WAVE OPTICS - drmgriyearbtech.com

UNIT – III

WAVE OPTICS

3.1. HUYGENS’ PRINCIPLE:

The Principle may be stated inthree parts:

(i) Every point on a given

wavefront may be regarded as the source

of a new disturbance, called secondary

wavelets.

(ii) The secondary spherical

wavelets from each point spread out in

all directions with the velocity of light.

(iii) The envelope of these

wavelets in the forward direction at

any instant gives the new

wavefront at that instant.

Consider a number of points say a,b, c, d, e ... on the spherical wavefrontAB (Fig. 3.1). With each point in turnas a centre and radius ct draw spheres.These spheres represent the secondarywaves originating from these pointsrespectively. The envelope or tangential surface A1 B1 touching all these spheres in

the forward direction is the new spherical wave front. At a large distance from thesource S, the wavefront AB because plane.

The Huygens’ construction gives a backward wavefront also (shown by dottedlines) which is contrary to observation. This result is avoided by assuming that theintensity of the secondary wavelets is not uniform, but varies continuously from amaximum in the forward direction to a minimum of zero in the backward direction.

Fig. 3.1

3.2. INTERFERENCE OF LIGHT:

Interference is the optical phenomenon in which brightness and darkness areproduced by two exactly similar light waves meeting. When two light waves, of samefrequency and having constant phase difference, coincide in space and time, there isa modification in the intensity of light. The resultant intensity at any point dependsupon the amplitudes and the phase relationships of the two waves. This modificationin the intensity distribution resulting from the superposition of two waves of lightis called interference and the pattern of bright and dark fringes produced is calledinterference pattern.

3.2.1. Principle of Superposition:

When two or more waves arrive at a point in space simultaneously, the netwave disturbance at that point and at any given time is the vector sum of all thewave disturbances at that point at that particular time. This is called the principle

of superposition. Let us apply the principle of superposition to the analyticaltreatment of interference of light.

Consider two waves of same frequency ω

2 π . Let a1 and a2 be the amplitudes of

the two waves. Let y1 be the displacement at the observation point due to one wave

at any instant t and can be represented by

y1 = a1 sin ωt.

Let the displacement of the second wave in the same direction be represented by

y2 = a2 sin (ωt + δ)

where δ is the phase difference between the two waves at the point underconsideration.

According to Young’s principle of superposition, the resultant displacement at thatpoint is given by

y = y1 + y2

= a1 sin ωt + a2 sin (ωt + δ)

= (a1 + a2 cos δ) sin ωt + a2 sin δ cos ωt.

Let a1 + a2 cos δ = A cos φ and

a2 sin δ = A sin φ

∴ y = A cos φ sin ωt + A sin φ cos ωt.

= A sin (ωt + φ)

3.2 ENGINEERING PHYSICS – I

where A is the amplitude of the resultant disturbance at that point.

Intensity of the resultant disturbance I = A2 = (a1 + a2 cos δ)2 + a2 2 sin2 δ

= a1 2 + a2

2 + 2 a1 a2 cos δ

Case 1: Constructive Superposition:

When the phase difference

δ = 0, 2π , 4 π . . . . = 2 n π , then

A2 = a1 2 + a2

2 + 2 a1 a2 = (a1 + a2)2

This is called the constructive (i.e. additive) superposition or constructive interferencein which the resultant intensity is maximum.

If a1 = a2 , then

A2 = 4 a1 2

Thus the resultant intensity becomes 4 times the intensity of the individual wave.

(a) Constructive Interference (b) Destructive interference

Fig. 3.2

Case 2: Destructive Superposition:

When the phase difference δ = π , 3 π , 5 π = (2n + 1) π , then

A2 = a1 2 + a2

2 − 2a1 a2 = (a1 − a2)2

This is called destructive superposition or destructive interference in which theresultant intensity is minimum.

If a1 = a2 , then A = 0.

Thus using the relation,

phase difference = 2 πλ

× path difference

we can conclude that

WAVE OPTICS 3.3

(i) When the phase difference between two waves δ = 2 n π or the path differencebetween them is n λ, then there is constructive superposition and

(ii) When the phase difference between two waves δ = (2 n + 1) π or the pathdifference between them is (2 n + 1) λ ⁄ 2, then there is destructivesuperposition.

Therefore the phenomenon of interference is nothing but a spatial redistributionof intensities of waves which results in the formation of a fixed pattern of interferencefringes.

3.2.2. Conditions for interference:

H There must be two sources which are coherent in nature (i.e.) They mustbe same wavelength, frequency and amplitude.

H The sources should have constant phase difference (or) zero phase difference.

H The sources should be closer to each other and the distance between thesources and the screen must be larger.

H For constructive interference (i.e.) for bright fringes to appear, the pathdifference between the two waves which are interfering should be equal ton λ .

Path Difference = n λ

H For detractive interference (i.e.) to get dark fringes.

Path Difference = (2n ± 1) λ ⁄ 2

3.3. INTERFERENCE DUE TO REFLECTED LIGHT:

Let us consider a transparent thin film of thickness ‘t’ and refractive index ‘µ’as shown in fig. 3.3. The ray AB incident on the upper surface of the film is partlyreflected along BH and partly refracted along BC. At ‘C’ part of the light is reflectedalong CD and finally emerges out along DK. The path difference between the tworays BH and DK can be calculated. The normals DG to the ray BH and BL to theray CD are drawn.

