engineering physics-01.p65

1.1 INTRODUCTION

The field of optics (study of light) is divided into three kinds:(a) Geometrical optics (Macroscopic optics) which is concerned with the behaviour of light on

a large scale (macro) and is treated by the method of light rays.(b) Physical optics (Microscopic optics) which is concerned with the study of the finer details

of light and involves theory of waves.(c) Quantum optics which is concerned with the interaction of light with the matter and its

treated by the method of quantum mechanics.Light is propagated (transported) in a straight line as long as there is no change of the medium or

no change in the properties of the same medium and is called a ray. Actually it is not possible toseperate a ray from a light wave. As we are concerned with the physical optics, which is required toexplain the finer details of light, we have to consider the wave picture of light.

The light is considered to be transverse wave which is defined as: when a wave passes in adirection the particles move at right angles to the direction of the propagation of the wave. Whenparticles move up and down in simple harmonic motion, a series of waves will be produced in whichthe displacement of the wave is proportional to the sine of the S.H.M. generating angle ‘�’

i.e., y ��sin �= a sin �

= a sin λπ2

Such waves are called sine waves and is shown in the Fig. 1.1.

��

2 Engineering Physics

Fig. 1.1

Here ‘a’ is the amplitude (the maximum displacement of the wave),‘�’ is the wavelength (lengthof complete wave from crest to crest or from trough to trough)

The wavelengths of visible light is in the range of 4000 Å to 7200 Å (i.e., 4 to 7.2 × 10–5 cm).Faraday and Maxwell showed that light is composed of electric and magnetic fields at right angles toeach other and again these fields are at right angles to the direction of propagation of light. This can beexplained as follows:

When a point charge is at rest, it produces electric field )(E�

and when it is in motion, the

magnetic field )(B�

is also produced is addition to the electric field. Further, if charge oscillates withsimple periodic motion an electromagnetic wave is produced like shown in the Fig. 1.2.

Fig. 1.2

Hertz demonstrated the existance of such electromagnetic waves.

InterferenceNewton could not observe the interference effect, so he held that light must be particle like in naturewhereas Huygen was able to observe the interference effect so he held that light must be wave-like innature. To explain the wave propagation, Huygen established a principle saying that “every point alonga wavefront serves as a source of secondary spherical wavelets. He also assured that the intensity of thespherical wavelts is not uniform in all directions but varies continuously from a maximum in theforward direction to a minimum of zero in the backward direction.

Wave Optics 3

For light waves, due to various process of emission, one cannot observe interference between thewaves from two independent sources although the interference does takes place. Thus, one tries toderive the interfering waves from a single wave so that the phase relationship is maintained. Themethods to achieve this can be classified under two broad categories.

Under the first category, a beam is allowed to fall on two closely-spaced holes/slits and the twobeams emerging from the holes interfere. Division of wavefront (E.g. : Young’s double slit experiment(or) Bi-Prism.

In the second category, a beam is allowed to fall on two or more reflecting surfaces/mediums andthe divided reflected beams interfere.

Division of amplitude (E.g. : Newton's rings experiment).

1.2 INTERFERENCE OF LIGHT

If two waves of the same frequency travel in approximately the same direction and have a phasedifference that remains constant with time, they may combine so that their energy is not distributeduniformly in space but is a maximum at certain points and a minimum (or zero) at other points. Thiseffect is called the Interference.

This was first demonstrated by Thomas Young in 1801 and wave theory of light was experimentallyproved. He was the first person to deduce the wavelength value from this experiment.

Young in his original experiment allowed sunlight to fall on a pinhole So punched in a screen Aas shown in Fig. 1.3.

Fig. 1.3 Young’s double slit experiment.

At a considerable distance away, the light is allowed to pass through two pinholes S1 and S2. Thetwo sets of spherical waves emerging from the two holes interfered with each other in such a way as toform a symmetrical pattern of varying intensity on the screen C.

To be convenient, at a later stage, the pin holes were replaced by narrow slits and a source givingmonochromatic light (light of a single wavelength ��) was used. And the condition a ��is fulfilled,where ‘a’ is the slit width. If the circular line (see Fig. 1.2) represent crests of waves, the intersectionsof any two lines represent the arrival at those points of two waves with the same phase or with phasesdiffering by a multiple of 2�� radians. Such points are therefore those of maximum disturbance orbrightness.


That means if crest falls on crest and trough falls on trough ��disturbance add upBut if crest falls on trough or trough falls on crest ��disturbances cancels outIn interference, we get equispaced fringes (dark and bright bands). This interference phenomenon

is not limited to light waves but applicable to all other kinds of waves like sound waves and otherelectromagnetic waves.

