notes of applied physics

Subject: Applied Physics [New] Code 300218 (15)

Aloke Verma, Department of Applied Physics, SRIT, New Raipur Page 1

1. Describe Michelson and Morley experiment with the help of neat diagram and discuss its negative result. Answer: In 1887 Michelson and Morley performed an experiment to find the velocity of earth relative to ether which was assumed to be stationary or an absolute frame of reference in which the speed of light is same (= c) in all direction

Fig.1. Experimental arrangement of Michelson-Morley experiment

Where, S = Monochromatic light source, P = partially silvered glass plate which inclined at 45 with the vertical, m1 and m2 = Perfectly reflected glass mirror, T = Telescope.

Fig.2. Theory of the Michelson- Morley experiment

Where, lMOPM;ctPMMP;vtMMPOOP 111111

In 11MPM ,

2

2

2

22

22

22

222222

2222222222

2222

11

22

1

c

v1c

lt

c

v1c

lt

vc

ltlvct

lvctctvltc

ctlctMMPMMP 1

---------- (1) Now the total time of travel of light from the plate P to the mirror M1 and the back from M1 to the plate P is



2

221

2

2

2

21

c2

v1

c

l2

c

v1

c

l2

c

v1c

l2t2t {By binomial expansion} ------ (2)

And the total time of travel of light from the plate P to the mirror M2 and the back from M2 to the plate P is

c2vc

lvcvc

vc

l)vc(vc

vc

l

vc

l

vc

lttt

222222bf2

2

21

2

2

2

2

2

22

222 c

v1

c

l2

c

v1

c

l2

c

v1c

l2

c

v1c

lc2

vc

lc2t ;

{By binomial expansion} ------ (3) The time difference in the time of travel of the light in the two mutually perpendicular directions

3

2

2

2

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

12

c

lv

c2

v

c

l2

c2

vv2

c

l2t

c2

v

c

v

c

l2

c2

v1

c

v1

c

l2

c2

v1

c

v1

c

l2

c2

v1

c

l2

c

v1

c

l2ttt

------------- (4) Hence, the path difference for the light rays travelling in the two mutually perpendicular directions is

2

2

3

2

c

lv

c

lv.ctcx -------- (5)

Now the entire apparatus is rotated by 90. Therefore the path difference is

2

2

c

lvx ------- (6)

Thus, the change in path difference between the light waves coming from the two directions due to rotation of the apparatus by 90 is

2

2

2

2

2

2

2

2

2

2

c

lv2

c

lv

c

lv

c

lv

c

lvxxx

---------- (7)

In the Michelson-Morley experiment .Sec/m103c.;Sec/m103v;m11l 84

m1022

103

10322

103

103112

c

lv2x 8

16

8

28

24

2

2

--------- (8)

For the yellow light ( mean wavelength =5500=5.510-7m ) the number of fringe shifted is

4.055

22

105.5

1022xn

7

8

----------- (9)



This result is called negative result of Michelson-Morley experiment because no fringe shift was observed by Michelson and Morley when the experiment was repeated after six month. Conclusions from the negative result- 1. The velocity of earth is zero relative to the ether. 2. The speed of light does not depend on the motion of source or on the motion of observer. 3. The concept of ether to be stationary is found to be wrong. 2. State the basic postulates of Theory of Relativity. Answer: The following are the two basic postulates of special theory of relativity proposed by Einstein- 1. All the frames of reference in relative motion with a constant velocity are equivalent for the description of the laws of physics. In other words, the laws of physics are the same in all inertial frames of reference moving with a constant velocity with respect to one another. 2. The speed of light is equal in all the inertial frames of reference. In other words, the speed of light in free space has the same value in all inertial frames of reference. The speed of light is 2.998 108 m/Sec. 3. Deduce the Lorentz transformation equation for space and time. Answer: The equations relating the coordinates of a particle in the two inertial frames on the basis of special theory of relativity are called the Lorentz Transformations.

