Waves Tutorial 8
- Diffraction
If we pass waves through a single slit, we observe that the waves spread out due to diffraction.
Notice:
If the slit is narrow, the diffraction is more marked
The wavelength remains the
same.
Diffraction does not need a slit.
Waves can bend round a barrier by diffraction.
Radio signals can be picked up behind hills for this reason.
The longer the wavelength, the more the waves will
diffract.
All waves diffract.
We
can show the diffraction of light due to a
single slit. (We must be
careful not to confuse this with Young's double slits.)
If we have a wide slit, we see just a single bright region with sharp
edged shadows.
If we make the slit narrower, we see a pattern emerging with a bright central region, and alternating light and dark bands. The narrower the slit, the marked the effect. The central bright region becomes dimmer as well because less light is transmitted.
If the light is monochromatic, the bands will be of the same colour. Red light has a broader pattern than blue light, suggesting that the diffraction effect increases with wave length. If we use white light, the central band is white, with the fringes being overlapped with the spectrum of colours. This is called Frauenhoffer diffraction.
We can plot a graph to show the intensity, and we see a bright central maximum, with subsidiary maxima either side. We can explain the effect of diffraction using the idea of secondary wavelets. In the middle these form a plane wave-front. At the edges, circular wave-fronts move into the shadow region. The maxima and minima are caused respectively by constructive and destructive interference.
We
can work out the angle of diffraction using a simple equation:
where q is the angle, l is the wavelength and a is the width of the aperture.
Read the question to see if it's single or double slit. The key thing is that the pattern from double slits is due to interference. From a single slit, it's due to diffraction. Single slit diffraction has NOTHING to do with Young's Double Slits. |
Worked Example A piano note of 256 Hz is played. It is heard through a door 200 cm wide. What is the maximum angle of diffraction that will occur if the speed of sound in air is 336 m/s? |
Answer
We
need to know the wavelength at first:
l
= c/f
Þ
l
=
336 m s^{-1}
= 1.31 m 256 Hz |
Now we can work out
q:
sin
q =
l
=
1.31 m
Þ
sin
q
= 0.655
Þ
q = 41^{o} a 2.0 m |
If
the door were less than 1.31 m wide, diffraction could not occur because
sin
q would be greater than 1, which is impossible.
The ASTRA satellite transmits radio waves (speed = 3 ´ 10^{8} m/s) at a frequency of 12.5 GHz. (1 GHz = 1 ´ 10^{9} Hz)
(a)
What is the wavelength of the signal?
(b)
The transmitting dish is 1.6 m in diameter.
Find the angle of diffraction. (c) What is the radius of the circular patch that receives the signal? The height of the satellite is 3.6 ´ 10^{7} m above the earth.^{ }(Ignore the curvature of the Earth) |
Central Bright Fringe
The width of the central bright fringe in single slit diffraction is shown in the diagram below:
From the diagram above, we can see:
For very small angles, tan q = sin q. Since:
we can write:
Rearranging gives us:
If white light is used, the pattern is rather messy:
The central bright spot is white. Then the next fringes show the spectra of visible light. Red light diffracts the most, and violet light the least. This diagram shows the first bright fringes only.
Resolution
The single slit diffraction equation can be used, with modification, to determine the limit to which an optical instrument can resolve. This is called the resolving power. (This is sometimes known as Rayleigh's Criterion.) For a light microscope, the theoretical limit is about 1 mm, so the microscope cannot be used to view atoms. A beam of electrons is regarded as having wave properties. So an electron microscope has a much bigger resolution, as the wavelength of the electron beam is much shorter.
Consider two objects very close together:
The red spot in the middle and the intensity peaks below show the central bright region. If the central bright regions are separate, they can be easily resolved. If they touch, they can be just resolved. If they overlap, they cannot be resolved.
Radio
waves diffract round hills, which is why we can pick up radio signals behind
hills, even though there is no direct line of sight between the transmitter and
the receiver.