1.3 YOUNG'S DOUBLE SLIT EXPERIMENT AND MAXIMA AND MINIMA CONDITIONS

Let us now analyse Young’s experiment quantitatively assuming that the incident light consists of asingle wavelength only. In the figure shown here P is an arbitrary point on the screen. It is at a distanceof r1 and r2 from the narrow slits S1 and S2, respectively. Draw a line from S2 to b in such a way thatthe lines PS2 and Pb are equal. If d, the distance between the two slits, is much smaller than the distanceD between the slit system and the screen S2, b is almost perpendicular to both r1 and r2. This means that

angle bSS 21ˆ is almost equal to angle OAP ˆ and is marked as ��(d << D and r1|| r2).

The rays arriving at O from S1 and S2 have same path lengths and there is no phase difference.

Fig. 1.4

The two rays arriving at P from S1 and S2 are in phase at the source slits, both being derived fromthe same wavefront in the incident plane wave. Because the rays have different optical path lengths,theyarrive at P with a phase difference. The number of wavelengths contained in S1b (path difference),determines the nature of the interference at P.

To have maximum (brightness) at P, S, b must contain an integral number of wavelengths, orS1b = m �� where m = 0, 1, 2, . . . . .but S1b = d sin �� (from the triangle S1bS2S1)∴ d sin � = m�� gives maxima. ...(1.1)

The central maximum is described by m = 0 and in between two maxima there should be aminima.

∴ S1b = d sin � = λ

+

2

1m gives minima ...(1.2)

where m = 0, 1, 2, . . . . .

Wave Optics 5

Now let us define the terms, phase and coherence. The term phase tells us what fraction of acomplete vibration (oscillation) the particle has executed at a given instant.

And phase difference ��path difference� ��(x2 – x1)

or � = K (x2 – x1)

= )(2

12 xx −πλ

...(1.3)

Fig. 1.5

1.4 COHERENCE

If the slits S1and S2 are illuminated by two completely independent light sources, no interferencefringes are observed. The phase difference between the two beams arriving at P varies with time in arandom way. At one instant the wave from source S2 may be in phase with that from source S1 ; but ina very short time (10–8sec), this phase relation may change from reinforcement to cancellation. Thesame random phase behaviour holds for points on screen, with the result that the screen is uniformlyilluminated. If instead, the light waves that travel from S2 and S1 to P have a phase difference ��thatremains constant with time, the two beams are said to be coherent.

Fig. 1.6


In the case of coherent waves, combine amplitudes vectorially and square the resulting amplitudeto obtain a quantity proportional to Luminous Intensity.

In the case of incoherent waves, square the individual amplitudes to obtain quantity proportionalto individual intensity and add the individual intensities.

1.5 INTENSITY IN YOUNG'S DOUBLE SLIT EXPERIMENT

We know that light wave is electromagnetic in nature and is represented by electric and magnetic field

vectors ( E�

and B�

). Let us assume that the electric field components of the two waves from slits S1 andS2 vary with time at point P (Fig. 1.7) as

E�

1 = Eo sin �t ...(1.4)

and 2E�

= Eo sin �t + �) ...(1.5)

where �� is the angular frequency and is equal to 2��and ��is the phase difference between the twowaves which depends on location of P which is described by �.

Eo is the initial amplitude.Let us assume that the slits are narrow (a < < �) and illuminates the central portion uniformly.But the resultant wave disturbance at P on the screen is found and is given by

21 EEE +=�

, and using eqns. (1.4) and (1.5), we have

= Eo sin �t + Eo sin (�t + �) = Eo [sin �t + sin (�t + �

Fig. 1.7

Using the trigonometric relation, sin B + sin C = 2 sin

−

+

2cos

2

BCCB we have

ω−φ+ω

φ+ω+ω=

2cos

2sin2

ttttEE o

�

Wave Optics 7

= 2Eo [sin (�t +��cos ��

��2Eo cos �/2 sin �/2) ��(put ��=��and 2 Eo = Em)

= Em cos ��sin (�t + �)

� � E�

�� )(sin β+ωθ tE

� E��= Em cos ��Eo cos � ��

where��is the amplitude of resultant wave disturbance which determines the I of interference fringes.Em is the maximum possible amplitude.We knew in the case of coherent waves, the intensity is proportional to square of the amplitude

i.e., I ��E2 where E is the electric field strength.If I��is the intensity of resultant wave at P and Io is the intensity that a single wave acting alone

would produce, thenI� ��E�

2 and I� ��Eo2

� 2

2

θ

θθ

==

kEI

kEI

o or

2

= θθ

oo E

E

I

I

�oI

Iθ= (2 cos ��from equation (1.6)

= 4 cos2 �

(or) I� = β2cos4 oI

(or) I� = Im cos2 � ...(1.7)

To compute I��as a function of �� we substitute ��

φ value in terms of ��from the relation

Phase difference ��Path difference (eqn. 1.3)

i.e., � ��

λπ2

d sin � ��

or � = �

φ =

λπ

d sin � ...(1.9)

The intensity pattern for the double slit interference is shown in the Fig. 1.8.

