Revisiting Fraunhofer Diffraction (2): The Diffraction Grating

  We now consider the case for diffraction with slit number > 2, which we describe as a diffraction grating.  Basically, we are dealing with the case of multiple slit interference patterns produced and it is of interest to examine the differences, e.g. from the single slit case, more closely.  The diagram (Fig. 1) below) is instructive:

Fig. 1: Multiple slit interference patterns

 In embarking on Part II we note first, from the diagram, that as the number of slits is increased, the primary maxima become narrower but remain fixed in positions (Check the abscissa values in relation to peaks). Note also that that the number of secondary maxima increases as the number of slits is increased.  The decrease in the intensity of the maxima - indicated by the dashed lines - is due to diffraction. 

In terms of practical applications, note  that any  arrangement of elements which is equivalent in its action to a number of parallel equidistant  slits of the same width is called a diffraction grating.   Not surprisingly, this feature makes the grating a powerful instrument for the application of spectral analysis, an aspect my 2nd year physics students enjoyed both at Harrison College and the Univ. of Alaska- Fairbanks.

In this case of a diffraction grating each of the slits gives rise to a separate diffracted beam – such as depicted in Fig. 2, but which are so narrow that the diffracted beam from each spreads out over a sufficiently wide angle to interfere with all other beams.

                                                       Fig. 2

 But it is precisely this property of dispersion that makes spectral analysis possible. In Fig. 2, suppose q is taken so that ab = l, the wavelength of the incident light. Then it follows that cd = 2l, ef = 3l and so forth. (Note the first order maximum occurs when ab = l.)  Note that the waves from all these elements, since they are in phase at the plane of the grating, are also in the plane along the oblique line AA’  and therefore all reach P in phase.

Now, if the angle q is increased slightly, wave fronts from the grating elements no longer arrive at AA’ in phase with each other. Note here that even a relatively small change results in almost complete destructive interference. Hence, the maximum at each q is an extremely sharp one.  If q is increased enough we find another maximum (2^nd order) where ab = 2l, cd = 4l, ef = 6l etc. The wave fronts at AA’ are again in phase. It follows from this that we can write:

ab = ml

where m = 0, +1,+ 2, +3,, …… where m  is the order of the diffraction grating.

Then the condition for a maximum is: d sin q =  ml/d

 Or:

 sin q =  ml/d

To see an example of such dispersion - how it works to separate different wavelengths, consider a grating which has  5 x 10 ⁵ lines per m, and monochromatic light of wavelength 6 x 10 ^-7 m is incident.  Then we first obtain d (the spacing) by taking:

d =   1m/   5 x 10 ⁵   =   2 x 10 ^-6 m

The various dispersed angles are than obtained by substituting for m, i.e. m = 0, m = 1. m = 2 etc.

Then: q _o = 0 ^o.  

 q₁ = 17 ^o        (e.g.  sin q₁  =  1 ( 6 x 10 ^-7 m)/ 2 x 10 ^-6 m )

q₂ = 37 ^o       (e.g.  sin q₂  =  2 ( 6 x 10 ^-7 m)/ 2 x 10 ^-6 m )

q ₃ = 64 ^o    (e.g.  sin q₃  =  3 ( 6 x 10 ^-7 m)/ 2 x 10 ^-6 m )

 And a simple sketch of the grating image can show the dispersed wavelengths. Also that there is no  q₄   given that the result is unphysical.  (I.e. this is impossible since no sine value for any angle can exceed 1).

Applying this condition across different orders m, one obtains the intensity graph shown below:

                                      Fig. 3: Intensity distribution for diffraction grating.

Example Problem:

 A grating with 1,000 lines/mm is illuminated by monochromatic light of wavelength 400 nm. How many diffraction images are produced?

Solution:

 We have:  l  = 4 x 10^-7 m and d =  1 mm/1,000 =

 10 ^-3 m/ 1,000 = 10 ^-6 m

Then: (10 ^{- 6} m) sin q =  = 4 x 10^-7 m

Therefore, the first diffraction image is at:

sin q₁ =   4 x 10^-7 m / (10 ^-6 m)  » 0.4

whence: arc sin (0.4) = q₁ = 23^o.6

The second diffraction image is at: sin q₂ =  2l/d

And: sin q₂ =  2(4 x 10^-7 m ) / (10 ^-6 m)  » 0.8

q₂ =   arc sin (0.8) = 53^o.1

 For third order:

sin q₃ =  3l/d =  3(4 x 10^-7 m ) / (10 ^-6 m)  »  1.2

 But, this is impossible since no sine value for any angle can exceed 1.  Hence, only two images are produced, i.e.  of orders 1, 2.

Intensity Distribution for an Ideal Grating:

To obtain the intensity distribution formula for an ideal grating  we employ the method of adding complex amplitudes.  Then, denote the number of slits by N, and the phase change - from one slit to the next by d .  The resulting complex amplitude is the sum:

A exp(i q) =  a [ 1 + e ⁱd   +  e ⁱ²d   +  e ⁱ³d  +.......     exp(i(N-1) d)

To find the intensity we must then divide by the complex conjugate, so that:

A ²   = a ²  {(1 - cos Nd )/  (1 - cos d)

Using:  1 - cos (x) =  2 sin ²  (x/ 2)    we write:

A ²   = a ²  { sin ² (Nd /2)/ sin ² (d /2)}  =  a ² sin ² Ng /sin ²(g )

Where - as in the double slit case- we let  g  =  d/ 2  =  ( p/ l) d sin q

Then a ²  represents the intensity  diffracted by a single slit.  After inserting its value:

a ²  =  A _o ²   (sin ² b/ b ²)

And we get:

I   =   A _o ²   (sin ² b/ b ²)  (sin ² Ng /sin ²(g )

The intensity in a Fraunhofer pattern for an ideal grating.  To find the minima for:   (sin ² Ng /sin ²(g ) we just note when the numerator = 0.  This occurs at:  Ng = 0, p,  2p,  3p, .....pp.  The condition for a minimum is in general:

g =  pp / N

Excluding p-values for which p = mN.

The   p-values correspond to path differences: 

d sin q   =  l/N,   2l/N,  3l/N, ........(N-1) l/N,  (N + 1) l/N, 

Omitting values:  0,  N l/N,  2N l/N,......

Which apply to principal maxima.

Angular dispersion:

The separation of any two colors (wavelengths): l1  and l2 ,  increases with the order number. We express this separation using a quantity called the angular dispersion.  It is obtained by differentiating the grating equation with respect to l.   Then:  d cos q  dq     =   m d l

Or:  dq    = m /  d cos q

Or, using finite differences: 

Dq /D l   =    m / d cos q

This equation shows for a given small differences in wavelength  D l  ,  the angular separation  Dq  is directly proportional to m, and inversely proportional to the grating spacing d.

Suggested Problems:

1) A diffraction grating has 600 lines/ mm. Find the angle q for the second order maximum if yellow light of l = 550 nm is used. 

 2) Two spectral lines in a mixture of hydrogen (H2) and deuterium (D2) gas  have wavelengths of 656.30 nm and 656.48 nm, respectively. What is the minimum number of lines a diffraction grating must have to resolve these two wavelengths to first order (m= 1)?

3) Two spectral lines at 6200  Å.  (doublet) are separated by 0.652  Å.    Find the minimum number of lines a diffraction grating must have to just resolve this doublet in the 2nd order spectrum.

Remember  1  Å  =   10^{- 8}  cm

See Also:

• Revisiting Fraunhofer Diffraction (1)