Fig.1. Two inertial frame in relative motion

In fig.1, S and S are the two inertial frames. Frame S is moving relative to the frame S with velocity v along the positive X-axis. Let there be a point source of light at the origin O of the frame S. the wavefront of the light emitted at t=0, when reaches a point P, the position and time observed by the observers at O and O are (x,y,z,t) and (x,y,z,t) respectively. According to the special theory of relativity, the speed of light is same (= c) in both the frames, hence the time taken by the light to reach from O to P as observed in the frame S is

222222

2222

222

zyxctc

zyxt

c

zyx

c

OPt

0ctzyx 22222 ----------- (1)

And the time taken by the light to reach from O to P as observed in the frame S is

22222

2

222

2

222

zyxctc

zyxt

c

zyx

c

POt



0ctzyx 22222 ----------- (2)

From eq.(1) and (2), we get

2222222222 ctzyxctzyx ------------------ (3)

Since the frame S is moving relative to the frame S along S-axis, the lengths in direction perpendicular to the direction of motion are un affected i.e., zzandyy , ----------------- (4)

therefore from eq.(3),

222222 ctxctx ----------- (5) The transformation between x and x coordinates is given by simple relation

vtxkx ------------ (6) Where k is constant, independent of x and t.

And tvxkx ------------- (7) Where k is constant, independent of x and t. Substituting the value x from eq. (6) in eq. (7), we get

kvtkxk

xtvtvkvtkx

k

xtvvtxk

k

xtvvtxkkx

kk

11

v

xtkt

kk

11xvtk

k

xkxkvttv ------------- (8)

Putting the value of x from eq. (6) and the value of t from eq. (8) in eq. (9) , we get

2

2222222

kk

11

v

xtkcvtxktcx

22

22222222222

kk

11

v

x

kk

11

v

xt2tkctvvt2xktcx

2

------- (9)

Equating the coefficients of t2 on both sides of the eq. (9), we have

2

22

2

22222222222222 kvc

cvckcvkkcckcvkc

2

2

2

2

2

2

2

2

2

2

2

c

v1

1

c

v1c

ck

c

v1

1

c

v1c

ck --------------- (10)

And similarly, equating the coefficient of 2xt on both sides of eq. (9), we get



kk

11cv

kk

11

v

kcvk

kk

11

v

kcvk0 22

22

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

c

v1

c

v1

c

v1

c

v1k

k

1

c

v1

kk

1

kk

11

c

v

2

2

c

v1

1k -------- (11)

Substituting the value of k and k in eq. (6) and (8), we get

2

2

c

v1

vtxx

and

2

2

2

c

v1

cxvt

t

------------ (12)

Thus the Lorentz transformations are

zz;yy;

c

v1

vtxx

2

2

and

2

2

2

c

v1

cxvt

t

------------ (13)

4. Show that the mass of a body depends on its velocity. Deduce an expression for the variation of mass with velocity. Answer: According to the Newtonian mechanics, mass is an invariant quantity. If a stationary body of mass m is acted upon by a finite and constant force F for a time t due to which it acquires a velocity v, then the gain in momentum of the body is tFmvP ------------ (1) The maximum momentum that can be gained by a body is

mcPmax ----------- (2)

Where c is the velocity of light. According to the special theory of relativity the mass of body is not constant, but it depends on the velocity of the body. When the body is at rest ( v=0 ), the mass of the body is m0 and when the velocity of the body becomes equal to the velocity of light ( v=c ), the mass of the body becomes infinite. Consider a frame S relative to which a particle is moving with the velocity v along the X-axis. The moving mass of the particle in frame S is m. Consider another frame S in which the particle is stationary. The rest mass of the particle in frame S is m0. If the displacement of the particle relative to the frame S in time t is y along Y-axis, then the velocity of particle along Y-axis in the frame S is

t

yvy

and its momentum is



t

ymP 0y

------------- (3)

Similarly, t

yvy

and

t

ymP 0y

------------- (4)

From the Lorentz transformations yy , hence yy and according to time dilation

2

2

c

v1

tt

t

y.

c

v1m

cv1t

ym

t

ymP

2

2

22

y

------------- (5)

Since the momentum of the particle is invariant, therefore Py=Py From eq. (3) and eq. (4), we get

2

2

0

2

2

02

2

0

c

v1

mm

c

v1mm

t

y.

c

v1m

t

ym

------------ (6)

The above expression gives the variation of mass with velocity. 5. Establish Einstein mass-energy relation. Answer: According to the Einstein, the mass and velocity are equivalent to each other. If a mass m is converted into energy, the energy produced is E=mc2, where c is the speed of light. This is called the mass energy equivalence relation. Let a particle of mass m moving with a velocity v is displaced by a distance dx by applying a force F on it. The work done by the force is

dx.dt

dmvdx

dt

dvmdx.