Fig. 1.8 Energy distribution in Young’s experiment.


1.6 BIPRISM - FRINGEWIDTH

After Young’s double slit experiment, objection was raised that the bright fringes may be due to somemodification of the light by the edges of the slits and not true interference. Thus wave theory of lightwas still questioned.

Fresnel brought forward several new experiments in which the interference of two beams of lightwas proved. One of them is the Fresnel bi-prism experiment.

The arrangement is as shown in the Figure 1.9. The bi-Prism consists of two prisms attachedback to back. The angle at the edges of the bi-prism is of the order of 30' and the other angle in the bi-prism is of the order of 179º. If a monochromatic source is placed in front of the bi-prism, the lightspreads out in the form of two beams which superimpose one over the other. Here the interferencecondition has been fulfilled. The two beams looks like originating from the imaginary source positionsS1 and S2. The location of S1 and S2 can be obtained by extending the two beam backwards. These arethe two virtual images of the source S and act as two slit sources in Young’s double slit experiment andproduce stationary interference pattern.

If � x is the distance between successive fringes and d is the distance between S1 and S2, then thewavelength value can be obtained from the equation.

��= ca

dx

+∆

= D

dx∆ ...(1.10)

Fig. 1.9 A Typical biprism arrangement.

Fig. 1.10 Determination of slit separation (d).

Wave Optics 9

The distances d and D can easily be determined by placing a convex lens between the biprism andthe screen/eyepiece. For a fixed position of the eyepiece, there will be two positions of the lens(L1 or L2) where the images S1 and S2 can be seen at the eyepiece. Let d1 be the distance between thetwo images where the lens is at L1 (b1 from eyepiece to L1). Similarly d2 × b2 for L2 position of thelens.

� d = 21dd

and D = b1 + b2

= a + c = distance from source to screen or eyepiece.

Interference in Thin Films - Reflected Light

Fig. 1.11 Interference in thin films.

Let XY and X'Y' be the two surfaces of a transparent film of uniform thickness t and refractiveindex µ as shown in figure 1.11. Suppose S is a monochromatic source of light. Suppose a ray SA isincident on the upper surface XY at an angle i. This ray is partly reflected along AR and refracted alongAB at an angle r. At B it is incident at angle r. Here it undergoes reflection along BC at an angle r andrefraction along BT at an angle i. At C also it undergoes refraction along CR1 and reflection along CD.This process will continue for a number of times until the intensity becomes very very small. The raysBR and CR1 are derived from the same ray SA and travel in the same direction they interfere. To findout the effective path difference between the rays AR and CR1 draw a normal CE on AR and normalAF on BC. Produce the normal at A and the ray CB in the backward direction until they meet. Supposethey meet at Q. From the geometry of the figure, iACE =∠ and .rCAF =∠ The optical path differencebetween the two reflected light rays (AR and CR1) is given by

� = Path (AB + BC) in film; Path AE in air= µ (AB + BC) – AE ...(1.11)


From triangles ACE and ACF we know thatµ = sin i/sin r = AE/AC � CF/AC = AE/CF

AE = µ CF ...(1.12)From equations (1.11) and (1.12), we can write

� ��µ (AB + BC) – µ (CF)= µ (AB + BF + FC) – µ (CF)= µ (AB + BF)= µ (QF) ...(1.13)

From triangle AQF, cos r = QF/AQor QF = AQ cos r = 2t = cos r ...(1.14)(Since AQ = AP + PQ = t + t = 2t)Substituting the value of QE from equation (1.14) in equation (1.13), we have

� = µ × 2t cos r = 2 µ t cos r ...(1.15)

It should be remembered that a ray reflected at a surface backed by a denser medium suffers anabrupt phase change of ��which is equivalent to a path difference �/2.

Thus the effective path difference between the two reflected rays is (2 µ t cos r ± ��2).We know that maxima occur when effective path difference = n �For interference maximum 2 µ t cos r ± ��2 = n �Or 2 µ t cos r = (2n ± 1) ��2 ...(1.16)

If this condition is fulfilled, the film will appear bright in the reflected light.The minima occur when the effective path difference is (2n ±1) ��2 i.e.,2 µ t cos r ± �/2 = (2n ±) �/2 or 2 µ t cos r = (2n ± 1) �/2 ± �/2 = n � ...(1.17)because (n + 1) or (n – 1) can also be taken as integer. Here n = 1, 2, 3... etc.When this condition is fulfilled the film will appear dark in the reflected light.Looking at the same point as we move our eye the angle of incidence and the corresponding angle

of refraction changes. Therefore the conditions of maxima and minima are changed alternately. Hencewe observe a number of bright and dark regions.

