Iprofphenc27t01st01.Notes.1p

1

Table of Contents

Interference _________________________________________________________________ 2

Analytical Interference ______________________________________________________________ 3

Condition for minimum Intensity _____________________________________________________ 4

Condition for Constrictive Interference _________________________________________________ 5

Condition for Destrictive Interference _________________________________________________ 5

Summary of Interference ____________________________________________________________ 8

Learning Points ______________________________________________________________ 14

Concept of Optical Equivalence _________________________________________________ 15

Optical Instruments __________________________________________________________ 22

Microscope _________________________________________________________________ 22

Uses ____________________________________________________________________________ 25

Compound Microscope ____________________________________________________________ 25

Telescope __________________________________________________________________ 28

Terrestrial Telescope ______________________________________________________________ 28

Galilean Telescope ________________________________________________________________ 30

Astronomical Telescope ____________________________________________________________ 32

Geometry of Fringes _______________________________________________________________ 35

Intensity Distribution ______________________________________________________________ 36

Black Line _______________________________________________________________________ 36

Coherent Sources ____________________________________________________________ 37

Youngs Double Slit Experiment _________________________________________________ 41

Fringe width () __________________________________________________________________ 43

Angular Fringe width () ___________________________________________________________ 44

Geometry of Fringes _______________________________________________________________ 44

Intensity Distribution ______________________________________________________________ 44

Examples ___________________________________________________________________ 47

2

Interference The phenomenon of redistribution of energy when two waves travelling in same

direction or rarely same direction meet each other is called interference.

Redistribution of energy

Constructive Interference

waves help each other.

Destructive Interference

waves oppose each other.

Interface

Law of Conservation

of energy

Principle of

Superposition of waves

Interference

Constructive Destructive

3

Analytical Interference

1 = a1 sin (wt-kx)

2 = a2 sin (wt-kx+)

W = 2f

1 = a1 sin wt

2 = a2 sin (wt+)

Phase difference = 2 - 1 =

From principle of superposition

= 1 + 2

= a1 sin wt + a2 sin (wt+)

= a1 sin wt + a2 sin wt cos + a2 sin cos wt)

= a sin (wt+)

A resultant Amplitude = 2 21 2 1 2

a a 2a a cos

1 2

1 2

a sintan

a a cos

When cos = 1

Amax = a1 + a2

When cos = 1

Amax = a1 a2

Let a1 = a2 = a

Amax = 2a

Amin = 0

4

Condition for Constructive interference

cos = 1

= phase difference = even multiple of

Condition for destructive interference

cos = - 1

= phase difference = odd multiple of

Intensity Relation

I a2

I = I1 + I2 +

Condition for maximum Intensity

cos = 1

Imin = ( + )2

Condition for minimum Intensity

cos = - 1

Imax = ( - )2

Let a1 = a2 I1 = I2

Imax = 2I

Imin = 0

Path difference = 2 Phase difference

5

x = Path difference

= Phase difference

x

2

Condition for Constrictive Interference

x = Path difference = n

= even multiple of /2

Condition for Destrictive Interference

x = odd multiple of /2

Let I1 = I2 = I0

IP = 4 I0 cos2(/2)

Let Imax = 4I0

IP = 4 Imax cos2(/2)

6

Example: A narrow monochromatic beam of light of intensity I is incident on a glass

plate as shown in fig. Another identical glass plate is kept close to the first oneand

parallel to it. Each glass plate reflects 25 percent of the light incident on it and transmits

the remaining. Find the ratio of the minimum and maximum intensities in the

interference pattern formed by the two beams obtained after one reflection at each

plate.

Solution

I1 = 0.25 I = I/4

I2 = (0.75)(0.25)(0.75I) 9I

64

2

1 2max

2

min1 2

I II 49Ans.

I 1I I

7

path difference 2 phase difference

x = path difference

= path difference

Condition for Constructive Interference:

= (even multiple)

x = n

= (integral multiple)

= (2n)

= (even multiple)

Condition of Destructive Interference:

x = (odd multiple)

Let I1 = I2 = I0

= phase difference between the arriving waves.

Imax = 4 I0

x =

IP = 4 I0 cos2

IP = 4 Imax cos2

8

Summary of Interference

(i) If amplitudes of waves arriving at point P on the screen are different then resultant

intensity is given by

I = 1 2 1 22 cosI I I I

Also, Imax = 2

1 2I I , when cos = 1

Imin = 2

1 2I I , when cos = 1

(ii) m axm in

I

I =

21 2

1 2

I I

I I

= 2

1 2

1 2

A A

A A

=

211

rr

where r = 12

A

A = 1

2

I

I

(iii) The phenomenon of interference is based on conservation of energy. There is no

destruction of energy in the interference phenomenon. The energy which

apparently disappears at the minima has actually been transferred to the maxima

where the intensity is greater than that produced by the two beams acting

separately.

Iav =

2

02

0

I d

d

= 2

1 2 1 2

0

1 ( 2 cos )2

I I I I d

= I1 + I2

2

0

cos d

= 0

As the average value of intensity is equal to the sum of individual intensities,

therefore the energy is not destroyed but merely redistributed in the interference

pattern.

9

(iv) All maxima are equally spaced and equally bright. This is true for minima as well.

Also interference maxima and minima are alternate. The intensity distribution in

interference pattern is shown in fig.

(i) If whole apparatus is immersed in liquid of refractive index then,

= Dd

i.e., fringes width decreases

(ii) Sometimes in numerical problems, angular fringe width () is given which is

defined as angular separation between two consecutive maxima or minima.