dt

dmv

dt

dvmdx.mv

dt

ddkdx.Fdwdk

dmvmvdvdk 2 ;

vdt

dx ----------- (7)

But 22022222

02

2

2

2

2

0cmvmcmm

c

v1m

c

v1

mm

------------- (8)

On partial differentiation of eq. (8), we get

vdvm2dmmv2dmmc20vdvm2dmmv2dmmc2 222222

mvdvdmvdmc 22 -------------- (9)



From eq. (7) and (9), we get

dmcdk 2 ------------ (10) Hence gain in kinetic energy in acquiring the velocity v from rest is

202

0

2

m

m

2 cmmcmmcdmck

0

------------ (11)

The eq. (11) gives the expression for the kinetic energy of the particle at relativistic velocity.

But at v=0, the rest energy of the particle = 20cm ------------ (12)

Total energy of the particle E= kinetic energy + rest energy

2

0

2

0

2 cmcmmcE 2mcE ---------- (13)

The above equation is called the Einstein equation of mass-energy equivalence. 6. What is chain reaction? Answer: When a neutron fissions a uranium nucleus then, besides the fission fragments a few fast neutrons are also emitted. If one or more of the emitted neutrons are used to fission other nuclei further neutrons are produced and the process is repeated. The reaction thus becomes self-propagating and is known as a chain reaction. The chain reactions may be two types:

I. Uncontrolled Chain Reaction and II. Controlled Chain Reaction.

Fig.1: Chain Reaction

1. Uncontrolled Chain Reaction: In this type of chain reaction, the number of neutrons is allowed

to multiply indefinitely and the entire energy is released all at once as a violent explosion. Such a chain reaction takes place in nuclear bomb.

2. Controlled Chain Reaction: In this type of chain reaction, the reaction is first acceleration so that the neutrons are built up to a certain level and thereafter the number of fission producing neutrons is kept constant by some means. Such a controlled chain reaction is used in nuclear reactors.

7. Explain the construction and working of Nuclear Reactor. Give well-labeled diagram of Nuclear Reactor. Answer: Nuclear reactors are the device or furnace to produce a large amount of energy through initiate and control a sustained nuclear chain reaction. The energy produced during nuclear reaction is also



termed as nuclear power and generally used for the production of electricity and for the propulsion of ships. The main components of any nuclear reactor are

1. Core: This is the main part of nuclear reactor, which contains the nuclear fuel and where nuclear reaction takes place. There are hundreds of thousands of individual fuel pins, which assembled in core of nuclear reactor.

2. Moderator: The moderator used to slow down the speed neutrons quickly as the fuel has high fission cross-section for low energy neutrons. Depends on the type of nuclear reactor, many compounds can be used as moderator like water, graphite rods, beryllium, heavy water as well as some organic compounds.

3. Reflector: Reflector is used to reduce the leakage of neutrons. Reflector reflects back the neutrons, which are escaping from the core. Generally, moderator can be used as reflectors in the case of thermal reactors. However, in the fast reactors nickel, stainless steel and molybdenum are used as reflectors.

4. Cooling System: Since nuclear fission occurs in nuclear reactor, which produces a large amount of heat, hence there must be some coolant to absorb this excess heat. Coolant absorbs the heat from the core, which consists of pipes through which the coolant is pumped. The heat absorbed by coolant can be transferred to another working medium through a heat exchanger and then returns to the reactor. Some common coolants are heavy and light water, and liquid metals like sodium, lithium, potassium etc.

5. Control System: This system is designed to control the number of neutrons involve in nuclear chain reaction which indirectly control the rate of the chain reaction as well as power level. Control system includes sensing elements, which can measure the number of neutrons in the reactor, and control rods, which are able to absorbed strong neutron like cadmium or boron, and other devices, which can regulate the position of the control rods. The rate of nuclear chain reaction can be controlled by inserting and ejecting these rods in reactor. In fact, the size of control rods also affects the rate of reaction.

6. Protective shield: With a large amount of energy, nuclear reactions are also a good source of radioactive rays like alpha, beta and gamma rays. These radioactive rays are very dangerous, so the exposure to these radiation must be prevent as it can affect the person working near the reactor. The nuclear reactor is cover by a protective shield, which made up of steel and concrete and the arrangement of protection is known as Radiation shielding.