The angle of incidence is ‘i’ and the angle of refraction is ‘r’. Also produce CDto meet BE produced at ‘P’. Here ∠ BPD = r.

The optical path difference between BH and BK is given by

∆ = Path (BC + CD) in film − Path BG in air.


Fig. 3.3

Since the refractive index for air = 1

We can write ∆ = µ (BC + CD) − BG .... (1)

We know, µ = sin isin r

From ∆ BDG sin i = BGBD

From ∆ BLD sin r = DLBD

∴ µ = BG ⁄ BDDL ⁄ BD

⇒ BGDL

∴ BG = µ (DL) .... (2)

Substituting eqn. (2) in eqn. (1), we can get

∆ = µ (BC + CD) − µ (DL)

= µ (BC + CD − DL)

Since in fig. 3.3. CD = CL + DL, we can write

∆ = µ (BC + CL + DL − DL)

= µ (BC + CL)

Since BC = PC, we can write

∆ = µ (PC + CL)

(or) ∆ = µ (PL) .... (3)

In ∆ BPL cos r = PLPB

(or) PL = PB cos r

WAVE OPTICS 3.5

Here PB = (PE + BE)

= t + t

= 2t

PL = 2t cos r .... (4)

Substituting eqn. (4) in eqn. (3), we get

∆ = µ 2t cos r

(or) ∆ = 2 µ t cos r .... (5)

This eqn. (5) is only the apparent path difference. In the case of reflects light,it doesn’t represent the correct path difference. On the basis in electromagnetic theorywe know that when light is reflected from the surfaces an optically denser medium(air-media) a phase change π equivalent to a put difference λ ⁄ 2 occurs.

The effective path difference between the two reflected rays is given by

∆ = 2 µ t cos r ± λ ⁄ 2

Case (i) Bright Fringes:

If the path difference ∆ = n λ, where n = 0, 1, 2 …, constructive interference

takes place and the film appears bright.

∴ 2 µ t cos r ± λ ⁄ 2 = n λ

2 µ t cos r (2n ± 1) λ ⁄ 2

If this condition is fulfilled, the film will appear bright in the reflected light.

Case (ii) Dark Fringes:

If the path difference ∆ = (2n ± 1) λ ⁄ 2 where n = 0, 1, 2, .... etc., destructive

interference takes place and the film appears dark.

Therefore 2 µ t cos r ± λ2

= (2n ± 1) λ2

2 µ t cos r = 2n λ

2 ±

λ2

+__ λ2

2 µ t cos r = n λ ± λ

(or) 2 µ t cos r = (n ± 1) λ


Since (n + 1) (or) (n − 1) can also be written as an integer, in general we canwrite

2 µ t cos r = n λ

When this condition is fulfilled the film will appear dark in the reflected light.

3.3.1. Interference due to Transmitted Light:

Let us consider a thin transparent film of thickness ‘t’ and refractive index ‘µ’as shown in fig. 3.4.

Fig. 3.4

A ray SA after refraction goes long AB. At ‘B’ it is partly reflected along BCand partly refracted along BR. The ray BC after reflection at ‘C’ finally emergesalong DQ. Here at B and C reflection takes place at the rate medium (medium −air interface). Therefore no phase change occurs. The normal BM to the ray CD andDN to the ray BR are drawn. The optical path difference DQ and BR is given by

∆ = µ (BC + CD) − BN .... (1)

We know µ = sin isin r

From ∆ BND sin i = BNBD

From ∆ BMD sin r = MDBD

∴ kmu = BN ⁄ BDMD ⁄ BD

= BNMD

(or) BN = µ ⋅ MD .... (2)

Substituting eqn. (2) in eqn. (1), we have

∆ = µ (BC + CD − MD)

WAVE OPTICS 3.7

Since CD = CM + MD we can write

∆ = µ (BC + CM)

Since BC = PC, we can write

∆ = µ (PC + CM)

(or) ∆ = µ (PM) .... (3)

In ∆ BPM cos r = PMBP

PM = BP cos r

= (PO + OB) cos r

PM = 2t cos r .... (4)

Substituting eqn. (4) in eqn. (3), we can write

The effective path difference between the transmitted rays given by

∆ = 2 µ t cos r .... (5)

Here it should be remembered that inside the film, reflection at different pointstake place at the surface backed by rarer medium (air) thus no abrupt these change( π ) takes place in this case.

Case (i) Bright Fringes:

For bright fringes, path difference ∆ = n λ

∴ 2 µ t cos r = n λ

Where n = 0, 1, 2, 3....

Case (ii) Dark Fringes:

For dark fringes, path difference ∆ = (2n + 1) λ ⁄ 2

∴ 2 µ t cos r = (2n + 1) λ

2

where n = 0, 1, 2, 3 ....

Thus the conditions for the bright and dark fringes of the transmitted light arejust reversed compared to the conditions for the reflected light.


3.4. FRINGES PRODUCED BY A WEDGE SHAPED THIN FILMS (AIR

WEDGE):

Let us consider two plane surfaces OAand OB inclined at an angle θ andreclosing a wedge shaped air film as shownin fig. 3.5. Let µ be the refractive index ofthe enclosing film in between the twoplanes OA and OB. The thickness of thefilm increases from O to A. When this thinair film is illuminated by themonochromatic light, interference occursbetween the rays reflected from upper andlower surface of the film. The interferingrays do not enter the eye parallel but theyappear to diverge from a point near thefilm.