Keeping the eye fixed if we change the point of observation then also we observe bright and darkregions. If the film is illuminated with white light, the maxima of different colours are observed atdifferent angles. Hence the film appears coloured.

1.7 NEWTON’S RINGS

When a plano-convex lens with its convex surface is placed on a plane glass plate, an air film ofgradually increasing thickness is formed between the two. The thickness of the film at the point ofcontact is zero. If monochromatic light is allowed to fall normally, and the film is viewed in reflectedlight, alternate dark and bright rings concentric around the point of contact between the lens and glassplates are seen.

Experimental Arrangement

The experimental arrangement of obtaining Newton's rings is shown in figure. L is a plano convex lensof large radius of curvature. This lens with its convex surface is placed on a plane glass plate G. Thelens makes contact with the plate at O. Light from an extended monochromatic source such as sodium

Wave Optics 11

lamp falls on a glass plate G' held at an angle 45° with the horizontal. The glass plate G' reflects a partof the incident light towards the air film enclosed by the lens L and the glass plate G. A part of theincident light is reflected by the curved surface of the lens L and a part is transmitted which is reflectedback from the plane surface of the plate. These two reflected rays interfere and give rise to an interferencepattern in the form of circular rings. These rings are localised in the air film, and can be seen with amicroscope focussed on the film.

Fig. 1.12a Newton’s rings apparatus. Fig. 1.12b Fig. 1.12c Plano-convex lens.

Explanation of the Formation of Newton’s Rings

Newton’s rings are formed due to interference between the waves reflected from the top and bottomsurfaces of the air film formed between the plates. The formation of Newton’s rings can be explainedwith the help of the Fig. 1.12c. AB is a monochromatic ray of light, which falls on the system. A partis reflected at B (glass-air boundary) which goes out in the form of ray R1 without any phase reversal.The other part is refracted along BC. At point C it is again reflected and goes out in the form of ray R2

with a phase reversal of �. The reflected rays R1 and R2 are in a position to produce interference fringesas they have been derived from the same ray AB and hence fulfill the condition of interference. As therings are observed in the reflected light, the path difference between them is (2µt cos r + �/2). For airfilm µ = 1 and for normal incidence r = 0. Hence in this case, path difference is (2 t + �/2). At the pointof contact t = 0, and the path difference is �/2, which is the condition of minimum intensity. Thus thecentral spot is dark.

For nth maximum, we have2 t + �/2 = n�

This expression shows that a maximum of a particular order will occur for a constant value of t.In this system, ‘t’ remains constant along a circle Thus the maximum is in the form of a circle. Fordifferent value of ‘t’, different maxima will occur. Hence we get a number of concentric bright circularrings. In a similar way, this can be shown that minima are also in the circular form.


Theory : Newton’s Rings by Reflected Light

Now we shall calculate the diameters of dark and bright rings. Let LOL' be the lens placed on a glassplate G. The curved surface LOL' is the part of spherical surface with centre at C. Let R be the radiusof curvature and r be the radius of nth bright ring corresponding to the constant film of thickness t. Asdiscussed above,

2 t + �/2 = n�

Or 2t = (2n – 1) �/2 for the bright ring where n = 1, 2, 3, ...etc.

For the property of the circleEP × PF = PQ × PQ

Substituting the valuesr × r = t × (2R – t) = 2Rt – t2 � 2Rt (approximately)

r2 = 2 R t or t = r2/2 R.

Thus for a bright ring2r2/2 R = (2n – 1) �/2

or r2 = 2

)12( Rn λ−

Replacing r by D/2, we get the diameter of nth bright ring as

4

2D=

2

)12( Rn λ− or Dn2 = 2(2n – 1)�R

or D = )12(2 −λ nR

or D ∝ A )12( −n

Thus the diameters of the bright rings are proportional to the square roots of odd natural numbersas (2n –1) is an odd number.

This shows that the difference in the squares of the diameters of the rings is constant.

Similarly for a dark ringor 2r2/2r = n�R

or D2 = 4 n�R

or D = nRn ∝λ2

Thus diameters of dark rings are proportional to the square roots of natural numbers.

If Dm and Dn are the diameters of the mth and the nth rings we have

22nm DD − = Rnm λ− )(4

This shows that the difference in the squares of the diameters of the rings is constant.

Wave Optics 13

1.8 DETERMINATION OF WAVELENGTH OF SODIUM LIGHTUSING NEWTON'S RING

Experimental Arrangement

The experimental arrangement of obtaining Newton's rings is shown in the Fig. 1.12. L is a planoconvex lens of large radius of curvature placed with its convex surface on a plane glass plate P. The lensmakes contact with the plate at O. Light from an extended monochromatic source such as sodium lampfalls on at glass plate G' held at an angle 45° with the horizontal. The glass plate G' reflects a part of theincident light towards the air film enclosed by the lens L and the glass plate P. A part of the incidentlight is reflected by the curved surface of the lens L and a part is transmitted which is reflected backfrom the plane surface of the plate. These two reflected rays interfere and give rise to an interferencepattern in the form of circular rings. These rings are localised in the air film, and can be seen with amicroscope focussed on the film.