= D d

In medium, other than air or vacuum,

= d

(iii) yd

xD

is valid when angular position of maxima or minima is less than 6

.

However sinx d is valid for larger values of provided d

10

uniform illumination due to overlapping of interference pattern of each

wavelength.

(vi) If the interference experiment is performed with bichromatic light, the fringes of

two wavelength will be coincident for the first time under following condition.

Y = n )Longer = (n + 1)Shorter or nLonger = (n + 1)Shorter

(vii) In many numerical problems we have to calculate number of maxima or minima.

We know that for maximum.

sin = nd or n = s ind

n d

( sin 1)

Suppose in some question d works out to be 2.3 so, permissible values of n are 0,

1, 2. Hence, total 5 maxima will be obtained on screen.

(viii) The original experiment performed by Young consisted two pin holes and sunlight

was used. But now-a-days pinholes are replaced by narrow identical slits which

increases brightness and sunlight by monochromatic light which increases number

of bands. Interference pattern consists of a large number of equally spaced

alternate bright and dark bands known as fringe running parallel to the length of

slits.

(ix) Fringes are locus of a point which moves in such a way that its path difference

from the two sources remains constant. For a given fringe, n is constant

S2P S1P = Constant

Under above restriction locus of P is a hyperbola with S1 and S2 as foci. Now if we

rotate a hyperbola about the line S1S2 we get hyperboloid of revolution. The

fringes seen on a screen are section of the these hyperboloids.

11

(x) The fringe visibility (v) is given by

v = max m inm ax m in

I I

I I

= 1 2

1 2

2 I I

I I = 1 2

1 2

2A A

A A

when I1 = I2

or A1 = I2

Imin = 0

Hence, v = 1 (best visibility)

(xi) If source is not placed symmetrically w.r.t. slits the path difference is given by :

13

Example: There are two glass plates and their thickness is negligible. These glass plates

are placed very close to each other.

Given: Reflection coefficient is 25%.

Assumption: Negligible absorption by glass plates.

Calculate the ratio of maximum and minimum intensity in the resulting interference

pattern if two rays are made to interfere.

Solution:

I1 = 0.25 I =

I2 = (0.75) (0.25) (0.75 I) =

Imax : Imin = 49:1

14

Light Sound

Learning Points 1.

This is due to geometry of This is due to difference in optical

arrangements i.e. relative properties of medium.

position of source & screen.

2. In case of Reflection\Transmission:

(i) Whenever there is a reflection from rigid body, a path difference of

is

introduced.

(ii) Whenever there is reflection from free boundary, no path difference is

introduced.

(iii) In situation of transmission, whether boundary is free or rigid, no path difference

is introduced.

(iv) Whenever a wave enters in a medium, where its speed decreases, it is called

rigid boundary. If speed of wave increases then interface is called free boundary.

3. A boundary which is optically free, will be acoustically rigid and vice-versa.

For example:

Path Difference

Geometrical Path Difference Optical Path Difference

Air-Water Boundary

Rigid Free

15

Concept of Optical Equivalence Two distances in two different media if light takes same time to travel, are

d1 = Distance travelled in medium (1)

d2 = Distance travelled in medium (2)

For optical equivalence,

=

=

Optically equivalent distances are not geometrically equivalent.

1.33 m of air 1 m of water

=

=

We know that, f1 = f2

=

Number of waves in medium (1) = Number of waves in medium (2)

Hence, two distances in different media are optically equivalent if they contain same

number of waves.

1 d1 = 2 d2

16

Example: In Youngs double slit experiment set-up with light of wavelength = 6000 ,

distance between the two slits is 2 mm and distance between the plane of slits and the

screen is 2 m. The slits are of equal intensity. When a sheet of glass of refractive index

1.5 (which permits only a fraction of the incident light to pass through) and thickness

8000 is placed in front of the lower slit, it is observed that the intensity at a point P,

0.15 mm above the central maxima does not change. Find the value of .

Solution:

In absence of glass steel, path difference at P,

x = yd

D = 3 30.15 10 2 102

= 1.5 107 m

Corresponding phase difference at P,

17

= 2 (x) = 7102 (1.5 10 )

26000 10

Intensity at P, I = 2

04 cos 2I

= 2I0

Phase difference when glass sheet is introduced,

= 2 ( 1)t

= 10

10

2 11(1.5 1)(8000 10 )2 66000 10

The intensity at P is now, I = 20 0 0 0112 cos 2

6I I I I

= 0.21 Ans.

Example: In Youngs experiment, the source is red light of wavelength 7 107 m. When

a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of

the interfering beams, the central bright fringe shifts by 103 m in the position

previously occupied by the 5th bright fringe. Find the thickness of the plate. When the

source is now changed to green light of wavelength 5 107 m, the central fringe shifts

to a position initially occupied by the 6th bright fringe due to red light. Find the refractive

index of glass for the green light. Also estimate the change in fringe width due to the

change in wavelength.

18

Solution:

given problem,

(R 1)t = 5R

or, t = 65 7 10

(1.5 1)

= 7 m

Now, when red light is replaced by green light

G 1) t = 6 R

So, 1

1R

G

=

56

or, G 1 = 6 (1.5 1)5

or G = 1.6

Further as 5R = 103

i.e., R = 2 104

So, G

R

= G

R

= 57

or, G RR

= 5 1

7 = 2

7

So, = 42 2 107

= 0.57 104 m

i.e., Fringe will decrease by 0.57 104 m when red light replaced by green light.

