Physical Optics

Second Year Physics study material 2015

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Wave Optics

1. Describe Various theories of light?

A) Descartes or Newton corpuscular theory :

1. According to this theory light is made up of stream of weightless particles called as

corpuscles.

2. Different colors of light are due to different size of corpuscles of light.

3. By means of corpuscle theory Newton explained Reflection and Refraction of light.

Reflection is due to force of repulsion between corpuscles and reflecting surface.

4. According to Newton , light travels faster in the denser medium compared to rarer

medium. But this theory was proved wrong by Focault that speed of light in rarer

medium is greater than denser medium.

Other properties of light like Interference, Diffraction , Polarization can not be

explained based on corpuscular theory of light.

Huygens Wave theory of light :

1. According to Huygens wave theory , light is travelled in the form of wave in a

medium called luminiferous ether.

2. According to Huygen light is travelling in the form of longitudinal wave like sound

waves. But later it was proved that light is a transverse wave.

3. The speed of light is very high. so according to equation c=. For such materials density is less, Elasticity is more. Such type of materials will never exist. more

importantly that type of matter is never detected.

4. This theory failed to explain rectilinear propagation of light.

Later this concept of Ether is rejected.

Maxwell Electromagnetic theory:

According to Maxwell light is an electromagnetic wave which consists of mutually

perpendicular electric and magnetic fields which are also perpendicular to the

direction of propagation of light.

Electromagnetic waves require no material medium to propagate. But this theory

cannot explain photoelectric effect and Compton effect.

So we can not consider this theory as final theory.


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Plancks quantum theory :

According to this theory light is propagated in the form of packet of energy called

quanta. The quanta of energy of light is called photon. The energy of photon is given

by E = h. where is frequency of radiation. Photoelectric effect and Compton effect

was successfully explained by this theory of light. It failed to explain interference ,

diffraction and polarization.

Debroglie dual nature of Matter :

In some phenomenon light behaving as wave , in some phenomenon light behaving as

a particle. Two properties are not created in the same experiment. so light has dual

nature. In the same way , Debroglie showed that matter also behaves as a wave.

2. Define Wave front? Explain various types of Wave front?

A) Wave front is defined as locus of all points in the medium vibrating in the same phase

at a given instant.

The shape of wave front depends on the source of disturbance.

1. Spherical wave front (if the source of disturbance is a point source)

2. Cylindrical wave front (if the source of disturbance is a slit)

3. Plane wave front (When spherical wave front is at infinity, part of spherical wave

front appears as plane wave front)

Note : Phase difference between two points situated on the same wave front is zero.

3. State and Explain Huygens principle?

A) According to Huygens principle

1. Each point on a wave front is the source of new disturbance called secondary

source that emits secondary wavelets.

2. The disturbance travels with the speed of light in that medium

3. The envelop of all the secondary wavelets draws in the direction in which the wave

travels is the position of new wave front.

In figure (a), if we know the shape of wavefront at time t =0, Huygens principle allows

us to determine shape of wavefront after time t = . From the old wave front now draw

a sphere of radius c from each point on the spherical wavefront. where C represent

speed of waves in medium. If we now draw common tangents to all these secondary

wavelets , we obtain a wavefront at new position after time t = . The new wave front


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is again spherical with point source. Similarly we can construct wavefront for planar

wavefront.

4. Explain the phenomenon of Refraction using Huygens wave front?

A) By using Huygens principle we can able to derive the laws of refraction. Let PP

represents surface separating medium 1 and medium 2. Let v1 and v2 represents

speed of light in medium 1 and medium 2.

Let us assume the plane wavefront AB propagating as shown in figure. Let i be angle

of incidence by this plane wavefront at the interface. Let be the time taken by

wavefront to travel distance BC , then BC = v1.

To determine the shape of refracted wavefront, draw a sphere of radius v2 from the

point A in the second medium. Let CE represents a tangent plane drawn from a point

C on to the sphere. Then AE = v2 and CE would represents refracted plane wavefront.

At point A draw a normal. In the triangle ABC , AB is perpendicular to incident ray

AA and AC is perpendicular to AN.