ProcedureFirst of all the eyepiece of the microscope is adjusted on its crosswires. Now the distance of themicroscope from the film is adjusted such that the rings with dark centre are in focus.

The centre of the crosswires is adjusted at the centre of the rings pattern. The microscope ismoved to the extreme left of the pattern and the crosswire is adjusted tangentially in the middle of aclearly nth bright or dark ring. The reading of micrometer screw is noted. The microscope is nowmoved to the right and the reading of micrometer screw are noted at successive rings etc., till we arevery near to the central dark spot.

Again crossing the central dark spot in the same direction, the readings corresponding to successiverings are noted on other side. Now a graph is plotted between number of rings n and the square of thecorresponding diameter. The graph is shown in Fig. 1.13. If Dm and Dn are the diameters of the mth andnth rings and R is the radius of curvature of curved surface of the lens the wavelength of the sodiumlight is given by

� = Rnm

DD nm

)(4

22

−−

= R4

Slope

Fig. 1.13

D 2

n

S lo p e


The radius R of the plano-convex lens can be obtained with the help of spherometer using the

following formula R = .26

2 h

h

l + Here l is the distance between the two legs of the spherometer and h is

the difference of the readings of the spherometer when it is placed on the lens as well as when placedon lens surface.

Let R be the radius of curvature of the surface in contact with the plate, ��the wavelength of lightused and Dm and Dn be the diameters of mth and nth bright rings respectively, then

2mD = 2(2m – 1) �R

and 2nD = 2(2n – 1) �R

or 2mD – 2

nD = 4 (m – n)�R

or � = 2mD – 2

nD / 4(m – n)R ...(1.18)

Using this formula, ��can be determined.To find the refractive index of a liquid, it is introduced between the lens and glass plate and the

experiment is repeated as before. If D'm and D'n are the diameters of mth and nth rings in liquid then the

refractive index µ can be calculated using µ = .'' 22

22

nm

nm

DD

DD

−−

...(1.19)

NUMERICAL EXAMPLES

1. A thin sheet of plastic of refractive index 1.6 is placed in the path of one of the interferingbeams in Young’s experiment using light of wavelength 5890 Å. If the central fringe shiftsthrough 12 fringes, calculate the thickness of the sheet. (OU, 2003)

Solution: Given Data� = 5890 × 10–8 cmrefractive index (µ) = 1.6order of the fringe (m) = 12optical path of beam with plastic sheet = x – t + µtand the path of another beam = x� path difference = x – t + nt – x

= (µ – 1) t = n� for maxima

� ��= m

tµ )1( − or

Wave Optics 15

t = )1( −µ

λn =

)16.1(

12

−×λ

= 6.0

12105890 8 ×× −

= 1178 ×10–8 cm.

2. The path of one of the interfering beams in biprism experiment, a thin sheet of mica ofrefractive index 1.55 is placed. A light of wavelength 5893 Å is incident on it. Calculate thethickness of the sheet if the central fringe shifts through 10 fringes.

(OU, 2000)Solution: Given Data

� = 5893 × 10–8 cmrefractive index (n) = 1.55order of the fringe (m) = 10optical path of beam with plastic sheet = x – t + µtand the path of another beam = x� path difference is = x – t + µt – x

= (µ – 1) t = n �

� t = )1( −µ

λn =

155.1

10589310 8

−×× −

= 55.0

105893 7−×

t = 1071.45 × 10–8 cm.3. In double slit arrangement, a strong green light of wavelength 5460 Å is used. The slits are

0.01 cm apart and the screen is placed 20 cm away. What is the angular position of the firstminima?

Solution: Given DataFor first minimum n = 0 anddistance between slits (d) = 0.01 cm.wavelength of light (�) = 5460 × 10–8 cm.

The condition for minima is d sin � = λ

+ .

2

1n

or sin � = d

nλ

+ .

2

1

= 01.0

105460

2

10

8−×

+

= 0.0027

Since when � is very small, sin � ≅ �

� � = 0.0027 radians = 0.16


4. A parallel beam of light (� = 5890 × 10–8cm) is incident on plate (��= 1.5) such that theangle of refraction into the plate is 60°. Calculate the smallest thickness of the glass plate,which will appear dark by reflection.Given that ��= 1.5, r = 60°, cos 60° = 0.5

n = 1, ��= 5890 × 10–8 cmApplying 2µt cos r = n�

We get t = r

n

cos2µλ

= 5.05.12

1058901 8

×××× −

��The minimum thickness of the film t = 4.207 × 10–5 cm

5. A soap film 4 × 10–5 cm thick is viewed at an angle of 35° to the normal. Find the wavelengthsof light in the visible spectrum which will be absent from the reflected light (��= 1.33).