7. Steam Generator Component: This component used in Pressurized water reactors to produce steam from water by absorption of heat produced in nuclear reactor core. Some typical nuclear reactors are as follows.

Fig.2: Nuclear Reactor

I. Boiling water reactor (BWR): It is a thermal reactor in which steam produced in the core,

which can be fed directly to the turbine and produce electricity. The fuel used in such type of reactor is uranium oxide and light water is used as the coolant and moderator. Example of boiling water reactor is nuclear power station of Tarapur, which used for power generation of 210 MW.



II. Pressurized water reactor (PWR): It is same as boiling water reactors but here water keeps at high pressure so it does not boil after absorbing heat also. It prevents the mixing of water from the reactor and the water in the steam generator; hence, most of the radioactivity cannot move out of the reactor area.

III. Fast Breeder Test Reactor: In this reactor, plutonium acts as nuclear fuel. This reactor involves artificial transmutation of Uranium-238 to Plutonium-239 by absorption of neutrons. The product of transmutation that is plutonium consumed in reaction and again creates or breeds from the uranium. Here liquid sodium acts as coolant with normal steam generator. Because of absence of moderator to slow neutrons, this reactor called as Fast Breeder Test Reactor.

8. Write a short note on Fusion Reaction in Stars. Answer: Nuclear fusion is the process by which two or more atomic nuclei join together, or "fuse", to form a single heavier nucleus. During this process, matter is not conserved because some of the mass of the fusing nuclei is converted to energy, which is released. Fusion is the process that powers active stars. There are extreme astrophysical events that can lead to short periods of fusion with heavier nuclei. This is the process that gives rise to nucleosynthesis, the creation of the heavy elements during events such as supernovae. Mark Oliphant first accomplished building upon the nuclear transmutation experiments by Ernest Rutherford, carried out several years earlier, the laboratory fusion of heavy hydrogen isotopes in 1932. During the remainder of that decade, Hans Bethe worked out the steps of the main cycle of nuclear fusion in stars. There are two types:

1. The protonproton cycle and 2. The carbonnitrogenoxygen cycle

1. The proton-proton cycle: Bethe, in 1939 suggested that the production of stellar energy is by thermonuclear reaction in which protons are continuously transformed into helium nuclei. In proton-proton cycle is given by-

1

1

4

2

3

2

3

2

3

2

1

1

2

1

0

1

2

1

1

1

1

1

HHeHeHe:3Step

HeHH:2Step

eHHH:1Step

So, that we have

MeV7.26114

2

1

1 H2HeH4

Fig.1: The proton-proton chain dominates in stars Fig.2: The CNO cycle in stars 2. The carbonnitrogenoxygen cycle: In this cycle, the synthesis of hydrogen nuclei with carbon takes place. The carbon nuclei absorb the protons in succession and ultimately discharge a particle becoming carbon nuclei again. The reactions are given below:



4

2

12

6

1

1

15

7

0

1

15

7

15

8

15

8

1

1

14

7

14

7

1

1

13

6

0

1

13

6

13

7

13

7

1

1

12

6

HeCHN:6StepeNO:5Step

OHN:4StepNHC:3Step

eCN:2StepNHC:1Step

The overall process may be written as

MeV27eHeHe40

1

4

2

4

2

9. What is mass spectrograph? Answer: A mass spectrograph is an instrument that measures the nuclear masses. Mass spectrographs were designed by F.W. Astons by A.J. Dempster and by K.T. Bainbridge usingdifferent properties. The initial mass spectrographs measured atomic masses with a precision of 1 part in 103 whereas the modern instruments can yield a precision of 1 part in 107. 10.Explain construction and working of Bainbridge Mass Spectrograph. How to relative proportion of isotopes are determined. Answer: Bainbridge mass spectrograph is the recent and advanced type mass spectrograph and having high resolving power, precise symmetric image and linear mass scale which could not be obtained in Astons or Dempsters spectrographs. This is possible because it uses velocity selector and power electromagnet. Principle: Whatever be the velocities of the positive ions in the process of their generation, which are to be investigated, they are made perfectly homogeneous in velocity by the use of a special device called velocity selector. They are subjected to an extensive transverse magnetic field, which deflects the positive ions along circular path of different radius. The deflected ions are then brought to focus on a photographic plate. The radius of the semi-circular path is directly proportional to the mass of ion describing the path. Construction & working: Fig.1. represent the schematic diagram of Bainbridge mass spectrograph. A vacuum chamber is placed in a acting perpendicular to its large surface. The given beam of ions is collimated by two narrow parallel slits S1 and S2. These ions are having different velocities. It is then allowed to pass through the velocity selector. The velocity selector is a device in which the electric and magnetic fields are applied in the same region but perpendicular to each other. These two fields are arranged in such a way that deflections produced by them are equal and opposite.