In triangle AEC , A + E + c = 180

(90-r) +90 + c = 180 , c = r

If we now consider the triangles ABC and AEC , we obtain

sin i = BC/AC = v1/AC and sin r = AE/AC = v2/AC

where i and r are angle of incidence and angle of refraction respectively.

Then sin i/sin r = v1/v2

If C represents the speed of light in vaccum, then n1 = c/v1 and n2 = c/v2 are known

as refractive indices of medium 1 and medium 2 respectively

n2/n1 = v1/v2 , Hence sin i/sin r = n2/n1

therefore n1sin i = n2 sin r

This is known as snells law of refraction.

5. Explain the phenomenon of Reflection using Huygens wave front?

A) Consider plane wave AB incident at an angle i on reflecting surface PQ. If v

represents the speed of wave in the medium and if represents the time taken by the

wavefront to advance from the point B to C then the distance BC = v. In order to

construct a reflected wave front we draw a sphere of radius v from the point A as


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shown. Let CE represent the tangent plane drawn from point C to this sphere.

Obviously AE = BC = v and CE represents reflected plane wavefront.

In triangle ABC , AB is perpendicular to AA and AC is perpendicular to AN.

In AEC , A + E+C = 180

(90 r) + 90 + C = 180 , C =r

From triangle ABC , sin i = BC/AC

and from triangle AEC , sin r = AE/AC = BC/AC (as AE = BC)

since sin i = sin r , therefore i = r

This is known as law of reflection.

6) Explain briefly Refraction and Reflection of Plane wavefronts?

A) When a plane wavefront passes through the thin prism , Speed of light waves is less

in glass , the lower portion of the incoming wavefront will get delayed resulting in tilt

of emerging wave front

When a plane wave front passes through the thin convex lens , the central part of

incident plane wave traverses the thickest portion of lenses and is delayed the most.

The emerging wavefront has a depression at the center. Therefore wavefront becomes

spherical and converges to the point F which is known as focus.

When a plane wave incident on concave mirror and on reflection we have a spherical

wave converging to the focal point .

7. Write a short note on Doppler effect in light?

A) When ever there is a relative motion between the source of light and the observer, the

apparent frequency of light observed is different from the actual frequency of light

emitted by the source of light. This effect is known as Doppler effect.

Suppose a source of light is stationary with respect to an observer. The frequency of

light received by the observer is given by =c/ , where c = 3 x 108 m/s.

Now if a source of light moves away from stationary observer , the time taken by the

wavefronts to reach the observer from the source will increase. Hence, the frequency of

the light received by the observer will be less than the actual frequency of light. How

ever , if the source of light moves towards the stationary observer, the frequency of

light received by observer will be more than the actual frequency of light.


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Frequecy of light received when source and observer are in relative motion is known as

apparent frequency.

When speed of light is very less, we use the same formula used for sound waves. The

fractional change in frequency / = -Vradial/C

where Vradial is the component of source velocity along the line joining the observer to

the source relative to the observer. Vradial is considered positive when the source

moves away from the observer.

Applications :

1. It is used to measure the speed of stars and galaxies

2. Doppler effect is used in RADAR and SONAR

8. Define Superposition principle?

A) According to this principle The resultant displacement of the particle at any instant

is the vector sum of individual displacements caused to the particle by the two or more

waves

let y1 and y2 be the displacements of particle respectively at any instant caused by the

two waves. Then the resultant displacement of particle at that instant is given by

y = y1 +- y2

The same principle can be applied for two or more waves.

They are two types of superposition

1. Constructive superposition : When two waves of same wavelength are superimposed

with each other in phase, then the superposition is constructive.

y = y1 + y2

2. Destructive superposition : When two waves of same wavelength are superimposed

with each other out of phase, then the superposition is destructive.

y = y1 - y2

9. Define Coherent addition of source of light waves?

A) Two sources of light are said to be coherent if they emit waves of same frequency (or

wavelength) and are either in phase or have a constant initial phase difference.

coherent sources are generally obtained from the single parent source.


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Slits A and B can be coherent sources because the light waves passing through them

are derived from single wavefront that illuminates the slits so they have same

frequency and may have constant initial phase difference or no phase difference.

The phenomenon of coherence can be explained by means of two vibrating needles.