Let i be angle of incidence and r the angle of refractionGive that i = 35° and µ = 1.33, r = ?

Applying µ = r

i

sin

sinor 1.33 =

rsin

35sin °

We get r = 25.55° and cos r = 0.90Apply the relation 2 µt cos r = n�

and taking t = 4 × 10–5 cm(i) For the first order, n = 1

� ��= 2 × 1.33 × 4 × 10–5 × 0.90 = 9058 × 10–5 cm

which lies in the infra-red (invisible) region.(ii) For the second order, n = 2

2�2 = 2 × 1.33 × 4 × 10–5 × 0.90 �2 = 4.79 × 10–5 cm

which lies in visible region.(iii) Similarly, taking n = 3

�3 = 3.19 × 10–5 cmwhich also lies in the ultraviolet range.Hence, absent wavelength in the reflected light is 4.79 × 10–5 cm

6. A parallel beam of light (��= 5890 Å) is incident on a thin glass plate (��= 1.5) such that theangle of refraction is 60°. Calculate the smallest thickness of the plate which will appeardark by reflection.Given that µ = 1.5 and r = 60°; cos 60° = 0.5, ��= 5890 Å or ��= 5890 × 10–10 mFor minimum thickness n = 1Applying 2µt cos r = n�

We have t = r

n

cos2µλ

Wave Optics 17

t = 5.05.12

1058901 10

×××× −

t = 3926 × 10–10 mt = 3.926 × 10–4 mm

7. A soap film of refractive index 1.33 is illuminated with light of different wavelengths at anangle of 45°. There is complete destructive interference for ��= 5890 Å. Find the thicknessof the film.Given that � = 1.33

r = 45°cos 45° = 0.707

� = 5890 Å = 5890 × 10–10 mn = 1 ; t = ?

Applying 2µt cos r = n�

Thickness of the film t = r

n

cos2µλ

t = 707.033.12

1058901 10

×××× −

t = 3.132 × 10–7mt = 3.132 × 10–4 mm

8. A thin film of soap solution is illuminated by white light at an angle of incidence, i =

sin–1

5

4. In reflected light, two dark consecutive overlapping fringes are observed

corresponding to wavelengths 5.1 × 10–7 m and 5.0 × 10–7 m. ��for the soap solution is 3

4.

Calculate the thickness of the film. Here n�1 = (n + 1)�2

n(5.1 × 10–7) = (n + 1) × 5 × 10–7

n = 50

sin i = 5

4

µ = 3

4 =

r

i

sin

sin

sin r = µ

isin =

3/4

5/4 = 0.6

cos r = 2/12 ]sin1[ r− = 0.8


Apply 2µt cos r = n�1

We have t = 8.0)3/4(2

101.550

cos2

71

×××=

µλ −

r

n

The minimum thickness of the film t = 1.2 × 10–5 m

9. In Newton’s rings experiment, what will be the order of the dark ring which will havedouble the diameter of that of 20th dark ring. The wavelength of incident light is 5890 Å.

(OU, 1999)Solution: Given Data

order of dark ring n2 = 20wavelength (�) = 5890 × 10–8 cmWe know that the radius of the ring is given by ‘r’.

r = mR ××λ

or Diameter (d) = 2r = 2 Rmλ or D2 = 4 �Rm

�� 21D = 4�R × n1

and 22D = 4�R × n2 = 4�R × 20

= 80 �R ...(1)Given that : D1 = 2D2

(or) �� 22D = 2

24D ...(2). On substituting we have

4 � Rn1 = 4 × 80 × �R�� n1 = 80

10. In Newton ring’s experiment, the diameter of the 5th and 15th rings respectively was0.336 cm and 0.590 cm. If the wavelength of light is 5890 Å. Find the radius of curvatureof lens surface in contact with plane glass plate. (OU, 2000)

Solution: Given dataDm = D15 = 0.590 cmDn = D5 = 0.336 cm� = 5890 × 10–8 cmm = 15 n = 5We know that in the Newton’s rings experiment (R)

R = )(4

22

nm

DD nm

−λ−

= )10(4

25

215

λ− DD

= 101058904

)336.0()590.0(8

22

×××−

−

= 99.8 cm

Wave Optics 19

11. Newton’s ring arrangement is used with a source emitting two wavelengths �1 = 6000 Å and�2 = 4500 Å and it is found that the nth dark ring due to �1 coincides with (n + 1) th dark ringfor �2. Find the diameter of nth dark ring of �1 if the radius of curvature of the lensR = 90 cm. (OU, 2002)

Solution: Given dataWavelength �1 = 6000 × 10–8 cm��2 = 4500 × 10–8 cmradius of curvature R = 90 cmLet dn be the diameter of nth ring corresponding to wavelength �1, then diameter of(n + 1)th dark ring corresponding to wavelength �2 will also be dn