Let ‘t’ be thickness of the film at a distance x from the edge, for normalincidence, then

The path difference = 2 µ t

If θ is very small then we can write

t = x θ .... (1)

∴ path difference = 2 µ x θ .... (2)

We know the condition for formation of bright fringes due to the reflected light is

2 µ t cos r = (2n + 1) λ2

Here for air m = 1 and cos r = 1

2t = (2n + 1) λ2

Substituting the value of ‘t’ from eqn. (1) we have,

or 2x θ = (2n + 1) λ2

.... (3)

Similarly, we know the condition for dark fringes is 2 µ t cos r = n λ

Since µ = 1 (for air) ; cos r = 1, we have

2t = n λ

Fig. 3.5

WAVE OPTICS 3.9

Substituting the value of ‘t’ from eqn. (1), we have

∴ 2 x θ = n λ .... (4)

Fringe width:

Case (i):

It we consider the dark fringe, then xn is the distance of the nth dark fringe

from the edge and x(n + 1) is the distance of the (n + 1)th dark fringe from the edge,

then from eqn. (4), we can write.

xn = n λ2 θ

.... (5)

and x(n + 1) = (n + 1) λ

20.... (6)

∴ Fringe width (β) which is the distance between any two consecute bright (or)dark fringes can be got by subtracting eqn. (5) from eqn. (6).

∴ x(n + 1) − xn = (n + 1) λ

2 θ −

n λ2 θ

(or) Fringe width β = λ

2 θ .... (7)

Case (ii):

Similarly if we consider any two consecutive bright fringe the fringe with βwill be the same. If x is the distance corresponding to fringes, then

The fringe width, β = x

n =

λ2 θ

.... (8)

Applications:

It is used to find the thickness of the thin piece of paper or wire or hair. Wedgeshaped air film is obtained by inserting the material between the ISS plates.

Eqn. (3) can be rearranged as θ = tx

.


Substitute for ‘θ’ in eqn. (7) we get,

β = x λ2t

t = x λ2 β

.... (9)

Using this equation we can measure the thickness of the given materialaccurately.

3.5. NEWTON’S RINGS:

Circular interference fringes can be produced by enclosing a very thin film ofair or any other transparent medium of varying thickness between a plane glassplate and convex lens of a large radius of curvature. Such fringes were first obtainedby Newton and are known as Newton’s rings.

Theory:

Consider a plano convex lens A O B of large radius of curvature placed overan optically plane glass plate M O N as in figure 3.6(b). In between the bottom ofsurface of the lens A O B and the glass plate surface an air film is enclosed. It canbe easily seen that it is of varying thickness as we proceed from A to O. The thicknessof the air film will be zero at the point O. The points at which the thickness of theair film is the same will lie along circles with O as centre. Therefore the fringesobtained are circular. Let light rays from a monochromatic source be incidentnormally on A B. The ray will be partly reflected back by the lens surface part ofit will travel down and get reflected at the surface M O N. Hence there will beinterference between the two rays one reflected from A and the other from M. Nowsince µ for air is unity and incidence is normal hence the angle of refraction r iszero.

∴ Path difference = 2t.

As one of the rays suffers reflection at a denser medium i.e. at M, a further pathdifference λ ⁄ 2 is introduced.

∴ Total path difference = 2t + λ ⁄ 2.

Therefore the points A and B equidistant from O will ie on a bright ring of diameterA B according as the path difference

2t + λ ⁄ 2 = n λ (or) 2t = (2n − 1) λ ⁄ 2

WAVE OPTICS 3.11

Similarly the points A and B will lie at the centre of a dark ring when the pathdifference

2t + λ ⁄ 2 = (2n + 1) λ ⁄ 2 (or) 2t = n λ

From the geometry of the circle, referring to figure 3.6(b), we have

AD × DB = OD × DE

= OD (2R − OD) = 2 R t − t2 = 2 Rt

where R is the radius of curvature of the lens and t is the maximum thickness of

air film. Since t is very small, t2 can be neglected as compared to 2 Rt.

If d is the diameter of the ring, then A D = DB = d ⁄ 2

∴ d2

4 = 2 R t (or) 2 t =

d2

4 R

If A and B lie on the nth bright ring of diameter dn, then dn

2

4 R = (2n − 1) λ ⁄ 2

and for the points to lie on the mth bright ring we have

dm 2

4 R = (2m − 1) λ ⁄ 2

dn 2 − dm

2

4 R = (n − m) λ

∴ R = dn

2 − dm 2

4 λ (n − m)

Results:

(1) Determination of R and λ:

The diameters dn and dm of any two bright rings are measured with the help

of a travelling microscope. Thus the radius of curvature R of the lower surface ofthe lens is measured knowing the wavelength of the monochromatic source. Or thewavelength of the monochromatic source is determined knowing the value of R. Usinga spherometer the value of R can be determined.


(2) Use of a plane mirror instead of glass plate:

If the glass plate is replaced by a plane mirror the fringes will disappear andwe shall get an uniform illumination. It is because the intensity of light reflectedfrom the surface of the mirror is very large as compared to the intensity of lightreflected from the curved surface of the air film. Hence the intensity at a dark fringe(difference of two intensities) and at the bright fringe (sum of the two intensities)will be practically the same.

(3) White light fringes:

The diameter of the Newton’s ring is proportional to the square root of thewavelength. Therefore if light is not monochromatic, but consists of more wavelengthslike white light there will be superposition of the rings due to different wavelengths.As a result if white light is used only a few rings will be observed and these willbe highly coloured except the central spot which even in this case will be dark sincethe path difference at the centre is only λ ⁄ 2.

