Diffraction

Diffraction of Light
The phenomenon of bending of light waves when it passes
through a corner or edge of an obstacle or the spreading of
light around the shadow region of obstacle, if the wave length
(λ) of light wave is comparable to the dimension or size of the
obstacle (d) is called diffraction.
The departure from the rectilinear property of the light wave
is called diffraction of light. During diffraction phenomenon,
geometrical shadows are formed on the screen.
Examples of diffraction:

Luminous border spreading over the profile

just before the rising of sun behind it.
Colored spectra observed when one sees the
distant source with the help of thin piece of
cloth.
Light streaks observed when one sees strong
light with half shut eyes.
Diffraction of sound.
Difference between interference and diffraction

Interference Diffraction
It arises due to the superposition of
It arises due to the secondary wavelets
secondary waves from two
separated wave fronts produced by originating from different points of the
two coherent sources. same wave front.
All the bright fringes are of same Intensity of central bright fringe is
intensity. maximum and goes on decreasing on
The minima of fringes are usually either sides.
almost perfectly dark. The minima of fringes never perfectly
Fringe width is same dark.
Fringe width is not same.
Differentiation between Fresnel’s diffraction and Fraunhofer’s
diffraction.

Fresnel’s diffraction Fraunhofer’s diffraction

In this diffraction the distance In this diffraction the distance between the
between the source and screen is source and screen is infinite, so another
finite. So another lens is not
required to focus the light on the lens is required to focus the diffracted rays
screen. on the screen.
The emitting wave front from the The emitting wave front from the source in
source in this diffraction is spherical this diffraction is plane.
or cylindrical. The theoretical treatment in this diffraction
The theoretical treatment in this is simple and accurate. Thus we prefer
diffraction is just approximate. using Fraunhofer’s diffraction.
Fraunhofer’s diffraction through a single slit

Fig:Fraunhofer’s diffraction through a single slit

Consider a parallel beam of light emitting from the source S passing through
the slit AB of width ‘a’ by the help of lens.
After passing through the slit AB, the diffraction pattern is observed on the
screen
According to the rectilinear propagation of light, the central maxima is seen
at the spot O.
Alternate dark and bright bands are seen on either sides of the central spot
O.
These bands are called secondary maxima and secondary minima.
It is observed that the width of secondary maxima is just half of the width of
central maxima.
The intensity of first secondary maxima is just 50% less than that of central
maxima.
The secondary minima are observed in between the maxima.
The intensity of secondary maxima goes on decreasing on both sides from
the central spot O.
From BAN, SinBN/AB =BN/a
Or, BN = a Sin
For diffraction to observe, BNλ
Therefore, the light coming out from edge point of the slit AB has its
path difference, λ when they meet at point P. For the resultant
amplitude at point P­­, lets divide AB into two halves AC and CB. Thus, the
path difference is λ/2 when light rays come out from the points A and C
if they meet at point P.
Thus phase difference between two rays coming out from A and C is
The resultant amplitude is zero as the wave diffracts destructively.
For destructive diffraction ( minima case )
Path difference = n λ
Or a Sinn = n λ
Or, Sinn = n λ/a ……………………..(1)
For n=1 Sin1 = λ/a
For n=2 Sin2 = 2 λ/a and so on.
For constructive diffraction ( maxima case)
Path difference = (2n+1) λ/2
Or a Sinn = (2n+1) λ/2
Or, Sinn = (2n+1) λ/2a ……………………..(2)
The intensity distribution for diffraction pattern is as shown below.

It can be concluded that the intensity of maxima goes on decreasing with increase of
order of diffraction pattern.

If the ratio λ/a is small then, the diffraction pattern is also small.

Thus, the ration λ/a gives the condition to what extent the light rays fail to travel in
straight line.
To show the width of the central maxima just two times the width of
secondary maxima, it is drawn as shown below.
The width of central maxima is defined as the distance between secondary minima on either side of the central
spot O as shown in figure. The angular width of central maxima is defined as the angle subtended by central
maxima at the lens or slit. It is denoted by 2.

From above figure,

Tan = y/D ; where x is the distance of first minima from central spot O.

Or,  = y/D ;for small , Tan  

Or, Sin = y/D ;for small ,   Sin

Or, λ/a = y/D ;as λ = a Sin

Or, y = λD/a ……………………….(3)

Width of central maxima is

central = 2x = ……………………………….(4)
Since we have, ; where yn is distance of nth minima from the
central spot O.
Similarly, for (n+1)th minima,
Therefore, width of secondary maxima is given by
secondary = yn+1 - yn
=
i.e.…………………………………(5)
From equations (4) and (5), we obtain
central = 2secondary
It is clear that the width of central maxima is two times the width of
secondary maxima. It is to be noted that the width of maxima
depends upon λ, D and a.
Diffraction grating