19

Example: A coherent parallel beam of microwaves of wavelength = 0.5 mm falls on a

Youngs double slit apparatus. The separation between the slits is 1.0 mm. The intensity

of microwaves is measured on a screen placed parallel to the plane of the slits at a

distance of 1.0 m from it as shown in the fig.

(a)If the incident beam falls normally on the double slit apparatus, find the y-

coordinates of all the interference minima on the screen.

(b) If the incident beam makes an angle of 30 with the x-axis (as in the dotted arrow

shown in the fig.), find the y-coordinates of the first minima on either side of the

central maximum.

Solution:

As shown in fig., the path difference between the two interfering waves reaching the

point P of the screen will be x = d sin and so the point P will beam interference

minima if

x = (2 1)2

n with n = 1, 2, .....

20

So d sin = (2 1)2

n

i.e., sin = (2 1)2nd

i.e., sin = (2 1)0.52 1n

= (2 1)

4n

and as sin 1

(2 1)4n 1

i.e., n 2.5

n = 1 or 2

When n = 1,

sin 1 = 14

so that tan 1 = 115

When n = 2

sin 2 = 34

so that tan 2 = 317

Now, the position of a point P on the screen which is at a distance D from the plane of

slits will be given by

y = D tan = tan ( D = 1 m)

21

So, the position of minima will be

y1 = tan 1 = 1

15 = 0.258 m

and y2 = tan 2 = 3

17 =

32.6

= 1.13 m

And as minima can be on either side of principal maxima, in the situation given there

will be 4 minima at positions 0.258 m and 1.13 m on the screen.

(b) In this situation as shown in fig., the path difference between the interfering wave

will be

For first minima, d [sin sin ] = 2

i.e., sin = sin 2d

Here, = 30 ; = 0.5 mm and d = 1 mm,

sin = 1 0.52 2 1

,

or, sin = 34

or 1 ,4

or, tan = 37

or 115

So, the position of first minima on either side of central maxima in this situation will be

22

y = D tan = tan ( D = 1 m),

y = 3

7 m and

1

15 m Ans.

Optical Instruments

Microscope

Principle

When an object is placed between a convex lens and its principal focus, an erect, virtual

and magnified image is formed on the same side as the object. A single convex lens, of

short focal length, thus used constitutes a simple microscope which is commonly known

as a magnifying glass or a reading glass.

Formation of Image

Fig. shows a small extended object AB which when viewed by an unaided eye cannot be

seen distinctly. A convex lens is then interposed between the eye and the object so that

the distance u of the object from the lens is less than the focal length of the lens. A

virtual, erect and magnified image AB will be produced. By adjusting u, the image is

set at least distance of distinct vision (D = 25 cm) from the eye so that the image

becomes most distinct. The rays of light forming the virtual and magnified image AB of

the small object AB as seen through the lens by placing the eye very close to it are

shown in fig.

Optical Instruments

Microscope Telescope

23

Magnifying power or Magnification

The magnifying power of a simple microscope is defined as the ratio between the angle

subtended by the image at the eye to the angle subtended by the object at the eye

when both the image and the object are situated at the distance of distinct vision.

Thus, if i is the angle which the image AB subtends at the eye and O is the angle

which the object AB (= CB) subtends at the eye when both are placed at the least

distance of distinct vision D, then

Magnifying power; M = i

O

= t a n

t a ni

O

(For small angle, tan

= //

AB OBCB OB

= AB OBOB CB

= OB

OB ( AB = CB)

= Du

= Du

Since the virtual image AB is formed at the least distance of distinct vision from the

lens, therefore,

v = D

Using lens formula, we get

1 1D u

= 1

... (1)

24

Multiplying both sides by D, we get

D DD u

= D

or 1 Du

= D

Du

= 1 D

... (2)

From equation (1) and (2), we get

M = 1 D

... (3)

Thus, a convex lens of shorter focal length yields higher magnification.

Remarks

(i) When final image is formed at infinity then magnification is given by

M = D

(ii) If the lens is kept at a distance x from the eye then

MD = Magnification for image formed at least distance of distinct

vision

= 1 D x

M = Magnification for image formed at infinity

= D x

25

Uses

A simple microscope is used by :

(i) Astrologers to read fate-lines of the hand,

(ii) Biology students to see the slides,

(iii) Watch repairers to locate defects, and

(iv) Detective department to match finger prints.

Compound Microscope

The magnification produced by a simple microscope is generally not sufficient to give a

detailed view of two very closely situated objects. A combination of two convex lenses is

employed to have a larger magnification. The instrument so designed is known as a

compound microscope. The convex lens O of short focal length and small aperture,

which faces the object to be viewed, is known as the objective. The convex lens E of

short focal length and comparatively large aperture, near the eye, is known eyepiece.

The two lenses are placed in two metal tubes so as to have a common principle axis. The

eyepiece is fitted in a draw-tube and can be slided within the main tube by means of

rack and pinion arrangement to focus the microscope upon the object.

Image Formation

26

Let AB be an extended object situated on the principal axis at a distance slightly greater

than the focal length of the objective. As refraction takes place through the objective O,

a real, inverted and magnified image A1B1 is formed.

The lens E is so adjusted that A1B1 falls within its focal length and so the final virtual

image A2B2 of A1B1 is obtained at the distance of distinct vision D from the eye. The

final image A2B2 is thus, highly magnified but is inverted with respect to the object AB.

The course of rays forming the final image is shown in fig.