Rndn 12 4 λ= ...(1) and Rndn 2

2 )1(4 λ+= ...(2)

Dividing eq. (2) by (1)

n

n 1+ =

2

1

λλ

= 8

8

104500

106000−

−

××

n

11+ =

45

60

n

1 = 1

45

60 − = 45

15 =

3

1

� n = 3. Putting the value of n in eq. (1)

2nd = 4 × 3 × 6000 × 10–8 × 90

� dn = 0.2545 cm

12. Newton’s rings are observed in reflected light of wavelength 5900 Å. The diameter of 10th

dark ring is 0.50 cm. Find the radius of curvature of the lens and the thickness of the airfilm. (OU 2001, 2003)

Solution: Given dataWavelength of light (�) = 5900 × 10–8 cmDiameter of the mth ring (D10) = 0.50 cmSay, Diameter of the nth ring (Do) = 0

Therefore radius of curvature (R), R = )(4

22

nm

DD nm

−λ−

= m

Dm

λ4

2

R = 101059004

50.050.08 ×××

×− = 510594

25.0−××

= 310594

25−××

= 105.96 cm.

� R = 105.93 cms.


13. Newton’s rings are formed with reflected light of wavelength 5.895 × 10–5 cm. with a liquidbetween the plane and the curved surface. The diameter of the 5th dark ring is 0.3 cm andthe radius of curvature of the curved surface is 1 metre. Calculate the refractive index of theliquid.

Solution: Wavelength of light (�) = 5.895 × 10–5 cmDiameter of 5th dark ring D5 = 0.3 cmRadius of curvature R = 1m = 100 cmFor the ring system to be dark, we have

R

dn

4

2µ = n�

µ = 2

4

nd

Rnλ =

3.03.0

10895.551004 5

××××× −

= 1310 ×10–5+2

= 1.31� Refractive index of liquid is 1.31

14. In Newton’s rings experiment the diameter of 10th ring changes from 1.40 cm to 1.20 cmwhen a liquid is introduced between the lens and the plate. Calculate the refractive index ofthe liquid.

For liquid medium 21D = µ

λRn4...(i)

For air medium 22D = 4n�R ...(ii)

Divide (ii) by (i)

µ =

2

1

2

D

D

Here D1 = 1.20 cm, D2 = 1.40 cm

� ��= 2

20.1

40.1

= 1.361

15. In a Newton’s rings arrangement, if a drop of water (µ = 4/3) is placed in between the lensand the plate, diameter of the 10th ring is found to be 0.5 cm. Obtain the radius of curvatureof the face of the lens in contact with the plate. The wavelength of light used is 6000 Å.

2nD = µ

4 Rnλ or R =

λµ

n

Dn

4

2

µ = nD,3

4 = 0.5 cm

n = 10, ��= 6000 Å = 6 × 10–5 cmHere R = ?

Wave Optics 21

R = 5

2

1061043

)5.0(4−××××

×

= 139 cm16. Newton’s rings are formed by reflected light of wavelength 5895 Å with a liquid between

the plane and curved surface. It the diameter of the 6th bright ring is 3 mm and the radius ofcurvature of the curved surface is 100 cm, calculate the reflective index of the liquid.Here, for the nth bright ring,

Given that n = 6, ��= 5895 × 10–8 cm, R = 100 cm, r = 2

3 mm = 0.15 cm

To find µ = ?

Applying µ = 22

)12(

r

Rn λ−, we get

µ = 2

8

)15.0(2

100105895)162( ×××−× −

µ = 1.44117. In a Newton’s rings experiment the diameter of the 15th ring was found to be 0.590 cm and

that of the 5th ring was 0.336 cm. If the radius of the plano–convex lens is 100 cm, calculatethe wavelength of light used.

Here D5 = 0.336 cm = 33.6 × 10–3 m

D15 = 0.590 cm = 5.90 × 10–3 m

R = 100 cm = 1m, � = ?

� = mR

DD nmn

4

)( 22 −+ = R

DD

××−

104

25

215

� = R××

×−× −−

104

)1036.3()109.5( 2323

= 5.880 × 10–7m� = 5880 Å

18. In a Newton’s rings experiment the diameter of the 12th ring changes from 1.50 cm to1.30 cm when a liquid is introduced between the lens and the plate. Calculate the refractiveindex of the liquid.

Given thatD1 = 1.50 cmD2 = 1.30 cm

For air medium21D = 4n�R ...(i)


For liquid medium

22D = µ

λRn4...(ii)

Dividing (i) by (ii)

µ =

2

2

1

D

D

µ = 2

30.1

50.1

,

� = 1.33119. Newton’s rings are observed in reflected light of ��= 5.9 × 10–5 cm. The diameter of the

10th dark ring is 0.52 cm. Find the radius of curvature of the lens and the thickness of the airfilm.