A diffraction grating can be used to
split light into different wavelengths with a high degree of accuracy, much more
so than glass prisms.
A diffraction grating usually consists of a piece of glass with very
closely spaced lines ruled on it. A
transmission grating has clear spaces
between the lines so that light can pass through it.
A reflection grating has a
shiny surface between the lines so that light gets reflected off it.
A compact disc acts as a reflection grating.
The
diffraction grating has the advantage over the double slit method of measuring
wavelength in that:
the maxima are more sharply defined;
the beam passes through more slits than two, so the intensity is brighter;
the angles are larger so that they can be measured with greater precision.
The derivation of the diffraction grating formula is on the syllabus, but experience has shown that most students struggle with it. It is more important that they know how to use the formula. If you want to see the derivation, click HERE. The formula is:
d sin q = nl
The term n is called the spectrum order. If n = 1, we have the first diffraction maximum. The other physics codes:
q is the angle,
l is the wavelength,
and
d
is the slit width
Sin q can never be greater than 1, so there is a limit to the number of spectra that can be obtained.
Worked Example A diffraction grating has 300 lines per mm. When it is illuminated normally by light of wavelength 530 nm, what is the angle between the first and second order maxima? What is the highest order maximum that can be obtained? |
Answer Formula first: nl = d sin q Þ sin q = nl d |
There are 300 lines per mm, so there are 3
´
10^{5} lines per metre.
Þ
d
= 1 ____ =
3.33
´10^{-6}
m 3 ´ 10^{5} m^{-1} |
Now put the numbers into the
equation to work out the angle of the first order maximum:
sin
q
=
nl
=
1 x 530
´ 10^{-9}m
= 0.159 Þ
q
= sin^{-1} (0.159) =
9.15^{o} d 3.33 ´ 10^{-6} m |
Now put the numbers into the
equation to work out the angle of the second order maximum:
sin
q =
nl
= 2
´
530
´10^{-9}
m = 0.318
Þ q
= sin^{-1}
(0.318) =
18.54^{o}
d
3.33
´10^{-6}
m
So the angle between the two maxima is 18.54 ^{o }- 9.15 ^{o} = 9.39^{o} |
Now we can work out the
highest order maximum by using
sin
q = 1:
1 =
nl
Þ
n = d
= 3.33
´
10^{-6} m
= 6.3
d
l
530
´ 10^{-9} m
Since the orders of maxima have to be whole numbers, the maximum order has to be 6. |
If
the answer to the problem had been 6.87, the maximum order would still be 6,
even though the nearest whole number was 7. |
When cadmium light is viewed through a diffraction grating having 500 lines per
millimetre, the following spectral lines were observed at the stated angles.
Angle (degrees)
Colour
18.78
red
14.74
green
13.89
light blue
13.53
dark blue
Find the wavelength of these lines. Find the other angles at which spectral lines would be observed. |
If we did further calculations we could see that the red light is diffracted more than blue light. The pattern would be like this:
Note that:
There is a central un-deviated maximum, which would be white light.
The pattern would be symmetrical with orders either side of the maximum.
We have not showed the ones below here.
The third order does not have a red light ray.
Why does the third order have no red ray? |
Diffraction Gratings and Spectra
The diffraction grating is a very good way of selecting light of a specific
wavelength. Chemists and
astronomers use diffraction gratings in
spectroscopy, which allows them to see the specific
spectra
given out by different elements.
Each element has its own individual spectrum. This allows astronomers to:
See what elements there are in stars.
If the
spectrum “fingerprint” is shifted at all, astronomers can tell that a star
is moving towards us (blue shift) or away from us (red shift).
This is due to the Doppler effect.
The pictures below show how spectra are used by astronomers:
Photo courtesy of NASA, Wikimedia Commons.
Picture courtesy of NASA, Wikimedia Commons
Picture by ESO, Wikimedia Commons