These needles are allowed to oscillate in phase. These needles generate two waves ,

each of amplitude A at any instant. Phase difference between their displacement does

not change with time. consider a point P such that S1P = S2P . At this point

, resultant displacement is the sum of individual displacements y1 and y2 of the two

waves respectively.

y = y1 + y2

But y1 = A sint and y2 = A sint

Therefore y = A sint + A sint = 2A sint

Since the intensity is proportional to square of amplitude , the resultant intensity will

be given by

I = 4I0

Where I0 is intensity produced by individual sources.

Consider another point P , where

S1P S2P

S2P - S1P = , where is wavelength of wave

Hence y1 = A sint and y2 = A sin(t + 2pi) (since phase difference corresponding to

is 2pi radian)

On substituting these values in equation y = y1 + y2 , we get y =2A sint . Amplitude

of resultant wave is double the amplitude of single wave. Hence Intensity in increased

and superposition is constructive.

Now consider one more point P , such that

S2P - S1P = /2

y1 = A sint , y2 = A sin(t + pi) = - A sint

On substituting these values in equation y = y1 + y2 , we get y =0. , giving zero

amplitude or destructive superposition.

Condition for constructive superposition (brightness) can be generalized as


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S2P - S1P = n , where n = 0,1,2.

Condition for destructive superposition (darkness) can be generalized as

S2P - S1P = (n + ) , where n = 0,1,2..

Note : Sources having different frequencies or the sources having same frequency but

no stable phase difference are known as incoherent sources.

10. Define Interference of light and Give some examples of Interferenceof light ?

A) The phenomenon of redistribution of light energy due to the superposition of light

waves from two coherent sources is known as interference of light.

When two sources of light having same frequencies and constant phase difference

then the redistribution of light energy is not uniform. At certain points the Intensity of

light energy is maximum and at certain points Intensity of light energy is minimum.

Thus the light energy is redistributed and this phenomenon is called Interference of

light.

Examples of Interference :

1. When oil spills on water we observe coloured streaks. These are formed due to

interference of light reflected from the oil film on water.

2. Soap bubbles appear coloured in sunlight due to interference of light reflected from

the soap bubble.

11. What is the relation between phase difference and Path difference of light?

A) Phase difference = (2pi/wavelength) X Path difference

12. Explain Briefly the Theory of Youngs Double slit Experiment?

A) The phenomenon of interference in light was demonstrated by Thomos young.

Dark and bright bands in the interference pattern is called interference fringes.

Consider two coherent sources s1 and s2 separated by a small distance d. Let D be the

distance between screen and the plane of slits s1 and s2.

Light waves emitted from the slits s1 and s2 reach point O on the screen after travelling

equal distances. So path difference and hence phase difference between these waves is

zero. Therefore they meet at O in phase and hence constructive interference takes

place. Thus O is the position of central bright fringe.


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Let the waves emitted by s1 and s2 meet at point P on the screen at a distance y from

the central bright fringe. The path difference between these waves at P is given by

x = s2P s1P ------------------ (1)

From right angled triangle S2BP,

S2P = [S2B2 +PB2]1/2 = [D2 + (y +d/2)2]1/2

= D [1 + (y +d/2)2/D2]1/2 -------------------(2)

Using Binomial theorem , and neglecting terms containing high powers

S2P = D [1 + (y +d/2)2/2D2]

= D + (y +d/2)2/2D

Similarly from right angled triangle S1AP , S1P = D + (y -d/2)2/2D----------------- (3)

substituting 2 and 3 equations in 1 , we get

x =( D + (y +d/2)2/2D) (D + (y -d/2)2/2D)

= 2yd/2D

x = yd/D

Condition for bright fringes

If the path difference is an integral multiple of , then bright fringe will be formed at P

yd/D = m or y = mD/d

which is the position of mth bright fringe from central bright fringe.

Fringe width () : The distance between any two successive bright fringes (or

successive fringes) is called Fringe width.

= y2 y1 = 2D/d - D/d = D/d

Condition for dark fringes :

If the path difference is an odd multiple of /2 , then dark fringe will be formed at P

yd/D = (m+1/2) , where m = 0,1,2.