Fig.1. Bainbridge Mass-spectrographs

The ion beam produced by discharge tube or a spark is collimated by slits S1 and S2 and allowed to enter the velocity selector. The electric field strength E and magnetic field strength B are adjusted such that only ions of given velocity v pass un-deflected experiencing equal and opposite electric and magnetic forces-



(1)B

EvqvBqE

Only the ions havening velocity B

E enter the evaluated chamber D through to the slit S3. The positive

ions entering into D are subjected to a strong uniform magnetic field of strength B perpendicular to their path. The force acting each ion is 'qvB and traverse circular paths of radius r given by

(2)qB'

mvrqvB'

r

mv 2

Thus is ions different masses m, but having same charge q, describe semi-circles of different radii

(3)rB'

v

m

q

Using value ofB

Ev , we get

rBB'

E

m

q

As E, B and B are constants,

r

1

m

q, if q is same for all ions, then mass rm . The ions different

masses strike the photographic plate of different points and we obtain a typical mass spectrum on the photographic plate, the mass scale being linear. 11. Explain construction and working of Aston Mass Spectrograph. How to relative proportion of isotopes are determined. Answer: In 1919, F.W. Aston designed a mass spectrograph. This spectrograph enables the measurement of the mass of single atomic ions and is useful for the investigation of isotopes. Principle: The deflection of the beam of positive rays in the electrostatic field depends on the velocity of ions and hence they are dispersed. This dispersed beam is then subjected to a magnetic field, which is perpendicular to electrostatic field. Construction, Working & Theory: The different parts of the apparatus are shown in fig.1. AO is direction of positive rays before entering the electrostatic field. S1 and S2 are slits which provide a fine pencil of positive rays. Plates P1 and P2 and the direction of the field being from P1 and P2 maintain the electrostatic field. The beam is deflected and dispersed downwards. Let and d be the angles of deviation and dispersion. Using a diaphragm D some of the rays are selected and are allowed to pass between the poles of an electromagnet. The magnetic field being perpendicular to the plane of the paper and inward. According to the Flemings left hand rule, the beam will be deflected upwards. This magnetic field nullifies the dispersion produced by electric field and recombines the particles, which are brought to focus in the form of sharp lines on a photographic plate CD. These lines are same as those of spectral lines.



Fig. 1. Aston Mass Spectrograph

Say, q= charge on positive ray particle; m'= mass of each particle; E= electrostatic field; B= magnetic field; = velocity of each particle; = angle of deviation produced by magnetic field; d= angle of dispersion produced by magnetic field; = angle of deviation produced by electric field; d= angle of dispersion produced by electric field. Taking into account that the deflection in electrostatic field is small, the curve near the vertex may be considered as circular of a radius r, we have

2

2

v'm

'Eq

r

1

r

v'm'Eq

Hence, the deflection , which is proportional to, r

1is given by

where C1 = CE, because E = constant.

)1(2

d

v

dv

v2

v'm

'qC2

dv

ddispersion

21

If r is the radius of curvature in magnetic field then

)2(v'm

'qC

v'm

'Bq'C

v'm

'Bq

'r

1

'r

v'mv'Bq 2

2

(Since B is constant)

)3(d

v

dv

vv'm

'qC

dv

ddispersion

22

From eqns. (1) and (2) we have,

)4(2

d

dd2

dd

2

dd

2

d

v

dv

Hence, if a deflection is given, the dispersion due to the electric field is twice that due to magnetic field. The small changes d and d refer to the particles with identical mass and charge but processing velocity differ by dv. In the absence of magnetic field, the dispersion produced in the beam for a distance (a+b) is given by

)5(d)ba(

212 v'm

'qC

v'm

'EqC



[here a= distance OO and b= distance OE] The magnetic field acts in a direction of perpendicular to the electric field and produces the same dispersion in a distance b but in the opposite direction. Dispersion produced by magnetic field )6(bd As all the ions are focused to the same position.

