DIFFRACTION -

Diffraction(Introduction)

The diffraction of light and its effect can be regularly seen in everyday life. The closely spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow pattern we see when looking at a disk. Diffraction in the atmosphere by small particles can cause a bright ring to be visible around a bright light source like the sun or the moon. A shadow of a solid object, using light from a compact source, shows small fringes near its edges. The principle can be extended to engineer a grating with a structure such that it will produce any diffraction pattern desired; the hologram on a credit card is an example. All these effects are a consequence of the fact that light is a wave. We will see the reason for all in this chapter. So here we are going to discuss the topic of Diffraction of light.

The phenomenon of bending of light around the corners of an obstacle or aperture is called the diffraction of light.

$diffraction of light$

Diffraction is a general characteristic exhibited by all types of waves, be it sound wave, lightwave, water wave or matter-wave.

Since the wavelength of light is much smaller than the dimension of most obstacles, we do not encounter the diffraction effect of light in everyday observation.

NOTE:-

(1) The colours that we see when a CD is viewed is due to the diffraction effect.

(2) In ray optics, we ignore diffraction and assume that light travel in a straight line.

(3) The smaller the size of the obstacle or aperture, the greater the bending at the corner and vice versa.

$diffraction of light$

THE SINGLE SLIT(DIFFRACTION)

When we use a single narrow slit and illuminated by a monochromatic source of light, a broad pattern with a central bright region is seen. On both sides, there are alternately dark and bright regions. The intensity becomes weaker away from the centre. The central bright is considerably wider than the slit.

$diffraction of light$

To understand this

$diffraction of light$

Consider a parallel beam of light falling normally on a single slit LN of width d. The diffracted light goes on to meet a screen. the midpoint of the slit is M.

CENTRAL MAXIMA:-

We know that every point of the wavefront LM is the source of secondary wavelets.

All the waves from different point arriving in phase at C, give rise to central maximum i.e central bright fringe is obtained at C.

The path difference NP-LP between the two edges of the slit can be calculated exactly as far young’s experiment.

$diffraction$

Where $diffraction$ is the angle due to which secondary wavelets diffracted

If $diffraction$ is small, then

Path difference $diffraction$

POSITION OF SECONDARY MINIMA:-

If the path difference NQ is $\fn_jvn \lambda$ (wavelength of light), then point P will have minimum intensity. i.e P is a point of first secondary minima.

This can be easily proved. The slit LN can be considered to be divided into two equal halves LM and MN.

$diffraction of light$

The path difference between L and M is $diffraction$ (i.e path difference is )

Also, the path difference between M and N is $diffraction$

This is true for any point in the upper half LM and the corresponding point in lower half MN.

Therefore secondary waves from the upper half of the slit interface interfere destructively with secondary waves from the lower half of the slit. Hence P is a point of first secondary minima.

$diffraction$

for 2nd secondary minima

$diffraction$

for nth secondary minima

$diffraction$

POSITION OF SECONDARY MAXIMA:-

If the path difference is $diffraction$ , then point P will have a maximum intensity and known as 1st secondary maxima.

This can be also proved easily. We can divide the slit into three equal parts as shown in fig.

$diffraction of light$

The path difference between the corresponding points of the first two parts of the slit will be $diffraction$ ( i.e phase difference of $diffraction$ ). Therefore they will give rise to destructive interference.

however, the wavelets from the 3rd unused part will reinforce to produce weak 1st secondary maxima.

$diffraction$

for 2nd secondary maxima

$diffraction$

therefore, for nth secondary maxima

$diffraction$

INTENSITY DISTRIBUTION CURVE:-

$diffraction of light$

Here for single slit diffraction, the intensity of pattern is a function of $diffraction$

Position of secondary minima

$diffraction$

Position of secondary maxima

$diffraction$

NOTE:-

1. The intensity of secondary maxima decreases with distance from the centre.

2. The width of central maxima is more than twice of secondary maxima.

3. The intensity at the first secondary maxima is less than 5% of the intensity in the middle of the central maxima.

4. The above diffraction pattern is for single slit due to monochromatic light. For white light diffraction pattern is coloured with central maxima is white.

In double-slit experiment if $diffraction$ (narrow slit) where a is the width of the slit.

The central maximum of the diffraction pattern of either slit covers the entire viewing screen. Moreover, the interference of light from the two slits produce bright fringe with approximately the same intensity as shown in fig(1)

$diffraction of light$

But in general $diffraction$ is often not met. i.e the interference of light from two slits produces bright fringes that do not all have the same intensity as shown in fig(2)

i.e the intensities of the fringes produced by double-slit interference are modified by the diffraction of the light passing through each slit.

In diffraction by a single actual slit; the diffraction pattern has a broad central maximum and weaker secondary maxima as shown in fig(3)

We can simply say that fig(3) as an envelope on the intensity plot in fig(1).

DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION

interference:-

1. Due to the superposition of light wave coming from two coherent source.

2. All bright fringe are the same intensity.

3. Width of fringes may or may not be same

4. dark fringe is perfectly black.

5. Bands are large in numbers.

Diffraction:-

1. Due to the superposition of secondary wavelets coming from different points of the same wavefront.

2. Intensity is decreasing as moves from the centre.

3. Width of fringe is never the same width.

4. Dark fringe is not perfectly black.

5. bands are few in numbers.

Note that we have assumed that the screen on which the fringe is formed is at a large distance. The two or more paths from the slits to the screen were treated as parallel. This situation also occurs when we place a converging lens after the slits and place the screen at the focus. Note that the lens does not introduce any extra path difference in a parallel beam.

$diffraction of light$

This arrangement of often used since it gives more intensity than placing the screen far away.