Magnifying Power or Magnification

The magnifying power of a compound microscope is defined as the angle subtended by

the final image at the eye to the angle subtended by the object at the eye when both

the object and the image are situated at the distance of distinct vision from the eye.

If the object was situated at B2, it would have occupied length B2C such that

B2C = AB

Let the angles subtended by the object B2C and final image A2B2 at the eye be O and

i respectively. The magnification M of the microscope is, then given by,

M = iO

= t a n

t a ni

O

( i and O are very small)

In O2B2C, tan i = 2 2

ABO B

In O2B2A2, tan i = 2 22 2

A B

O B

M = 2 2A BAB

= 2 2 1 11 1

A B A B

A B AB

But 1 1A BAB

= M1 and 2 21 1

A B

A B = M2

Where M1 and M2 are magnifications produced by the two lenses O and E respectively.

M = M1 M2 ... (1)

For the lens O, = Dis t a n ce of im a ge

Dis t a n ce of object = v

u

27

Again, since the lens E acts like a simple microscope, its magnification M2 is given by

M2 = 1e

D

Where e is the focal length of the eyepiece. Substituting for M1 and M2 in equation

(5.4), we get

M = 1e

v Du

... (2)

If the object AB is situated very near the principal focus of the objective, the image

A1B1 will be far removed from the lens O.

In that case, u = O = Focal length of the objective

v = L = Length of the microscope tube

M = 1O e

L D

... (3)

It is clear from equation (3) that in order to have a microscope of large magnification,

the two lenses used should have small focal lengths.

A good microscope has magnification as high as 2000 or even more.

Remarks

If final image is formed at infinity then

M = vu e

D

Length of tube = v + e

where u and v are distance of object and image from objective lens respectively.

28

Telescope

Terrestrial Telescope

As astronomical telescope is used to view heavenly objects since the inversion of their

images does not produce any complication. While viewing objects on earth we would

prefer to have their images erect. Therefore, astronomical telescope is not suitable in

such cases. By using an additional convex lens O in between O1 and O2 of an

astronomical telescope we can have the final erect image. The lens O is called erecting

lens, while the improved version of the telescope is called terrestrial telescope.

Rays from the distant object get refracted through the objective O1, giving a real

inverted image A1B1 fig. The erecting lens O is so adjusted that its distance from A1B1 is

equal to twice its (erecting lens) focal length. An image A2B2 having same size as that of

A1B1, inverted w.r.t. A1B1 and hence erect w.r.t. the object is obtained at a distance 2 on

other side of O. A2B2 acts as an object for lens at O and final erect and magnified image

is obtained after refraction through O2. If the distance O2B2 is equal to focal length e,

of the eye lens O2, final image is formed at infinity figure and the telescope is said to be

Telescope

Terrestrial Telescope

Galilean Telescope

29

in normal adjustment. If the distance O2B2 is less than e then corresponding to a

certain value of this distance, a virtual and magnified image is obtained at the distance

vision as shown in fig.

Since the sizes of A2B2 and A1B1 are same, introduction of the erecting lens O has not

produced any charge in its magnifying power but has helped in getting the final image

erect only. It may also be noted that the use of erecting lens O results in a slight increase

(equal to four times the focal length of erecting lens) in the length of tube of telescope.

Remarks

(i) MD = 1O ee D

M = OO

(ii) LD = O + 4 + ue

L = O + 4 + e

30

Galilean Telescope

Instead of using a combination of two lenses O1 and O2 for getting an erect image,

Galileo used only one concave lens to get the final erect image.

Parallel beam of incident rays from infinity are focused by the objective O. An inverted

image A1B1 would have been formed after refraction through O1. Before the rays meet

at A1 a concave lens (at O2) intercepts them figure. The beam diverges and the final

erect image A2B2 is obtained. The distance O2B1 is so adjusted that final image is formed

at the distance of distinct vision.

If O2B1 is equal to the focal length e of eye lens at O2, final image is formed at infinity

and the telescope is said to be set in normal adjustment. In such a case the length of the

tube is equal to the difference between the focal length of two lenses. The field of view

of this telescope is small because of the use of concave lens.

When set in normal adjustment, its magnifying power M is given by

M = i

O

= i

o

tan

tan

=

1 1

1 2

1 1

1 1

A B

B O

A B

B O

=1 1

1 2

B O

B O

or, M = Oe

=

F oca l len gt h of object ive

F oca l len gt h of eye len s ... (1)

31

Remarks

(i) Magnification when final image is formed at least distance of distinct vision is given

by

MD = 1O eO

D

(ii) Length of tube is given by

LD = O ue (For least distance)

L = O e (For normal adjustment)

M= OO

32

Astronomical Telescope

It is an optical instrument used to have clear and detailed view of heavenly objects. In

simplest form if consists of two convex lenses separated some distance apart and

mounted in two metal tubes so as to have a common principal axis. The lens O which

faces the object is known as objective and has a large aperture and a large focal length

O. The lens E which faces the eye is known as eyepiece and has a small aperture and a

smaller focal length e. The tube carrying eyepiece can slide into the tube carrying

objective by means of rack and pinion arrangement. A parallel beam of light coming

from a distant object suffers refraction through objective and produces a real, inverted

and diminished image at a distance O from O1. This image then acts as an object for

the eyepiece E and so the final magnified image is obtained after refraction through the

eyepiece.

33

(i) Normal adjustment : An astronomical telescope is said to be in normal adjustment

if the final image formed after refraction through both the lenses is situated at

infinity. The course of rays, in such an adjustment, is shown in fig.

