There are two types of diffraction: Fraunhofer and Fresnel, and we consider Fraunhofer first because it is much easier conceptually. We use the diagram below to understand single slit diffraction first:

** Illustrating
basis for single slit diffraction**.

In
the diagram we take the distance BO to be *very large* and the light rays
incident on the single slit AB of width d, to come from a large distance so
they’re essentially parallel. We therefore consider the slit AB to act as a
source of secondary disturbance, resulting in two wave fronts.

Let O be the center of a zero-order central maximum of maximum intensity so that all the waves arrive in phase. By the geometry of the situation:

sin
q = CD/ AC
where CD = l/2
and AC = d/2

Then:
CD = AC sin q

And: l/2 = d/2 (sin q) or **l**** = d (sin ****q****)**

This is for the first order minimum for which:

sin q = l/ d

This can be generalized to: sin q =m l/ d

Or
: d sin q = m l

e.g. for a diffraction grating, where: m = 0, 1, 2, 3, etc.

* Example
Problem*:

Find the angular position of the first minimum of a single slit diffraction grating of slit width 0.20 mm if yellow light of l = 550 nm is incident on the slit.

*Solution:*

sin
q = l/ d = (5.5 x 10 ^{-7} m)/ (2 x 10 ^{-4}
m) = 2.75 x 10 ^{- 3}

or
q = 0.0027 radians

*Fraunhofer Diffraction:
Double Slit*

We
now look at diffraction for the case of a *double slit*, as opposed to
single. The geometry of the set up and the phenomenon are illustrated below:

*Basis for double-slit Fraunhofer diffraction*In
this case, the Electric field at P is a vector superposition of *two E-waves*
from slits S1 and S2

1)
E1
= E_{o} sin wt and

2)
E2
= E_{o} sin (wt + f)

So: *E1 + E2 = E _{o} [sin *

*w*

*t + sin (*

*w*

*t +*

*f*

*)]*Where f is the phase difference. Though the waves have equal phases at the slits their phase difference at P depends on the path difference, d. (d = r2 – r1 = d sin q). Then we may write:

d/ f = l/2 p

And**: **f =
(2 p/ l) d = (2 p/ l) d sin q

To
get the average light intensity at P (I _{av}) note that:

the *time
average of sin ^{2} (*

*w*

*t +*

*f*

*/2) = ½.*Therefore: I _{av} = 4 E_{o }^{2}
cos^{2}( f/2)
· ½ = 2 E_{o }^{2}
cos^{2}( f/2)

Whence:
I _{av }= I _{o }^{2} cos^{2}( f/2)

Where: I _{o } = 2 E_{o
}^{2 }= I _{max }/2

Now: f = (2 p/ l) d sin q

** Therefore:
** f/2 = p d sin q/ l

What do we find if d sin q = l? Then:

f/2 = p and I _{av }=
I _{o }^{2} cos^{2}(p) = I _{o }^{2} (1) = I _{o }^{2}

What
do we find if d sin q
= **2****l**?
Then:

f/2 = 2p and I _{av }=
I _{o }^{2} cos^{2}(2p) = I _{o }^{2} (1) = I _{o }^{2}

We can proceed with all other values of d sin q and find that for each maximum the intensity is the same. This is illustrated below:

*Intensity pattern for double slit diffraction.** Suggested Problems*:

1) Sketch the intensity diagram for a single slit diffraction grating and assign relative amplitudes and intensities for the first maximum.

2) Show
that for the three first odd (half integral) wave positions in double slit
diffraction, that the amplitude I _{av
}= 0.

## No comments:

Post a Comment