= y2 y1 = 3D/2d - D/2d = D/d

Fringe widths of bright and dark are equal.

13. Explain briefly factors depending on fringe width?


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A) Fringe width directly proportional to wavelength of light. Larger will be the fringe

width and vice versa. It means fringe width will be greater for red colour than violet

colour.

Fringe width directly proportional to D , Larger is the distance of the screen from the

slits , greater will be the fringe width.

Fringe width inversely proportional to d , smaller is the distance between the coherent

sources , larger will be the fringe width.

14. List out some important observations regarding interference (general

reference)

A) If one of the slits is closed , Interference pattern will disappear.

If instead of Monochromatic light , if we use white light we will get coloured fringes of

various widths . But central spot is white.

If two independent sources are used instead of coherent sources , Interference pattern

is not observed.

15. What is Diffraction of light?

A) The phenomenon of bending of light around the corners of an obstacle or an

aperture into the region of geometrical shadow of obstacle is called diffraction of light.

Diffraction of light is more pronounced when size of the obstacle/aperture should be

the order of wavelength of wave.

They are two types of diffraction

1. Fresnel diffraction ( Source or screen both are at finite distance from obstacle or

aperture)

2. Fraunhoffer diffraction (Source or screen both are at infinite distance from obstacle

or aperture)

16. Explain theory of diffraction due to a single slit?


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A)

Light which is diverging from a monochromatic source S is made parallel after

refraction through convex lens L1 . The refracted light from L1 is propagated in the

form of wavefront WW . The plane wavefront WW in incident on the slit AB of width

d . According to Huygens principle, each point on the AB acts as a source of

secondary disturbance or wavelets.

Second lens L2 is used in converging the parallel beam. Now consider a point O on

screen, which is placed at a distance D from the slit AB . Since point O is equidistance

from A and B , therefore the secondary wavelets from A and B reach the point O in the

same phase and the constructive interference takes place at O . In other words point O

is the position of central maximum.

Now Let the light is diffracted through an angle . So the secondary wavelets will also

be diffracted through an angle . Let these wavelets meet the screen at point P. The

point P will be of maximum or minimum intensity depending on the path difference

between secondary wavelets (reaching point P) originating from corresponding points

of wavefront.

To find the path difference between secondary wavelets originating from the

corresponding points of A and B of the same plane wavefront , draw AN perpendicular

to BB . The path difference between these wavelets originating from A to B is BN.

From BAN , BN/AB = sin

or BN = AB sin

BN = dsin = d (if is small)

for minima :


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If path difference is equal to one wavelength that is BN = dsin = , then divide the slit

into two equal halves that is AC and CB . Now the path difference between the

secondary wavelets originating from A and C is equal to /2. So these wavelets will

meet at point P out of phase (phase difference pi) and hence destructive interference

will take place at P . similarly the path difference between the wavelets originating

from C and B is /2 and hence these will also produce destructive interference at P.

Thus Position P will be of minimum intensity.

Hence for first minimum = dsin1 =

sin 1 = /d

1 = /d ( if angle is very small)

similarly , if BN = 2 , then the slit AB can be imagined to be divided into four equal

halves . Then the path difference between the secondary wavelets originating from the

corresponding from of each half = (2/4) = /2 . Thus these wavelets produce

destructive interference and point P is of minimum intensity.

Thus for second minimum dsin2 = 2

sin2 = 2/d

2 = 2/d (for smaller angles)

In general , for minima dsinm = m , sinm = m/d

m = m/d (for smaller angles)

Where m is the angle giving direction of the mth order minimum and m = 1,2,3---- an

integer.

For Maxima :

If path difference BN = d sin is an odd multiple of /2 , then the constructive

interference takes place at P , Hence point P is position of secondary maxima

dsinm = (m+1/2)

m = (m+1/2)/d

Where m is the angle giving direction of mth order secondary maximum and m =

1,2,3----


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Width of central maximum or principal maximum :

Central maximum occurs at O on screen, where angle is zero. that is path difference

is zero so all points in the slit being in same phase lead to maximum brightness.