LINEAR WIDTH OF CENTRAL MAXIMA

the linear width of the central maxima is the distance between the 1st secondary minimum on the two sides of the centre O of the central maxima.

$diffraction of light$

Assume lens is very close to slit i.e $diffraction$

for 1st secondary minima

$diffraction$

if $\fn_jvn \theta$ is very small $diffraction$

$diffraction$

from (1) and (2)

$diffraction$

Therefore linear width of the central maxim $diffraction$

It is

1. It is directly proportional to the wavelength $\fn_jvn \lambda$ of light

2. It is inversely proportional to the width d of the slit

3. It is directly proportional to the distance D between slit and screen.

ANGULAR WIDTH OF CENTRAL MAXIMA

Angular width= $diffraction$

we know that $diffraction$

if $\fn_jvn \theta$ is very small, $diffraction$

$diffraction$

if the width of the slit d is large compare to $diffraction$ , then angular width is approx zero

i.e no spreading occurs. this explains the rectilinear propagation of light.

SEEING THE SINGLE SLIT DIFFRACTION PATTERN

RESOLVING POWER OF OPTICAL INSTRUMENTS

The image of a point object formed by an optical instrument (lens, human eye etc) is not a sharp point but a spot because due diffraction pattern.

$diffraction of light$

Hence when two point objects very close to each other are seen through an optical instrument, then their diffraction pattern of images will also be very close and will overlap each other.

If the overlapping is small, then both images are seen separate in the optical instrument. i.e the optical instrument is able to resolve the objects. If the overlapping is large, the objects will not be seen as separate, they will be seen as one. In this case, the optical instruments is not resolving the two objects.

$diffraction of light$

i.e” The resolving power of an optical instrument is the power or ability of the instrument to produce distinctly separate images of two close objects.“

According to Rayleigh, the two point objects A and B are just resolved when the central maximum of the diffraction pattern of one fall over the 1st secondary minimum of the diffraction pattern of the other.

$diffraction of light$

The minimum distance/angle between two objects when they can be seen as separate by an optical instrument is called the limit of resolution of the instrument.

It has been found that for a circular aperture

$diffraction$

where $diffraction$ wavelength of light and $diffraction$ aperture of the opening

The smaller the limit of resolution, the greater in its resolving power. i.e resolving power of an optical instrument is the reciprocal of the limit of resolution of the instrument i.e

$diffraction$

NOTE:- For the human eye, the limit of resolution is one minute (1′) or $diffraction$

RESOLVING POWER OF TELESCOPE

$diffraction of light$

The Resolving power of a telescope is its ability to show as separate the images of two distant objects laying to each other.

We know that the limit of resolution is

$diffraction$

where a is the aperture of the objective lens.

Therefore, Resolving power of telescope= $diffraction$

$diffraction$

Here R.P is greater if a is large and $diffraction$ is small.

Here we have no control on $diffraction$ . Therefore, the resolving power of a telescope can be increased by increasing the aperture of the objective lens.

RESOLVING POWER OF MICROSCOPE

The resolving power of a microscope is its ability to show as separate the images of two point objects lying close to each other.

$diffraction of light$

Here the object is placed just before the focus.

i.e we can write $diffraction$

for small aperture, we can write

$diffraction$

$diffraction of light$

here a is optical aperture

now,

$diffraction of light$

from fig, $diffraction of light$

we know that

$diffraction of light$

Also, magnification of the lens is

$diffraction of light$

from (1) and (2)

$diffraction of light$

Therefore, R.P of the microscope is

$diffraction of light$

Here R.P is increased if $diffraction of light$ is large( This is possible if the focal length of the lens is taken small because we cannot increase the aperture of the objective lens). Also, R.P is large, if $diffraction of light$ is taken smaller. In practice if we want to reduce $diffraction of light$ , we must pass light through a liquid of R.I $diffraction of light$

i.e $diffraction of light$

$diffraction of light$

i.e we can increase the resolving of a microscope by using blue light instead of ordinary light.

NOTE:-

1. Usually, an oil having a R.I close to that of the objective glass is used( oil immersion objective)

2. It is good to remember that a property of telescope is resolving whereas a microscope magnifies.

THE VALIDITY OF RAY OPTICS

The ray optics is based on the assumption that light travels in a straight line but in wave optics( specially diffraction effect) is based on the assumption that light doesn’t travel in a straight light, it bends around the edge.

We know that the linear width of central maxima from the center is

$diffraction of light$

It is clear from relation, if x is small, then ray optics is valid. Here x must be small when D should be small and d should be large.

Here in case of D

” The distance at which the diffraction spread (x) of a beam of light is equal to the size of the aperture is called Fresnel Distance. It is denoted by $diffraction of light$

now, $diffraction of light$

$diffraction of light$

Fresnel Distance

$diffraction of light$

If $diffraction of light$ , then diffraction effects can be neglected and ray optics is valid. If $diffraction of light$ the spreading due to diffraction is very pronounced and ray optics is not valid.

RELATED LINKS OF WAVE OPTICS
Introduction	Huygens Principle
Refraction and Reflection of Plane waves using Huygens Principle	Coherent and Incoherent Addition of Waves
Interference of Light Waves and Young’s Experiment	Diffraction
Polarisation

IMPORTANT QUESTIONS FOR PRACTICE( DIFFRACTION)

1. In question 1 from previous topic 10.5, what should the width of each slit be to obtain 10 maxima of the double-slit pattern within the central maximum of the single slit pattern?

2. Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inches?

3. For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?

4. Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.

5. Answer the following questions:

(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other
properties of images in optical instruments. What is the justification?

6. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
7. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
8. Answer the following questions:

(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

9. In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.