A1B1 is the image of distant object situated at infinity. If A1B1 happens to be

situated at the principle focus of the lens E, the rays will emerge in parallel

direction and a final virtual but magnified image is formed at infinity.

Magnification M, in normal adjustment, is defined as the ratio of the angle i,

subtended by the final image at the eye as seen through the telescope to the angle

O subtended by the object at the unaided eye when both the image and the

object are situated as infinity.

Thus, M = iO

= t a n

t a ni

O

( O and i are small)

Now, in A1O1B1 tan O = 1 11 1

A B

O B

and in A1O2B1, tan i = 1 12 1

A B

O B

M = 1 1 1 12 2 1 1

A B O B

O B A B = 1 1

1 1

O B

O B

But O1B1 = O = focal length of the objective

And O2B2 = e = focal length of eyepiece

M = Oe

... (1)

Thus, the magnification of astronomical telescope in normal adjustment is equal to ratio

of the focal length of objective to the focal length of eyepiece. The distance between

Astronomical Telescope Adjustment

Normal Adjustment Final image at a distance of

distinct version

34

the two lenses in this case will be (O + e). The strain on the eye will be least if the object

is viewed for a long period.

(ii) When the final image is formed at the distance of distinct vision : The course of

rays in such an adjustment is shown in fig. A parallel beam of light coming from an

object situated at infinity gets refracted through the objective O and produces a

real, inverted and diminished image A1B1 at a distance O from O1. If A1B1 happens

to be within the focal length e of the eyepiece, a final virtual but magnified image

A2B2 is observed. The position of eyepiece E is so adjusted that final image is

obtained at a distance of distinct vision D from the eye. Magnification in this case

is defined as the ratio of the angle i subtended at the eye by the final image

formed at the distance of distinct vision to the angle O subtended at the unaided

eye by the object situated at infinity.

Thus, M = iO

= t a n

t a ni

O

( O and i are small)

Now, in A1B1O1, tan O = 1 11 1

A B

O B

And in A1B1O2, tan i = 1 12 1

A B

O B

M = 1 1 1 12 1 1 1

A B O B

O B A B = 1 1

2 1

O B

O B = O

u ... (2)

Where u is the distance of A1B1 from the lens E. Considering refraction through lens. E,

1e

= 1 1v u

35

Here, v = D

1u

= 1 1e D = e

e

D

D

We get e

e

Du

D

(3)

Substituting for in equation (2), we get

M = eOe

D

D

... (4)

Out of the two adjustments discussed above the second adjustment gives a higher

magnification since the factor O ee

D

D

> 1. Equation (1) and (4) also show that for

greater magnification,

(i) the focal length O of the objective should be large, and

(ii) the focal length e of the eyepiece should be small.

The objective of an astronomical telescope should have a large aperture so as to allow a

large number of rays to fall on it. In that case, the intensity of the image is large. This

becomes even more essential in the case of heavenly bodies which are situated at large

distances. A powerful astronomical telescope placed at Yerkes observatory at Lake

Geneva has an objective aperture about one metre and of focal length about 18 metre.

Geometry of Fringes

For any fringe

x = constant

S1P S2P = constant

The locus of any point which moves in such a way so that difference of its distance from

the two paired point is constant is known as Hyperbola.

36

Intensity Distribution

IS1 = IS2 = I (Say)

2P

I 4Icos ( / 2)

Where = Phase difference

Hyperbola approaching straight line for large values of x.

R > V

R > V

= Wavelength of light in air

= Wavelength of light in medium

=

Black Line

This is defined as points where minima are due to two wavelengths coincide.

37

Example: Two sources S1 and S2 give out waves of same frequency separated by 3 as

shown in fig. Find the number of maxima and minima recorded on straight screen

passing through S1 and perpendicular line joining S1 and S2 from to +

Solution:

xP = Path difference at P

= = S1A S2A

2 29 x x

As x p, XP 0

No. of maxima = 5

No. of minima = 6

Coherent Sources Coherent sources are those which are derived from same one source (i.e., atomic

oscillator) and the light emitted by them have same frequency. Amplitude of emitted

wave may or may not be same. The phase relationship does not change with time. The

initial phase difference at the time of emission between the waves from coherent

sources is either zero or constant. Thus, coherent sources give out waves which have

point to point phase correspondence (or relationship).

Production of Coherent Sources

By division of wave

front

By division of amplitude

38

(a) By division of wave front: In this method the wave-front is divided into two or

more parts by the use of mirrors, lenses and prisms. The well known methods are

Youngs double slit arrangement. Fresnels biprism and Lloyds single mirror etc.

(b) By division of amplitude: In this method the amplitude of the incoming beam is

divided into two or more parts by partial reflection or refraction. These divided parts

travel different paths and finally brought together to produce interference. This type of

interference needs broad source of light. The common examples of such interference of

light are the brilliant colours seen when a thin film of transparent material like soap

bubble or thin film of kerosene oil spread on the surface of water is exposed to a broad

source of light.

Ways of obtaining a pair of Coherent Sources

(i) Double slit method: Light from a source S is limited to a narrow beam (not a narrow

beam but diverging source) with the help of a slit (Fig.). The emergent light is made to

fall upon a screen containing two slits S1 and S2 placed symmetrically with respect to

the slit. In that case both S1 and S2 are illuminated by the same wave-front. Therefore,

the beams of light coming out from S1 and S2 have no phase difference.

Thus, S1 and S2 can be treated as the coherent sources. Young made use of them in

his famous youngs double slit experiment.

