Waves Tutorial 8 - Diffraction



Diffraction Effects

Single Slit Diffraction

Central Bright Fringe


The Diffraction Grating

Diffraction Gratings and Spectra


Diffraction Effects

If we pass waves through a single slit, we observe that the waves spread out due to diffraction.





Single Slit Diffraction

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?


We need to know the wavelength at first:


            l = c/f    ̃ l = 336 m s-1  = 1.31 m

                                        256 Hz

Now we can work out q:


            sin q = l = 1.31 m    ̃ sin q = 0.655  ̃ q = 41o

                         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.


Question 1

The ASTRA satellite transmits radio waves (speed = 3 ´ 108 m/s) at a frequency of 12.5 GHz.

(1 GHz = 1 ´ 109 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 ´ 107 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.



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.


The Diffraction Grating

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 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:


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?


Formula first:      nl = d sin q ̃ sin q = nl


There are 300 lines per mm, so there are 3 ´ 105 lines per metre.


̃ d =     1 ____   =  3.33 ´10-6 m

            3 ´ 105 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 ´ 10-9m  = 0.159 ̃ q = sin-1 (0.159) = 9.15o

                   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 ̃ q = sin-1 (0.318) = 18.54o

                   d        3.33 ´10-6 m


So the angle between the two maxima is 18.54 o - 9.15 o = 9.39o

Now we can work out the highest order maximum by using sin q = 1:


            1 = nl        ̃ 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.



Question 2

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:


Question 3

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:


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



Strictly speaking, there is a difference between spectroscopy and spectrometrySpectroscopy is the study of how a material radiates energy when it interacts with other matter.  Spectrometry is the application of spectroscopy to obtain quantative results in a particular spectrum.  In this section, the section of the spectrum we are interested in this the visible light spectrum.  At this level, I would doubt very much that you would lose marks for calling a spectrometer a "spectroscope".  


In the school physics laboratory, the most common spectroscope is like this:



They are a simple tube with a diffraction grating at one end and an eyepiece at the other.  Using these, we can observe emission lines from a glowing gas like hydrogen, or neon.


A more sophisticated instrument like the one shown below allows us to measure angles of diffraction:


This one uses a glass prism, although it can use a diffraction grating.  The diagram below shows a spectroscope seen from above:



Light from the source emerges through a slit.  It passes through a tube to the transmission diffraction grating.  The light is diffracted according to the wavelength.  The eyepiece can travel around the table until the first diffracted ray is observed.  The angle can be read off the scale.  The vernier scale allows an accurate angle to be determined.  Then the eyepiece is moved around the table until the next diffracted ray is seen.  The data recorded in Question 2 would have been obtained with a spectroscope like this.  The pattern of diffracted rays would be this:




The wavelength of each diffracted ray would determined by:

   nl = d sin q  



The pattern in the diagram above is shown on one side only, but is in reality symmetrical.  Therefore the eyepiece can move either side of the central undeviated maximum.


It is possible to have a spectroscope where the eyepiece, or photo-detector is fixed and the diffraction grating is movable, like this:



In this case, the diffracted ray is selected by the diffraction grating and passes to the photo-detector when the angle of the diffraction grating is correct.  In some instruments, the knob that controls the rotating table is calibrated in nanometres (wavelength) rather than degrees.


The instrument in the diagram uses a transmission grating, but many would use a reflection grating.