3.5.1. Experimental set upto Observe Newton’s Rings:

Fig. 3.6: Experimental setup to observe Newton’s rings

S is a monochromatic source of light (fig.3.6(a)). L1 is a converging lens placed

in such a way that a parallel beam emerges from it. G2 is a glass plate kept at 45°

to the horizontal. The parallel beam from the lens L1 is incident on this plate and

gets reflected vertically down and reaches the lens L placed over an optically plane

glass plate G1. The convex lens L has a large radius of curvature (R −~ 1 metre). A

WAVE OPTICS 3.13

thin film of air is enclosed in between G1 and L. The rays reflected by the bottom

surface of the lens and the top surface of the plane glass plate ‘G1’ interfere and

produce Newton’s rings. The fringes are circular because the air film is symmetricalabout the point of contact of the lens L and the plane glass plate G1. The fringes

can be observed by means of a low power travelling microscope M and their diameterscan also be measured using this microscope.

(a) Determination of the radius of curvature of a given convex lens:

The given convex lens is placed over the optically plane glass plate G1 and the

position of the microscope is adjusted so that the centre of the ring system is observedin the field of view of the microscope. A good number of clear rings must be availableon either side of the central spot. By proper adjustment of the arrangement andsource this can be effected. It is seen that the cross-wire of the eyepiece of travellingmicroscope is tangential to the rings. The microscope is moved to one side from thecentre and fixed at a position at which the cross-wire is tangential to the fourthbright ring. Let the order of the ring be ‘n’. The reading of the position of themicroscope is noted on the horizontal scale. The microscope is moved away from thecentre further and when the vertical cross-wire is again tangential to the (n + 3)bright ring the reading is again found. Thus by moving the microscope the readingscorresponding to n + 6, n + 9, n + 12, n + 15, etc., rings are taken. This is repeated onthe other side of the central spot. For a given order of the ring, the difference betweenthe readings on the right and left gives the diameter of the ring of that order. Thereadings are tabulated as below:

Order of

Ring

Reading of microscope Difference =

Diameter ‘d’d2 dn + 15

2 − dn 2

Left Right

nn + 3n + 6

.

.

.

Average =

Now R = dn + 15

2 − dn 2

4 × 15 × λ

If sodium light is used, then λ = 5893 × 10− 10 metre and R can be found out by

substituting the average value of dn + 15 2 − dn

2 in the above formula.


Note:

(i) If R is known (using a spherometer) the wavelength of the light can be foundout using the above formula.

(ii) If focal length of the lens is known then the refractive index of the materialof the given lens can be found using the formula

1f = (µ − 1)

1R1

+ 1

R2

Here R1 and R2 are the radii of curvatures of the given convex lens and their

values can be determined using Newton’s rings.

(iii) If the refractive index of the material of the given lens is known, then thefocal length of that given lens can be found determining R1 and R2

experimentally.

(b) To determine the refractive index of a liquid supplied in small quantity:

The experiment as detailed above is performed first with air film. The average

of dn + 15 2 − dn

2 is found. Let this be S. A small drop of given liquid is placed over

the glass plate. The given liquid is uniformly spreaded by turning the lens. Care istaken to see that there is no air bubble in the liquid film so formed. Now the

experiment is repeated as before and the mean value of dn + 15 2 − dn

2 is noted as S1.

Now when air film is present,

dn + 15

2 − dn 2 air

= S = 4 × 15 × R × λ

When the liquid is used

dn + 15

2 − dn 2 liquid

= S1 = 4 × 15 × R × λ

µ

∴ µ = SS1

= (dn + 15

2 − dn 2)air

(dn + 15 2 − dn

2)liquid

Thus the refractive index of a liquid is found out.

WAVE OPTICS 3.15

3.6. MICHELSON’S INTERFEROMETER:

Principle:

The amplitude of the light beam from a light source is divided into two partsof equal intensity by partial reflection and refraction. These two beams are sent intwo directions at right angles and are brought together after reflection from planemirrors to produce interference fringes.

Fig. 3.7: Michelson’s Interferometer

Construction:

It consists of two highly polished front silvered plane mirrors M1 and M2 and

two plane parallel glass plates G1 and G2 of the same thickness. The mirror M1 is

mounted on a carriage and can be moved exactly parallel to itself with the help of

a micrometer screw. The least count of the micrometer screw is about 10− 5 cm. Themirrors M1 and M2 are provided with leveling screws at their back. Using these

screws the mirrors can be made perfectly perpendicular to the direction of the twobeams. The plate G1 is semisilvered at the back so that the incident beam is divided

into a reflected and transmitted beam of equal intensity. The plates G1 and G2 are

held parallel to each other and are inclined at an angle of 45° to the mirrors M1

and M2 which are mutually perpendicular to each other. The interference fringes are

observed in the field of view of the telescope ‘T’.


Working:

Light from the source S is rendered parallel by means of a collimating lens and

is made to fall on the plate G1. It is partly reflected at the back surface of G1 along

AC and partly transmitted along AB. They ray AC is received by the plane mirror

‘M1’ normally, so that it is reflected along the same path and emerges out along AT.

The transmitted ray AB is received by mirror M2 normally and it is reflected along

the same path and then moves along AT after reflection at the back surface of G1.