An arrangement consisting of a large no. of parallel

slits of equal width ‘a’ and separated from one
another by equal opaque spaces ‘b’ is called
diffraction grating.
The opaque spaces act as opacities or obstacles
whereas the spacing between them is called
transparencies.
The value a + b = d is known as grating element.
Let N be the number of lines per inch of the grating.
Then grating element is
d= a + b =
Fig: Diffraction Grating
Consider a parallel beam of
monochromatic light normally
incident on a plane grating.
The diffracted lights are focused by
the lens on the screen kept at the
focus of lens L. The patterns seen on
the screen is called Fraunhofer
diffraction patterns due to N-parallel
equidistant slits
It is observed that the central
maximum is formed at point O in
accordance with rectilinear property
of light. On either sides of O,
alternate dark and bright bands are
observed, called secondary minima
and secondary maxima.
If 1 is the angle of diffraction for first secondary maxima, then
d Sin1 = λ
Or (a+b) Sin1 = λ
Similarly, for 2nd order maxima,
(a+b) Sin2 = 2λ and so on.
In general, for nth order maxima,
(a+b) Sinn = nλ ………………………..(1) where, n = 1, 2, 3, ………….
This equation (1) is called Famous grating equation.
For n = 0, central maxima is formed. The intensity variation of
diffraction pattern on the screen of a grating against angle  is as
shown in figure.

It is to be concluded that the intensities if secondary maxima goes on

decreasing from the central spot O.
If white light is used to illuminate the slit, diffraction grating is widely used in
spectrometry instead of prism.
Resolving power of optical instrument

When two objects lying very close to each other form an angle greater than 1’
or (1/60)o with the eye, then we can see them distinctly.
But if the angle is less than 1’, we can’t see them as separate. Thus they are
seen as separate from each other by using lens or microscope or telescope or
grating or prism. The method of seeing such objects as separate is called
resolution.
The ability of an optical instrument to produce distinct images from the two
objects lying very close to each other is called resolving power of the
instrument.
The term “resolving power” is used for
Seeing two objects close together as separate or seeing the fine structure in a
microscope or telescope.
Seeing two spectral lines as distinct from each other.
Resolving power of telescope

It is the reciprocal of the angular serration between

two objects lying very close to each other and it is
given by
R.P. of telescope =
Where D is the diameter of objective lens and λ is
wavelength of illuminated objects.
Thus, resolving power of telescope increases with
decreasing wavelength of light illuminated on the
objects and increasing diameter of the objective lens.
Resolving power of microscope

The resolving power of microscope is the reciprocal of

distance between two objects seen through the
microscope. It is given by
R.P. of microscope =
Where, µ is the refractive index of medium between
objects and objective.
 is the half angle of the solid angle made by the
object with objective.
And λ is the wavelength of the light illuminated on the
object.
1) How wide is the central diffraction peak on a screen
3.5 m behind a 0.010 mm slits illuminated by 500 nm
light?
2) Red light of wavelength 633nm from helium-neon laser passes
through a slit 0.350 mm wide. The diffraction pattern is observed
on a screen 3.00 m away.
a) What is the width of the central bright fringe?
b) What is the width of the first bright fringe on either side of the
central one?
[ 10.85 mm, 5.4mm ]
3) A screen is placed 2m away from the single narrow slit.
Calculate the slit width if the first minimum lies 5mm on either
side of the central maximum. Incident plane waves has a
wavelength of 5000 Å.
[2 x 10-4 m]
4) Monochromatic light from a distant source is incident on
a slit 0.750 mm wide. On a screen 2.00 m away, the
distance from the central maximum of the diffraction
pattern to the first minimum is measured to be 1.35 mm.
Calculate the wavelength of the light.
[506nm]
5) Two spectral lines of sodium D1 and D2 have wavelengths of
approximately 5890 Å and 5896 Å. A sodium lamp sends incident
plane wave on to a slit of width 2 micrometre. A screen is located
2m from the slit. Find the spacing between the first maxima of
two sodium lines as measured on the screen.
[9x10-4 m]
6) A plane transmission grating having 500 lines per mm is
illuminated normally by light source of 600 nm wavelength. How
many diffraction maxima will be observed on a screen behind the
grating.
7) A parallel beam of sodium light is incident normally on a
diffraction grating. The angle between two first order spectra on
either side of the normal is 27. Assume that the wavelength of
light is 5.893 × 10-7 m, find the number of rulings per mm on the
grating.
8) A plane transmission grating having 400 lines per mm is illuminated normally by light
source of 600nm wavelength. Calculate
a) the grating spacing
b) the angle at which the first maxima is seen
c) How many diffraction maxima will be observed on a screen behind the grating?
[0.25 x 10-5m] [13.900] [4]
9) Plane monochromatic waves with wavelength 520 nm are
incident normally on a plane transmission grating having 350
slits /m. Find the angles of deviation in the first, second and
third orders.
[10.50 , 21.30, 33.10 ]
10) A slit 4.0 cm wide is irradiated with microwaves of
wavelength 2.0cm. Find the angular spread of central
maximum, assuming incidence normal to the plane of the slit.
[600]
Conceptual
Questions
1. Why we cannot observe the diffraction pattern in a wide slit
illuminated by monochromatic light?
2. Diffraction of radio waves and sound waves is much more
pronounced than light waves. Explain.
3. Diffraction grating is better than a two slit set up for measuring the
wavelength of a monochromatic light. Why?
4. In a single slit experiment, the width of the slit is made double the
original width. How does this affect the size and intensity of the
central diffraction maximum?
5. The resolving power of a microscope increases when red
light illuminating the object is replaced by the blue light.
Explain.
6. Write difference between interference and diffraction.
7. What happens to the Fraunhofer single slit diffraction if
the whole apparatus is immersed into water?
8. Coloured spectrum is seen when we look through a
muslin cloth. Why?
Muslin
Cloth