Let f be the focal length of lens L2 (placed very close to slit) and the distance of first

minimum on either side of the central maximum be x

tan = x/f

Since the Lens L2 is very close to the slit , so f = D , tan = x/D

since is very small , so tan sin , sin = x/D ---------(1)

Also, for first minimum , d sin = or sin = /d -----------(2)

from equations 1 and 2 x/D = /d or x = D/d

This is the distance of first minimum on either side from the center of central

maximum. width of central maximum is given by

2x = 2D/d

17) What are the factors effecting width of central maxima ?

A) It is directly proportional to wavelength of light used. so it is small for violet colour

and maximum for red colour.

Width of central maximum is directly proportional to distance D between plane of slit

and screen, so width will increase with increase in D

if width of slit (d) is small , width of central maximum is large and viceversa


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18) What is the difference between Interference and diffraction of light?

A)

Interference Diffraction

It is the result of superposition of waves from two wavefronts originating from the same source

It is due to superposition of wave from different points of the same wavefront

Interference fringes are all of same width Fringes are not of the same width

All Maxima are bright and all minima are perfectly black

Maxima are of decreasing intensity , minima are not perfectly dark

Over a large area on screen interference is observed

Over a small region on screen diffraction is observed.

19) Define Resolving power of an optical instrument?

A) The ability of an optical instrument to produce distinctly separate images of two

objects located very closely to each other is called its Resolving power.

Resolving power can also be defined as the reciprocal of limit of resolution.

smaller is the limit of resolution , larger is its resolving power.

20) Explain briefly Resolving power of an Microscope?

A) Let PQ be object of size d placed in front of objective lens AB of the microscope. The

central maxima of the images of P and Q are P and Q respectively. As the size of

object PQ decreased, d decreases , central maximum P moves closer to the central

maximum Q.

According to Rayleigh criterion , two points P and Q are resolved if central maximum

P coincides with the minimum N near to Q. The corresponding d = dmin gives the

resolving limit of microscope.

Since N is the first minimum around Q so ,

(QA + AN) (QB + BN) = 1.22

or (QA QB) + (AN BN) = 1.22 -----------------(1)

Also P coincides with N , so

PA + AN = PB + BN

(AN BN) = (PB PA) --------------------- (2)

using equation 2 in equation 1


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(QA QB) + (PB PA) = 1.22 ----------------- (3)

if is the semi angle subtended by the objective AB of the object PQ , then

(QA QB) + (PB PA) =2 dmin sin

Hence equation 3 becomes

2 dmin sin = 1.22

dmin = 1.22 /2 sin

Generally the high resolving microscope used in laboratories has liquid between the

object and objective of microscope. If n is the refractive index of the liquid , then the

wavelength in liquid becomes /n

and the resolving limit of

microscope is given by

dmin = 1.22 /2 nsin

where nsin is called numerical

aperture of the objective lens of the

microscope.

Resolving power of the microscope

is defined as the reciprocal of the

resolving limit of the microscope,

therefore

Resolving power of microscope = 2nsin /1.22

21) Explain briefly Resolving power of Astronomical telescope?

A) The minimum angle subtended at the objective lens of the telescope by two distinct

objects (Lying close to each other) whose images are just resolved is called limit of

resolution of the telescope. The reciprocal of limit of resolution is called resolving

power of telescope.

Let d be minimum angle subtended at the objective lens L of the telescope by two

distant objects O and O. The images of these objects are formed at P1 and P2

respectively. According to Rayleigh criterion, the two images P1 and P2 will be just

resolved if the central maximum of the image P1 of the object O coincides with the

first minima of the image P2 of object O or vice -versa

The path difference of between BP2 and AP2 is given by , x =AC

In ABC , sin d = AC/AB


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x = AB sin d

if D (= AB) be the diameter of the objective lens L of the telescope then

x = D sin d

The angular separation of the center of the central maximum and the first minima for

a circular aperture ( of a lens) can be obtained

x =

or D sin d =

since d is very small angle , so sin d = d

D(d) =

d = /D

But for a circular aperture , d = 1.22 /D

Which is the limit of resolution of the telescope

Resolving power of a telescope = D/ 1.22

it depends on diameter of the aperture of objective lens and wavelength of light used.


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22) Write a short note on Validity of Ray optics ?