)7()ba(

b

d

dbdd)ba(

From eq. (4) and (7),we have

)8(2

2

b

a1

2b

a

21

b

a

2b

)ba(b2)ba(

)ba(

b2

)ba(

b2

This is the required condition for focusing. If b , then

)9(2202

20

a

This gives the position of photographic plate. The modified apertures gives an accuracy of 1 part in 104 and resolvability of 600. 12. What is ultrasonic? Answer: Ultrasound is a cyclic sound pressure wave with a frequency greater than the upper limit of the human hearing range. Ultrasound is thus not separated from "normal" (audible) sound based on differences in physical properties, only the fact that humans cannot hear it. Although this limit varies from person to person, it is approximately 20 kilohertz (20,000 hertz) in healthy, young adults. Ultrasound devices operate with frequencies from 20 kHz up to several gigahertzes. Ultrasonics is the application of ultrasound. Ultrasound can be used for imaging, detection, measurement, and cleaning. At higher power, levels ultrasonics are useful for changing the chemical properties of substances. 13. Explain Magnetostriction Method of producing ultrasonic waves and hence describe its advantages. Answer: Production of ultrasonic waves:- There are two major types of method-

I. Magnetostriction oscillator and II. Piezo-electric oscillator

Magnetostriction oscillator (Pierce Oscillator):- Magnetostriction is a property of ferromagnetic materials that causes them to change their shape or dimensions during the process of magnetization. The variation of material's magnetization due to the applied magnetic field changes the magnetostriction strain until reaching its saturation value, . James Joule when observing a sample of iron first identified the effect in 1842.



This effect causes losses due to frictional heating in susceptible ferromagnetic cores. The effect is also responsible for the high-pitched buzzing sound that can be heard near transformers on alternating current carrying pylons. Principle: The general principle involved in generating ultrasonic waves is to cause some dense material to vibrate very rapidly. The vibrations produced by this material than cause air surrounding the material to begin vibrating with the same frequency. These vibrations then spread out in the form of ultrasonic waves. When a magnetic field is applied parallel to the length of a ferromagnetic rod made of material such as iron or nickel, a small elongation or contraction occurs in its length. This is known as magnetostriction. The change in length depends on the intensity of the applied magnetic field and nature of the ferromagnetic material. The change in length is independent of the direction of the field. When the rod is placed inside a magnetic coil carrying alternating current, the rod suffers a change in length for each half cycle of alternating current. That is, the rod vibrates with a frequency twice that of the frequency of A.C. The amplitude of vibration is usually small, but if the frequency of the A.C. coincides with the natural frequency of the rod, the amplitude of vibration increases due to resonance. Construction: the coils L1 and L. wind the ends of the ferromagnetic rod A and B. The coil L is connected to the collector of the transistor and the coil L1 is connected to the base of the transistor as shown in the figure. The condenser C can adjust the frequency of the oscillatory circuit (LC) and the milliammeter connected across the coil L. can note the current. The battery connected between emitter and collector provides necessary biasing i.e., emitter is forward biased and collector is reverse biased for the NPN transistor. Hence, current can be produced by applying necessary biasing to the transistor with the help of the battery. Working: The rod is permanently magnetized in the beginning by passing direct current. The battery is switched on and hence current is produced by the transistor. This current is passed through the coil L, which causes a corresponding change in the magnetization of the rod. Now, the rod starts vibrating due to magnetostriction effect.

Fig.1.

When a coil is wounded over a vibrating rod, then e.m.f. will be induced in the coil called as converse magnetostriction effect. Due to this effect, an e.m.f. is induced in the coil L1. The induced e.m.f. is fed to the base of the transistor, which act as a feedback continuously. In this way the current in the transistor is built up and the vibrations of the rod is maintained. The condenser C adjusts the frequency of the oscillatory circuit and when this frequency is equal to the frequency of the vibrating rod, resonance occurs. At resonance, the rod vibrates longitudinally with larger amplitude producing ultrasonic waves of high frequency along both ends of the rod. Condition for resonance: Frequency of the oscillatory circuit = Frequency of the vibrating rod

E

lLC 2

1

2

1

where, l=the length of the rod. E=the youngs modulus of the material of the rod. = is the density of material of the rod.