39

(ii) A source and its own virtual image: Light from a source S is made to fall on a plane

mirror M (Fig.). Point of observation P on a screen AB receives direct light as well as

reflected light. For an observer reflected light appears to come from a source S (virtual

image of S). So, interference at p takes place between waves coming from S and S.

Since S is not an independent source, being the virtual image of S, it will have the same

phase as S.

Hence, the two can be treated to be coherent source. Lloyd made use of this

arrangement in Lloyd single mirror experiment.

(iii) Bi-prism method : Light from a source S is made to fall on an assembly of two right

angled prisms A and B joined base to base as shown in fig. S1 and S2 are the virtual

images of S produced by refraction through prisms A and B respectively. Being virtual

images of the same source, S1 and S2 have same phase and hence can be treated as

the coherent sources. This type of arrangement is made use of in Frensels bi-prism

experiment.

40

Remarks

(i) Two independent sources similar in every respect cannot act as coherent sources.

The reason lies in the origin of light which is due to random emission of radiation

from excited atom. When the atom is excited, say, by heating or electric discharge,

etc, then the revolving electrons absorb energy and go to outer orbits which are

unstable. Hence, soon the electrons spontaneously fall back to inner orbits thus

emitting energy in the form of radiation. When the frequency of these radiations

lies within the range of visible spectrum, we have the emission of light, which

consists of a broken chain of wave-trains, accompanied with sudden and abrupt

changes in phase occurring in very short intervals of time (of the order of 108

sec.). Consequently with two independent sources or with two separate portions

of the same source, the phases of waves emanating from them will be changing

independently of each other so that the phase difference between two such

sources cannot be constant.

(ii) Two identical laser sources can act as coherent sources.

41

Youngs Double Slit Experiment

x= Path difference at P

= = S2P S1A

= d sin

42

Positions of maxima

x= Path difference at = n

d sin = n

sin = n

d

n= Order of Maxima

y n

D d

Yn = nth maxima = nD

d

Zero Order

Maxima

(Central Maxima)

First Maxima Second Maxima Third Maxima

43

Positions of Minima

x= Path difference at

(2n 1)

2

d sin =

(2n 1)2

sin =

(2n 1)

2d

Fringe width ()

This is defined as linear separation between two consecutive maxima or minima

n n 1

Dy y

d

(2n 1)

2d

n=0,1,2

(2n 1)

2d

n=0,1,2

First

maxima Second

maxima

First

maxima Second

maxima

44

Angular Fringe width ()

This is angular separation between two consecutive maxima or minima

D d

Geometry of Fringes

For any fringe

x = constant

The locus of any point which moves in such a way so that difference of its distance from

two fixed point is constant, is known as hyperbola.

Intensity Distribution

Is1 = Is2 = I (say)

Where = phase difference

= wavelength of light in air

1 = wavelength of light in medium =

S2P S1P = constant

Ip = 4 I cos2

R > V

R > V

45

Displacement of fringes due to introduction of thin transparent medium in the path of

one of the interfering beam.

Consider two coherent sources sending waves to reach P (Fig.) with a path difference x

given by

S2P S1P = dD

If n is the order of bright fringe coinciding with vertical cross-wire of the field of view at

P,

dyD = n

Introduce a thin plate of transparent material (refractive index ) of thickness t in one

of the beams say S1P (fig.). Since velocity of light in the denser medium is lesser, light

will take comparatively greater time to go from S1 to P in the presence of plate. Thus,

net path difference between S2P and S1P will decrease.

Net Path difference = S2P [(S1P t)air + tsheet]

= S2P [(S1P t) + t]

= (S2P S1P) ( -1)t

= dyD ( -1)t

yd

x ( 1)t nD

New fringe width () =

New angular fringe width () =

46

yn = Position of nth maxima in presence of thin transparent sheet.

nD D

( 1)t nd d

For central maxima

x = 0

yd

( 1)t 0D

D

y ( 1)t shift(s)d

47

Examples Example: A beam of light consisting of two wavelength 6500 and 5200 is used to

obtain interference fringes in a Youngs double slit experiment.

(i) Find the distance of the third bright fringe on the screen from the central

maximum for the wavelength 6500 .

(ii) What is the least distance from the central maximum when the bright fringes due to

both wavelengths coincide?

The distance between the slit is 2 mm and the distance between the plane of the slits

and the screen is 120 cm.

Solution : Given, 1 = 6500 = 6.5 107 m

2 = 5200 = 5.2 107 m

d = 0.2 cm = 2 103 m

D = 120 cm = 1.2 m

(i) For nth bright spot yn = Dn d

Here, 1 = 6.5 107 m

y3 = 7

3

3 6.5 10 1.2

2 10

= 1.17 103 m Ans.

(ii) Since 2 < 1, fringe width for 2 is smaller. If two bright fringes due to 1 and 2 are

to coincide, then minimum distance from the central spot will be where nth order bright

spot due to 1 and (n + 1)th bright spot due to 2 coincide.

1D

nd

=

2( 1)n D

d

n 6.5 107 = (n + 1) 5.2 107 or n = 4

y = n Dd

= 7

3

4 6.5 10 1.2

2 10

= 1.56 103 m Ans.

48

Example: In Youngs double slit experiment using monochromatic light, the fringe

pattern shifts by a certain distance on the screen when a mica sheet of refractive index

1.6 and thickness 1.964 micron is introduced in the path of one of interfering waves. The

mica sheet is then removed and the distance between the slits and the screen is

doubled. It is found that the distance between successive maxima (or minima) now is

the same as the observed fringe shift on the introduction of mica sheet. Calculate the

wavelength of the monochromatic light used in the experiment.