A) Fresnal distance is defined as the distance of the screen from the slit or aperture

when the spreading of light due to diffraction from the center of screen is equal to size

(width) of the slit . It is represented by ZF .

ZF = d2/

Fresnal distance depends on Square of size of slit and wavelength of light

23) What is polarization of light ? Explain briefly experimental polarization of

light ?

A) The phenomenon of restricting the vibrations of light vector or electric field vector in a

particular direction in a plane perpendicular to direction of propogation of light is called

polarization of light.

Tourmaline crystal is used to polarize the light and hence it is called Polarizer.

Ordinary light is represented as

Polarized light can be represented in two ways . One way where vibrations parallel to

plane of paper and another way where vibrations perpendicular to paper.


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Consider two polaroids , Polarizer P and Analyzer A . Let both polaroids P and A are

placed in such a way that their pass axes or polarizing directions are parallel to each

other. When unpolarized light incident on Polaroid P , it is found tha intensity of light

transmitted by the Polaroid is reduced to Half tohe original value of intensity of

incident light.

When ordinary light is incident on Polaroid P , electric vector vibrating parallel to the

pass axis of Polaroid P passes through it , while the electric vector vibrating

perpendicular to the pass axis of Polaroid is blocked. so the light transmitted through

the Polaroid P has only electric field vector parallel to the axis.

Now place the two polaroids P and A such that their axes are perpendicular to each

other. when ordinary light is incident on the Polaroid P , it is found that the intensity

of light transmtted thorugh the Polaroid is zero.

24) State and Explain Malus law ?

A) This law states that Intensity of the polarized light transmitted through the

analyzer varies as square of the cosine of the angle between the plane of transmission

of the analyzer and the plane of the polarizer.

Let E be the amplitude of the light transmitted

through the analyzer and be the angle between the

planes of polarizer and analyzer.

Resolve E into two components

1) E cos along OP (parallel to plane of transmission

of analyzer)

2. E sin along OV (perpendicular to plane of

transmission of analyzer)

only E cos component is transmitted through the analyzer.

we know Intensity (Amplitude)2

Intensity of transmitted light through the analyzer is given by

I (E cos)2

I = kE2 cos2

I = Io cos2 which is known as Malus law

25) State and Explain methods for creating Plane polarization light?

A) A Plane polarized light can be created by means of two methods


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1. Polarization of light by scattering

2. Polarization of light by reflection

Polarization of light by Scattering :

Let a beam of unpolarized light be incident on a small dust particles or air molecules

(scatterer) along X axis. Since light is transverse in nature , therefore all the possible

directions of vibrations of electric vector E in the unpolarized light lie in Y-Z plane .

Hence the electrons associated with the scatterer will be set with vibrations along Y

and Z axis.

A vibrating electrons behaves as an oscillating dipole and emits radiations in all

directions expects its own line of vibration. when an observer observe the scatterer

along Y axis then the light reaching the observer comes from the electrons vibrating

along Z axis. This light is linearly polarized.

Polarization of light by reflection :

When ordinary light or unpolarized light falls obliquely on a glass slab , it is partially

reflected and partly refracted. Both the reflected and refracted beams of light are

partly polarized. The incident ordinary light has two components , one component of

incident light lies in the plane parallel to the surface of the glass slab and this

component is reflected.

The other component lies in the plane perpendicular to the glass surface is refracted.

Both the reflected and refracted components of light are partly polarized.

At a particular angle of incidence , the reflected light is completely plane polarized. The

angle of incidence at which the reflected light is completely plane polarized is known

as polarization angle . This is also known as Brewster angle. when angle of incidence

ordinary light is equal to polarization angle , then reflected and refracted beams of

light are perpendicular to each other.

26) What is Brewster law ?

A) The refractive index of refractive medium (n) is numerically equal to the tangle of

angle of polarization.

n = tan iB

Experimentally reflected component and refracted component of light are mutually

perpendicular then i = iB

from figure BOY + COY = 90


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(90 r) + (90-r) = 90

(90-IB) + (90-r) = 90

r = 90- IB

using n = sin i/sin r

= sin iB/sin (90 P) = sin iB/cos P

n = tan iB

Brewster law is used to calculate polarizing angle for any two media in contact when

their refractive indices are already known.

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Physical Optics