Thus two beams along AT are produced from a single source by division of the

amplitude. These two beams produce interference under suitable conditions. Since

the ray starting from the source and suffering reflection at the mirror M1 traverses

the glass plate G1 thrice where as the ray reflected from the mirror M2 traverses

the glass plate G1 only once, a compensating glass plate G2 which is exactly similar

to G1 in thickness and in size is introduced in the path AB parallel to glass plate

G1.

Adjustments:

First the distance G1 M2 must be made very nearly equal to G1 M1 and then

the image of M2 must be made parallel to M1. If this is not done the angle between

the two faces of the air film will be large and extremely narrow fringes will be

formed which may be quite invisible.

This can be carried out by the following way: A tin sheet with a fine hole in

it is placed in front of the source ‘S’ just opposite to the bright part of the flame.

The hole and the flames are adjusted in line with the centre of the glass plates and

mirror M2. A lens is then placed between G1 and tin sheet. A plane mirror is also

placed between G1 and the lens normally. The position of the lens is adjusted till

the image of the hole falls back on the tin sheet very close to the hole. The light

beam from the lens is now rendered parallel. After removing the plane mirror if we

look in the telescope four images of the hole will be seen, one pair formed by reflection

from M1 and other from M2. Each pair consists of one bright image due to the

reflection from the silvered surface of G1 and one rather faint image due to reflection

from the unsilvered surface.

WAVE OPTICS 3.17

The adjustment of the mirrors is made till the images coincide two by two. Ifthe tin sheet is removed and the paths are exactly equal the field of view will bedark. A slight motion of M1 parallel to itself will produce circular fringes. By tilting

the mirror M2 slightly the fringes can be made straight.

3.6.1. Theory behind the formation of different kinds of fringes:

If we look in the direction of the mirror M1, the eye will see the mirror M1

directly and also a virtual image of M2 is formed by reflection in the glass plate

G1. Therefore one of the interfering beams come from M1 and other appears to come

by reflection from the virtual image of mirror M2 which appears to be at M2′ . The

system is therefore similar to the interference from an air film enclosed between the

two mirrors M1 and M2′ . The air film is of uniform thickness if M2 is exactly

perpendicular to M1 and G1 is inclined at an angle of 45° to the direction of the

beam. The two interfering beams appear to come to the eye from two virtual imagesS1 and S2 of the point source S as shown in figure. Let us call S1 and S2 are the two

virtual sources. The distance between the virtual sources is 2d. As the ray AB comingafter reflection from M2 suffers reflection at the silvered surface of glass plate G1,

additional path difference of λ ⁄ 2 is introduced between the two.

Therefore total path difference = 2d + λ ⁄ 2.

If the eye looks in the direction of E making an angle ‘r’ with the normal tothe mirrors then the total path difference = 2d cos r + λ ⁄ 2.

For maximum intensity in the fringes

2d cos r + λ ⁄ 2 = nλ

where n = 0, 1, 2, ..... etc.

Case 1:

When M2′ coincides with M1 i.e. air paths AC and AB are equal, then path

difference is only λ ⁄ 2 and the field of view is totally dark.


Fig. 3.8: Formation of interference fringes

Case 2:

When M1 is moved either way parallel to itself widely spaced circular fringes

are produced. If the path difference is increased further, the width of these fringes

Fig. 3.9: Formation of different types of fringes

WAVE OPTICS 3.19

decreases. If the source is an extended one and the light used is monochromatic fora given value of n, r is constant, therefore the locus of a fringe is a circle. Hencewe can see the circular bright and dark fringes with dark spot at the centre.

Case 3:

When the mirror M1 and M2′ are not exactly parallel the air path between the

two mirrors is a wedge shaped. With such a film, the locus of points of equalthickness is a straight line parallel to the edge of the wedge. Therefore the fringesare straight ones.

3.6.2. Applications of Michelson’s interferometer:

(1) Measurement of wavelength of a monochromatic source:

When the mirror M1 is moved through a distance λ ⁄ 2 the path difference

changes by λ and position of a particular bright fringe is taken by the next brightfringe. If ‘n’ is the number of fringes that moved across the field of view when themirror is displaced through a distance “L” then

n λ ⁄ 2 = L

λ = 2Ln

The interferometer is adjusted to obtain circular fringes. The position of the mirrorM1 is adjusted till a particular bright fringe in the field of view of the telescope with

its centre coinciding with the cross-wire. Now note the micrometer screw reading.Now slowly move the mirror M1 and count the number of fringes ‘n’ that moved

across the field view of the telescope. This may of the order of 20 or 30 fringes. Nownote the micrometer screw reading. Find the difference between initial and finalmicrometer screw readings which will give the value of ‘L’. Hence one can determine

the wavelength using the formula λ = 2Ln

very accurately upto five decimal places in

centimeter.

(2) Determination of thickness of a transparent medium or ‘µ’:

If we introduce a transparent medium with thickness ‘t’ and refractive index ‘µ’in between G2 and M2, then the increase in optical path due to the introduction of

this transparent medium is given by (µ − 1) t. Here if we know µ, then one cancalculate ‘t’ (or) vice versa in the following way: First adjust the Michelsoninterferometer for white light fringes such that dark straight fringe coincides with


the crosswise of the telescope. In this position, the paths AB and AC are equal. Nowthe thin film is introduced between the glass palate G2 and the mirror M2 such that

light passes through this normally. Then the increase in optical path is (µ − 1) t. Dueto that the position of the dark fringe is shifted to somewhere else. Now note themicrometer screw reading. Adjust M1 using micrometer screw such that the position

of dark fringe coincides with the crosswise of the telescope and note the micrometerscrew reading. Find the difference of the two readings. Let it be ‘L’ metre.

