wave crests to either side of this, at sine theta equal to minus 2 lambda D, minus lambda D, lambda D, 2 lambda D and so on.
A graph of single-slit diffraction intensity showing the central maximum to be wider and much more intense than those to the sides. In fact, the central maximum is six times higher than shown here.

Calculating single-slit diffraction

Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0 ° relative to the incident direction of the light, as in [link] . (a) What is the width of the slit? (b) At what angle is the first minimum produced?

Figure shows a vertical line on the left. This has a slit in the middle, of length D. A ray labeled lambda passes horizontally through the slit. This splits into 5 dotted lines that fall on a screen. The screen is shown as a vertical line. Of the five dotted lines, two make angles theta 1 and theta 2 with the horizontal. Theta 2 is 45 degrees. Theta 1 is smaller than theta 2 and is unknown. Intensity on the screen is shown as a vertical wave. The crest at the center, where a horizontal dotted line from the center of the slit falls on the screen, is the largest. The wave attenuates at the top and bottom. The remaining four dotted lines correspond to troughs in the wave.
In this example, we analyze a graph of the single-slit diffraction pattern.

Strategy

From the given information, and assuming the screen is far away from the slit, we can use the equation D sin θ = m λ first to find D , and again to find the angle for the first minimum θ 1 .

Solution

  1. We are given that λ = 550 nm , m = 2 , and θ 2 = 45.0 ° . Solving the equation D sin θ = m λ for D and substituting known values gives
    D = m λ sin θ 2 = 2 ( 550 nm ) sin 45.0 ° = 1100 × 10 −9 m 0.707 = 1.56 × 10 −6 m .
  2. Solving the equation D sin θ = m λ for sin θ 1 and substituting the known values gives
    sin θ 1 = m λ D = 1 ( 550 × 10 −9 m ) 1.56 × 10 −6 m .

    Thus the angle θ 1 is
    θ 1 = sin −1 0.354 = 20.7 ° .

Significance

We see that the slit is narrow (it is only a few times greater than the wavelength of light). This is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects such as this single-slit diffraction pattern. We also see that the central maximum extends 20.7 ° on either side of the original beam, for a width of about 41 ° . The angle between the first and second minima is only about 24 ° ( 45.0 ° 20.7 ° ) . Thus, the second maximum is only about half as wide as the central maximum.

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Questions & Answers

as a free falling object increases speed what is happening to the acceleration
Success Reply
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 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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