Given that � = 5.9 × 10–5 cm = 5.9 × 10–7 m

n = 10

Radius of the ring r = 5.2 × 10–3 m

Apply r2 = n�R

��We get R = 7

23

109.510

)102.5(−

−

×××

(ii) Thickness of the air film = t2t = n�

t = 2

λn

= 2

109.510 7−××

t = 2.95 × 10–6 m

EXERCISE PROBLEMS

1. In Newton’s ring experiment, diameter of the 5th ring is 0.336 cm and that of the 15th ringis 0.590 cm. Find the radius of curvature of the plano-convex lens if the wavelength of lightused is 5890 Å.

2. Light of wavelength 6 × 10–5 cm falls on a screen at a distance of 100 cm from a narrow slit.Find the width of the slit if the first minimum lie 1mm on either side of the central maxima.

3. The diffraction maxima due to single slit diffraction is at � = 30° for a light of wavelength5000 Å. Find the width of the slit.

Wave Optics 23

4. Monochromatic light of wavelength 6.56 × 10–5 cm falls normally on a grating 2 cm wide;the first order spectrum is produced at an angle of 18° 14' from the normal. What is the totalnumber of lines on the grating.

5. Newton’s rings are observed in reflected light of wavelength 5900 Å. The diameter of 10th

dark ring is 0.50 cm. Find the radius of curvature of the lens and the thickness of the airfilm.

6. Green light of wavelength 5100 Å from narrow slit is incident on a double slit if the overallseparation of 10 fringes on a screen 200 cm away is 2 cm. Find the slit separation.

7. A biprism is placed at a distance of 5 cm in front of a narrow slit and is illuminated bysodium light of wavelength 5890 Å. The distance between the virtual sources is found to be0.05 cm. Find the width of the fringes observed in a light eyepiece placed at 75 cm from thebiprism.

8. Newton’s rings are observed in a reflected light of wavelength 6000 Å. The diameter of 15th

dark ring is 7 mm. Find the radius of curvature of the lens.

9. Find the ratio of intensity at the center of a bright fringe to the intensity at a point 1/4th ofthe distance two fringes from the center.

10. In Newton’s rings experiment, the diameter of 10th ring changes from 1.5 to 1.3 cm when aliquid is introduced between the lens and the plane. Calculate the refractive index of theliquid.

11. In Young’s double slit experiment two parallel slits 1 mm apart are illuminated bymonochromatic light. If the width is 0.50 mm and the screen is held at a distance of 80 cmfrom the slits, what is the wavelength of light?

12. In Newton’s rings experiment the diameter of 5th and 15th rings are 0.336 and 0.590 cmrespectively. If the wavelength of light is 5896 Å, find the radius of curvature.

13. A thin film 4 × 10–5 thick is illuminated by white light normal to its surface. Its refractiveindex is 1.5. What wavelength within the visible spectrum will be intensified in the reflectedbeam?

14. White light falls normally on a film of soapy water whose thickness is 5 × 10–5 cm andrefractive index is 1.33. Which wavelength in the visible region will be reflected morestrongly?

15. Newton's rings are formed with reflected light of wavelength 5895 Å with a liquid betweenthe plane and the curved surface. The diameter of the 5th dark ring is 0.3 cm and the radiusof curvature of the curved surface is 100 cm. Calculate the refractive index of the liquid.

16. A thin parallel liquid film of refractive index 1.28 and thickness 0.48 µm is illustrated byvisible light and observed at an angle of 35°. Determine the wavelengths(s) which are absentin reflected light.

17. In Newton’s rings experiment, the wavelength of light used is 600 nm and radius of curvatureof lens is 1 m. Determine the diameter of the tenth dark ring.

18. In Newton’s rings pattern, the diameter of a certain ring is 0.38 cm. A liquid of refractiveindex 1.38 is introduced between the glass plate and the lens in the same set up. Determinethe diameter of the same ring.


19. In Newton’s rings pattern, the diameter of the fifth ring is 0.3 cm and that of the tenth ringis 0.5 cm. Wavelength of light used is 589 nm. Determine the radius of curvature of the lens.

20. Newton’s rings are observed in reflected light of wavelength 589 nm with the liquid filmformed between plane glass plate and a plano-convex lens. The diameter of 9th bright ring is105 cm. Determine the refractive index of the liquid.

QUESTIONS

1. Explain constructive and destructive interference. Explain in detail Young's double slitexperiment.

2. Discuss the necessary theory of interference in thin films.

3. Discuss the formation of Newton’s rings and calculate their diameters. How do you determinethe wavelength of monochromatic light using Newton’s rings ?

4. What is interference of light ? Deduce the conditions for maxima and minima of interferencefringes formed by thin films.

engineering physics-01.p65

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