Now (µ − 1) t = L

Therefore t = L

(µ − 1) metre, if we know µ and L

(or) µ = L + t

t if we know t and L.

(3) Standardisation of a metre in terms of a standard wavelength:

We know that n λ2

= L where n is the number of fringes moved across the field

of view when the mirror M1 is moved through a distance ‘L’. So one can calculate

the distance in terms of standard wavelength.

But we cannot directly standardise the metre since the movement of 1 metreis not possible because (1) we can get fringes for a path difference of 24 cm. Thiscorresponds to a movement of mirror M1 = 12 cm. Further (2) counting the fringes

for 1 metre distance is somewhat difficult since for 10 cm distance, 6 × 105 fringesshifted. So we must modify the Michelson’s interferometer to standardise a metre.

3.6.2.1. Applications of Michelson Interferometer:

Michelson interferometer can be used to determine

(i) The wavelength of the monochromatic light.

(ii) The difference between the two neighbouring wave resolution of the spectrallines.

(iii) The refractive index and thickness of the various thin materials and

(iv) For the measurement of the standard metre in terms of light. First threeapplications are discussed here.

WAVE OPTICS 3.21

3.7. DIFFRACTION:

Phenomenon of bending of light waves around comers and their spreading intothe geometrical shadow of an object is called diffraction.

To measure the diameter of a wire: A narrow vertical slit, a narrow wire and amicrometer eyepiece are arranged on an optical bench. The centres of thesecomponents are arranged to lie in the same horizontal line. When the slit isilluminated with monochromatic light, interference fringes of equal width are seenwithin the geometrical shadow of the wire. The fringe width β is measured. D, thedistance between the slit and the cross-wires is also measured. λ is known.

∴ The diameter of the wire = d = D λ ⁄ β.

Example: If the diameter of a wire be 2 × 10− 4 m, calculate the separation betweenthe fringes formed on a screen placed at 0.5 m from the wire. The light source usedhas a wavelength of 5 × 10 m and is placed at a finite distance from the wires.

Solution:

The two edges of the wire act as if two small coherent sources are situatedthere. They, therefore, produce equidistant interference bands.

Diameter of the wire − Distance between coherent sources = 2 × 10− 4 m

D = distance between wire and screen = 0.5 m ; λ = 5 × 10− 7 m ; β = ?

β = D λd

= 0.5 × (5 × 10− 7)

2 × 10− 4 = 1.25 × 10− 3 m.

3.7.1. Fraunhofer Diffraction at a single slit:

Fig. 3.10


Let a parallel beam of monochromatic light of wavelength λ be incident normallyupon a narrow slit of width AB = a (Fig. 3.10). The diffracted light is focused by aconvex lens on a screen placed in the focal plane of the lens. Th diffraction patternobtained on the screen consists of a central bright band, having alternate dark andweak bright bands of decreasing intensity on both sides.

As a plane wavefront is incident on the slit AB, each point on the wavefrontbecomes a source of secondary wavelets. The rays diffracted along the direction ofincident rays are focused at O. The point O is optically equidistant from all pointson the slit AB. Therefore, all the secondary wavelets from AB reach O in the samephase. Hence there is maximum intensity at O.

Let us consider the intensity at any point P1 above O. The secondary waves

travelling at an angle θ with the normal are focused at P1. Let AC be drawn

perpendicular to BC. The optical lengths of all the rays from the plane AC to thefocal point P1 are the same. The path difference between the secondary waves

originating from extreme points A and B is

BC = AB sin θ = a sin θ .

Let this path difference be one wavelength. If now we imagine the aperture ABto be divided into two equal halves, then the wavelets from corresponding points ineach half will differ in phase at P1 by λ ⁄ 2. They would mutually interfere and cancel

out each other’s effect thus producing the first minimum at P1.

∴ BC = a sin θ = λ

or sin θ = λa

or θ = λa

( . . . θ is very small)

Hence the first minimum on either side of O will occur in a direction given by

θ = λa

.

Suppose for another point P2, the path difference BC is 2 λ. Now the slit can

be supposed to be divided into four equal parts. The rays from corresponding pointsseparated by a distance a/4 in the two halves of each half of the slit will have apath difference of λ ⁄ 2. These rays will mutually interfere and cancel out each other.A second minimum, therefore, occurs at P2 in a direction θ given by

BC = a sin θ = 2 λ

In general the various secondary minima will occur when the path differencebetween the extreme rays is an even multiple of λ ⁄ 2 or an integral multiple of λ .

WAVE OPTICS 3.23

∴ For minima BC = a sin θ = 2n (λ ⁄ 2) = n λ .... (1)

Besides the central maximum at O, there are secondary maxima which lie inbetween the minima on either side of the central (principal) maximum. These aresituated in directions in which the path difference BC is an odd multiple of λ ⁄ 2.

∴ For secondary maxima, BC = a sin θ = (2n + 1) λ ⁄ 2 .... (2)

The intensity of these secondary maxima is much less and falls off rapidly aswe move outwards.

Thus the diffraction pattern due to asingle slit consists of a central brightmaximum at O followed by secondary maximaand minima on both the sides. Fig. 3.11

shows the variation of intensity with distancefrom the centre of the central maximum.

Let f be the focal length of thecollimating lens. Let x = distance of the firstminimum from O. If the collimating lens isvery near the slit or the screen is far awayfrom the lens, then

sin θ = λa

= xc or x =

f λa

∴ Width of thecentral maximum

= 2x =

2 f λa

Hence the width of the central maximum is proportional to the wavelength oflight λ.