Solution : Shift y in the fringe system is

y = ( 1)t

when distance between slits and screen is doubled,

= 2

Given = y

( 1)t

= 2

= ( 1)2

t

Here, = 1.6, t = 1.964 106 m

= 6(1.6 1) 1.964 10

2

= 0.3 1.964 106 m

= 5892 Ans.

49

Example: In a modified Youngs double slit experiment, monochromatic uniform and

parallel beam of light of wavelength 6000 and intensity 210 Wm incident normally on two circular operation A and B of radii 0.001 m and 0.002 m respectively. A perfectly

transparent film of thickness 2000 and refractive index 1.5 for the wavelength 6000

is placed in front of aperture A (fig.). Calculate the power (in watt) received all the focal

spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume

that 10% of the power received by each observer goes in the original direction and is

brought to the focal spot.

Solution : Let I1 and I2 be the intensities at A and B

I1 = I2 = 10 Wm2

Area of cross-section of aperture A, A1 = 21r = (0.001)2 = 106 m2

Area of cross-section of aperture B, A2 = 22r = (0.001)2 = 4 106 m2

Let P1 and P2 be the powers of incident radiations at A and B respectively.

P1 = 610 10

= 106 = 105 W

P2 = 610 4 10

= 4 105 W

Introduction of a transparent medium in one of the beams produces some path

difference x.

x = ( 1) t

Here, = 1.5 and t = 2000

x = (1.5 1) 2000 = 0.5 2000

or, x = 107 m

Let = phase difference between the two beams

= 2 x

50

or, = 7

10

2 106000 10

= 3

radian

If a1 and a2 are the amplitudes of light from apertures A and B, net amplitude R at F is,

R2 = 2 21 2 1 22 cosa a a a

Power = Intensity Area of cross-section

= I A2

or, P = KR2 A2 = KR2

P1 = 21K a

and P2 = 22K a ... (1)

Multiply equation (1) by K throughout

KR2 = 2 21 2 1 22 cosK a K a K a K a

or, P = 1 2 1 22 cosP P P P

Substituting for P1, P2 and , we get

P = 5 5 5 5(10) 4 10 2 10 4 10 cos 3

or, P = 7 105 W Ans.

51

Example: In fig. shown, S is a monochromatic point source emitting light of wavelength

= 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical

halves L1 and L2 by a plane passing through a diameter. The two halves are placed

symmetrically about the central axis SO with a gap of 0.5 nm. The distance along the axis

from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 m. The screen at O is

normal to SO.

(i) If the third intensity maximum occurs at the point P on the screen, find distance

OP.

(ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the

distance OP increase, decrease or remain the same?

Solution : (i) As shown in fig. each part of the lens will form image of S which will act as

coherent sources.

From lens equation, we can write

1 115v

= 110

or, v = 30 cm

m = vu = 2

d = 3 0.5 mm = 1.5 mm

Also, D = 1.30 0.30 = 1 m

Now, from the theory of interference the distance y of a point P on the screen is given

by

52

y = ( )D xd

and as point is third maximum

x = 3

So, y = (3 )Dd

or, y = 7

3

3

5 10 10 m0.5 10

= 1 mm Ans.

(iii) If gap between L1 and L2 is reduced then d will decrease. As Dd and OP = 3,

therefore OP will increase.

Example: A glass plate of refractive index 1.5 is coated with a thin layer of thickness t

and refractive index 1.8 Light of wavelength travelling in air is incident normally on the

layer. It is partly reflected at the upper and the lower surfaces of the layer and the two

reflected rays interfere. Write the condition for their constructive interference. If =

648 nm, obtain the lease value of t for which the rays interfere constructively.

Solution : The ray reflected from upper surface suffer a phase change of due to

reflection, at denser media, so the condition of constructive interference for normal

incidence is given by

2 t 2

= n or 2 t = (2 1)

2n

53

For minimum value of t, n = 0

tmin = 4 = 90 nm Ans.

Example: Fig. shows three equidistant slits being illuminated by a monochromatic

parallel beam of light. Let BP0 AP0 = /3 and D >> . (a) Show that in this case

2 /3d D . (b) Show that the intensity at P0 is three times the intensity due to any of

the three slits individually.

Solution: (a)

BP0 AP0 = 3

or, d sin = 3

or, d tan = 3

/2 3ddD

(For small angle tan sin )

or, d = 23D

54

(b) xA/B = path difference between waves coming from A and B = 3

A/B = phase difference

= /2

A Bx

= 23

Similarly, xB/C = d sin

= 3 / 2dd D = 23

2dD

=

B/C = 2

Now, phase diagram of the waves arriving at P0 is as shown below:

Amplitude of resultant wave is given by

A = 2 2(2 ) 2( )(2 )cos120A A A A

= 3A

As intensity (I) A2

Intensity at P0 will be three times the intensity due to any of the three slits

individually.

55

Example: In a Young experiment the light source is at distance l1 = 20 m and l2 = 40 m

from the slits. The light of wavelength = 500 nm is incident on slits separated at a

distance 10 m. A screen is placed at a distance D = 2 m away from the slits as shown in

fig. Find:

(a) The values of relative to the central line where maxima appear on the screen?

(b) How many maxima will appear on the screen?

(c) What should be minimum thickness of a slab of refractive index 1.5 be placed on

the path of one of the ray so that minima occurs at C?