3.8. RESOLVING POWER OF OPTICAL INSTRUMENTS:

The capacity of an instrument to show two close things separately is called‘resolution’. The ability of an optical instrument to produce distinctly separate spectral

lines of light having two or more wavelengths very close to each other or to resolve

the images of two nearby points is called its resolving power.

Rayleigh’s criterion for resolution: Rayleight proposed the following criterionfor resolution:

Fig. 3.11


Two spectral lines of equal intensities are just resolved by an optical instrument

when the principal maximum of the diffraction pattern due to one falls on the first

minimum of the diffraction pattern of the other.

Consider the intensity distribution curves of two wavelengths λ and λ + d λ(Fig. 3.12)

Fig. 3.12

The principal maximum of one coincides with the first minimum of the other.The resultant intensity curve shows a distinct dip in the middle indicating thepresence of two different wavelengths. The lines are said to be ‘just’ resolved.

A grating or a prism gives us spectral resolution. If an optical instrument ‘just’resolves two spectral lines of wavelengths λ and λ + d λ, then λ ⁄ d λ is a measure ofthe ‘resolving power’ of the instrument.

3.9. DIFFRACTION GRATING:

Fig. 3.13(a) Double slit interference factor (b) Double slit diffraction factor and(c) their product

WAVE OPTICS 3.25

A diffraction plane transmission grating is an equivalent to a large number ofequidistant slits of equal width. Further this is an extension of double slit diffraction.Each slit forms its diffraction. Maxima and minima of the various slits overlapresulting in an intensity, proportional to the square of the number of slits, at themaxima. Figure 3.14 compares the intensity patterns for N = 2 and N = 5. From thisfigure, we know that increasing N, (a) does not change the spacing; between the(principal) interference fringe maxima provided d, the spacing between the centresof the adjacent slits and λ remain unchanged (b) sharpens the (principal) maximaand (c) introduces small secondary maxima between the principal maxima.

A principal maximum will occur when the path difference between rays fromadjacent slits is equal the integral multiples of the wavelength of light.

i.e. d sin θ = m λ where m = 0.1, 2 .... (principal maxima)

where m is called the order number. Therefore the locations of the principal maximaare thus determined only by the ratio λ ⁄ d and are independent of N. Further theintensity at principal maxima is proportional to the square of the number of slitsand the intensity at the secondary maxima decreases and they become more crowdedas the number of slits is increased. Generally if there are N slits, then there areN − 1 minima and N − 2 secondary maxima between two consecutive principalmaxima.

Fig. 3.14: Diffraction patterns due to multiple slits


3.9.1. Grating:

A grating is a plane sheet oftransparent material like glass with alarge number of opaque rulings madeon it. There will be about 600 rulingsper mm, on the grating. Thetransparent spaces and the rulingswill be equally spaced. Thusdiffraction due to a planetransmission grating is an equivalentto a large number of equidistant slitsof equal width. In the grating, thespace between the rulings will betransparent and they act as slits.

Theory:

A B C D E F represents a section of the grating normal to the plane of thediagram. BC and DE are the slits allowing light to be transmitted and AB, CD andE F are the opaque rulings (fig. 3.15). The width of a ruling and a space togetheris called the grating element.

If the ruling is of width ‘a’ and the space is of width ‘b’ then (a + b) gives thegrating element. As a plane wavefront is incident normally on a grating the particlesalong A B C D E F will act as centres of disturbances and secondary wavelets fromthem will proceed in all directions. The secondary wavelets at the ruling are stoppedand the wavelets through the transparent spaces or slits alone travel. Let us considerthe transmission of these wavelets in a direction inclined θ to the direction of theincident beam of light (θ is called angle of diffraction).

BE1 is a line drawn perpendicular to the direction of the diffracted light. B and

D are corresponding points of the first two slits BC and DE. The wavelets startingfrom B and from D have a path difference given by DD1.

DD1

BD = sin θ

∴ DD1 = (a + b) sin θ , since BD = (a + b)

If DD1 is equal to an odd multiple of λ ⁄ 2 then the wavelets from BC and DE will

cancel each other. Since the entire grating can be considered to be made up of anumber of pairs of adjacent slits like this, the net intensity obtained by focussingall these waves (by a lens L) at P will be zero.

Fig. 3.15: Diffraction grating

WAVE OPTICS 3.27

But if DD1 is equal to an even multiple of λ ⁄ 2 then the wavelets from adjacent

slits will reinforce each other. Intensity of P will the be maximum. i.e. P will bebright if DD1 = (a + b) sin θ = 2 n λ ⁄ 2 = n λ (or) (a + b) sin θ = n λ

If there are N rulings per metre then N (a + b) = 1 or sin θ = N n λ is thecondition to obtain brightness at P. In the above discussion light passes through thegrating and hence the latter is called a transmission grating. Diffraction can also beobserved when light is reflected from the surface of a grating. Such gratings arecalled reflection gratings and usually have concave surfaces.

Results:

1. As one proceeds from the central position to the exterior θ will increase.There will be bright bands when sin θ1 = N.1.λ ; sin θ2 = N. 2 λ etc. if the light is of

a single wavelength. But if the light is composite then λ changes from colour tocolour so that even when n = 1,

sin θ1 = N.1 λ1, for one colour and

sin θ2 = N.2 λ2, for another colour, and so on.

or we get a series of bright colours at different angles or using a diffraction gratingthe composite light can be split into its constituent colours. This is called a gratingspectrum and ‘n’ is called order of the spectrum.