Solution: (a) The optical path difference between the beams arriving at P,

x = (l1 l2) + d sin

The condition for maximum intensity is,

x = n

Thus, sin = 1 21 [ ( )]x l ld

= 1 21 [ ( )]n l ld

= 9 6

6

1 [ 500 10 20 10 ]10 10

n

= 2 140n

Hence, = 1s in 2 1

40 n Ans.

56

(b) |sin | 1

1 2 1 140n

or, 20 (n 4

or, 20 n 60

Hence, Number of maxima = 60 20 = 40 Ans.

(c) At C, phase difference, = 2 12 ( )l l = 692 20 10500 10

= 80

Hence, maximum intensity will appear at C. For minimum intensity at C,

( 1)t = 2

or, t = 2( 1)

= 9500 10

2 0.5

= 500 nm Ans.

Example: Light of wavelength = 500 nm falls on two narrow slits placed a distance d =

50 104 cm apart, at an angle = 30 relative to the slits shown in fig. On the lower slit

a transparent slab of thickness 0.1 mm and refractive index 3/2 is placed. The

interference pattern is observed on a screen at a distance D = 2 m from the slits. Then

calculate :

(a) Position of the central maxima?

(b) The order of minima closest to centre C of screen?

(c) How many fringes will pass over C, if we remove the transparent slab from the lower

slit?

57

Solution:

(a) Path difference, x = d sin + d sin ( 1) t

For central maxima, x = 0

sin = ( 1)

s int

d

= 3(3/2 1)(0.1)

s in 3050 10

= 12

= 30 Ans.

(b) At C, = 0,

Therefore, x = d sin ( 1) t = (50 103) 12 (3/2 1) (0.1)

58

= 0.025 0.05 = 0.025 mm

Substituting x = n,

We get n = x

= 60.025

500 10

= 50

Hence, at C, there will be maxima. Therefore, closed to C order of minima is 49.

Ans.

(c) Number of fringes shifted upwards = ( 1)t

= 6

(3/2 1)(0.1)

500 10

= 100 Ans.

Example: Consider the situation shown in fig. The two slits S1 and S2 placed

symmetrically around the central line are illuminated by a monochromatic light of

wavelength . The separation between the slits is d. The light transmitted by the slits

falls on a screen 1 placed at a distance D from the slits. The slit S3 is at the central line

and the slits S4 is at a distance z from S3. Another screen 2 is placed a further distance D

away from 1. Find the ratio of the maximum to minimum intensity observed on 2 if z is

equal to

(a) 2Dd

(b) 4Dd

Solution:

(a) z = 2Dd

59

Let I is intensity due to slits S1 and S2 1. Further, intensity at any point on

1 is given by

IP = 24 cos 2I

At slit S3, = 0

3S

I = 4I

At slit S4, x = d zD

= 2

= or 4SI = 0

Now on screen 2

Imax = 3 42

S SI I = 4I

Imin = 3 42

S SI I = 4I

m axm in

I

I = 1 Ans.

(b) z = 4Dd

3S

I = 4I

At slit S4, x = d zD = 4

= 24

= 2

4S

I = 24 cos 24I I

Imax = 3 42

S SI I = 2

4 2I I = 2

2 2I

Similarly, Imin = 3 42

S SI I = 2

4 2I I = 2

2 2I

m axm in

I

I =

22 2

2 2

Ans.

60

Example: A convex lens of focal length is used as a simple microscope. Find its

magnifying power if the final image is at the distance of distinct vision. Find also the

magnification and the distance of the object = 5 cm, D = 25 cm.

Solution : The object is within the focal distance

i = 1

A Bv =

1

ABu

and O = ABD

M = iO

= 1 D

2515

= 6

Magnification = A BAB = 1

1

v

u =

1

25u

u1 = 11

v

v = 25 5

25 5

= 12530

= 256

cm

Magnification = 25 625 = 6 Ans.

61

Example: The focal lengths of the objective and the eyepiece of a compound

microscope for 4 cm and 6 cm respectively. If an object is placed at a distance of 6 cm in

front of the objective, what is the magnification produced by the microscope? Distance

of distinct vision is 25 cm.

Solution :

Ma gn ifica t ion of m icr oscope

= Ma gn ifica t ion produced by t he object ive

An gu la r ma gn ifica t ion pr odu cedby t h e eyepiece

m = 1e

v Du

u = 6 cm, O = 4 cm, e = 6 cm, D = 25 cm

v = OO

u

u = 6 4

2 = 12 cm

m = 12 2516 6 = 3126

= 10.33 Ans.

62

Example: A telescope has an objective of focal length 50 cm and an eye piece of focal

length 5 cm. The least distance of distinct vision is 25 cm.

The telescope is focussed for distinct vision on a scale 200 cm away from the objective.

Calculate:

(i) The separation between the objective and the eyepiece.

(ii) The magnification produced.

Solution : Let AB be the position of the object

u = 200 cm, = 50 cm

The image AB formed by the objective is a distance v from it

v = u

u = 200 50200 50

= 1000

15 =

2003

cm

This serves as an object for the eyepiece. The distance between AB and the eyepiece is

u = 2003

l , where l is the separation between the lenses. The images (final) distance is

v = 25 cm from the eyepiece and = 5 cm.

u = v v

= 25 5

25 5

= 256

cm

l = 25 2006 3

= 4256

= 70.83 cm

Total magnification mO me = v vu u

= 200 1 625

3 200 25 = 2 Ans.

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Iprofphenc27t01st01.Notes.1p