For a given colour of wavelength λ,

sin θ = λ N when n = 1.

By measuring θ, the angle of diffraction and knowing N, the number of rulingsper metre, the wavelength ‘λ’ can be calculated.

2. Considering the expression (a + b) sin θ = nλ, suppose (a + b) is less than λ.Then n has to be less than sin θ. That is n has to be zero because sin θ can neverbe greater than one. Hence there will be no diffraction by transmission. This is whathappens when light passes through a crystal. But when X-rays pass through a crystalthey are diffracted because λ of X-rays is smaller than (a + b).

(3) Maximum number of orders in a grating:

The maximum angle of diffraction possible is 90°. If there are 600000 lines/metreand a source of light of wavelength 5000 Å is used then

sin θ = 1 = 600000 × n × 5000 × 10− 10

Therefore n is greater than 3 but less than 4. So it is not possible to get morethan 3 orders with such a grating.


(4) conditions for absent spectra:

The conditions for the principle maxima are

(a + b) sin θ = n λ ... (i)

Similarly the condition for minimum intensity is given by

a sin θ = λ ... (ii)

When the conditions (i) and (ii) are simultaneously obeyed, the beams from allthe slits reinforce each other but the resultant intensity is zero. Hence the spectrumis absent.

Dividing (i) by (ii) we have

a + ba

= n

The condition for any spectrum to be absent can be obtained by giving variousvalues of n.

That is if the second order spectrum is to be absent substituting n = 2, we havea + b = 2a

or a = b.

i.e. the width of the transparency should be equal to the width of the opacity.

(5) Resolving power of a grating:

Resolving power of a grating is defined as the ratio of the wavelength of a linein the spectrum to the least difference in the wavelength of the next line that canjust be seen as separate. Thus it is equal to λ ⁄ dλ where dλ is the difference betweenthe two lines of wavelengths which can just be seen as separate. So if two lines ofwavelengths λ and λ + dλ are to be just resolved the first minimum accompanyingthe principal maximum of λ should fall on the principal maximum of λ + dλ.

The nth order principal maximum for a plane diffraction grating for normalincidence is given by

(a + b) sin θ = n λ

where (a + b) is the grating element and θ is the angle of diffraction correspondingto the nth order spectrum.

WAVE OPTICS 3.29

3.9.2. Resolving Power of a Telescope:

The telescope is used to see distant objects. Therefore the amount of detailswhich the telescope reveals depends on the angle subtended at its objective by thetwo point objects. The RP of a telescope is therefore defined as the reciprocal of thesmallest angle subtended at the objective by the two distant object points which canbe just seen as separate ones through the telescope.

Let D be the diameter of thetelescope objective AB (Fig. 3.16).Two distant point objects O and O′subtend an angle d θ at theobjective. The image of each objectis a diffraction pattern. P1 and P2

are the positions of the centralmaxima of the two images. Accordingto Rayleigh, the two objects O and O′will be just resolved when the central maximum of the diffraction pattern of onecoincides with the first minimum of the diffraction pattern of the other.

The path difference between AP1 and BP1 is zero. Hence they reinforce with one

another at P1. Thus P1 corresponds to the position of the central maximum of the

first image. Let ∠ P2 AP1 = d θ.

The path difference between BP2 and AP2 = BC

Consider ∆ ABC.

BC = AB sin d θ = AB ⋅ d θ = D ⋅ d θ (for small angles)

If D ⋅ d θ = λ then P2 corresponds to the first minimum of the first image. But

P2 is also the central maximum of the second image. Thus, Rayleigh’s condition of

resolution is satisfied if

D ⋅ d θ = λ or d θ = λ ⁄ D .... (1)

This condition holds good for rectangular apertures.

Fig. 3.16


For circular apertures, Airy showed that

d θ = 1.22 (λ ⁄ D) .... (2)

The reciprocal of d θ measures the RP of the telescope.

∴ RP = 1

d θ =

D1.22 λ

.... (3)

RP depends upon the diameter of the objective and the wavelength of light used.

RP ∝ D and RP ∝ 1 ⁄ λ. Hence RP can be increased by using objectives of largediameters.

Let r be the radius of the first dark ring and f, the coal length of the objective.Then,

d θ = rf =

1.22 λD

or r = 1.22 f λ

D.... (4)

Eq. (4) shows that if f and λ are small, and D is large, then the radius of the centralbright disc is small. Thus the diffraction pattern will appear sharper and the angularseparation d θ between two just resolvable point objects will be smaller. So the RPof the telescope will be higher.

Relation between Magnifying Power and Resolving Power of a Telescope:

The magnifying power of a telescope is

M = Diameter of the objectiveDiameter of the exit pupil

= Dd

The magnifying power is said to be normal if diameter of the exist pupil is

equal to the diameter de of the pupil of the eye. Hence normal magnifying power

= Dde

Limit of resolution of thetelescope objective of diameter D

= d θ =

1.22 λD

Limit of resolution of the unaided eye = d θ′ = 1.22 λ

de

WAVE OPTICS 3.31

∴ Limit of resolution of the eye

Limit of resolution of the telescope =

d θ′d θ

= 1.22 λ ⁄ d1.22 λ ⁄ D

= Dde

= Normal magnifying power of the telescope.

Thus, (normal magnifying power of telescope) × (its limit of resolution) = limitof resolution of the unaided eye.

*********


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