Merits: 1. Magnetostriction materials are easily available and inexpensive. 2. Oscillatory circuit is simple to construct. 3. Large output power can be generated. Limitations 1. It can produce frequencies upto 3 MHz only. 2. It is not possible to get a constant single frequency, because rod depends on temperature and the degree of magnetization. 3. As the frequency is inversely proportional to the length of the vibrating rod, to increase the frequency, the length of the rod should be decreased which is practically impossible. 14.Explain Piezoelectric Method of producing ultrasonic waves and hence describe its advantages. Answer: Jacques and Pierre Curie discovered piezoelectricity in 1880. Paul Langevin first investigated quartz resonators for use in sonar during World War I. The first crystal-controlled oscillator, using a crystal of Rochelle salt, was built in 1917 and patented in 1918 by Alexander M. Nicholson at Bell Telephone Laboratories, although Walter Guyton Cady disputed his priority. Cady built the first quartz crystal oscillator in 1921. Other early innovators in quartz crystal oscillators include G. W. Pierce and Louis Essen. Quartz crystal oscillators were developed for high-stability frequency references during the 1920s and 1930s. By 1926 quartz crystals were used to control the frequency of radio broadcasting stations and were popular with amateur radio operators. Piezoelectric effect: When crystals like quartz or tourmaline are stressed along any pair of opposite faces, electric charges of opposite polarity are induced in the opposite faces perpendicular to the stress. This is known as piezoelectric effect.

(A) ( B) Piezoelectric effect- Mechanism: Piezoelectric and inverse piezoelectric effects are only exhibited by certain crystals, which lack centre of symmetry. In a piezoelectric crystal, the positive and negative electrical charges are separated, but symmetrically distributed, so that the crystal overall is electrically neutral. Each of these sides forms an electric dipole and dipoles near each other tend to be aligned in regions called Weiss domains. The domains are usually randomly oriented, but can be aligned during



poling, a process by which a strong electric field is applied across the material, usually at elevated temperatures. When a mechanical stress is applied, this symmetry is disturbed, and the charge asymmetry generates a voltage across the material. For example, a 1 cm cube of quartz with 2 kN (500 lbf) of correctly applied force can produce a voltage of 12,500 V. Piezoelectric materials also show the opposite effect, called converse (inverse) piezoelectric effect, where the application of an electrical field creates mechanical deformation in the crystal. Inverse piezoelectric effect: When an alternating e.m.f is applied to the opposite faces of a quartz or tourmaline crystal, it undergoes contraction and expansion alternatively in the perpendicular direction. This is known as inverse piezoelectric effect. This is made use of in the piezoelectric generator. Piezoelectric generator: A slab of piezoelectric crystal is taken and using this parallel plate capacitor is made. Then with other electronic components an electronic oscillator is designed to produce electrical oscillations >20 kHz. Generally one can generate ultrasonic waves of the order of MHz using piezoelectric generators. Quartz slabs are preferred because it possesses rare physical and chemical properties. A typical circuit diagram is given below. The tank circuit has a variable capacitor 'C' and an inductor 'L' which decides the frequency of the electrical oscillations. When the circuit is closed current rushes through the tank circuit and the capacitor is charged, after fully charged no current passes through the same. Then the capacitor starts discharging through the inductor and hence the electric energy is in the form of electric and magnetic fields associated with the capacitor and the inductor respectively. Thus, we get electrical oscillations in the tank circuit and with the help of the other electronic components including a transistor, electrical oscillations are produced continuously. This is fed to the secondary circuit and the piezoelectric crystal (in our case a slab of suitably cut quartz crystal) vibrates, as it is continuously subjected to varying (alternating) electric field, and produces sound waves. When the frequency of electrical oscillations is in the ultrasonic range then ultrasonic waves are generated. When the frequency of oscillation is matched with the natural frequency of the piezoelectric slab then it will vibrate with maximum amplitude. The frequency generated is given as follows:

E

l

P

LCf

22

1

where E= The Youngs modulus of the piezoelectric material and = The density of the piezoelectric material. 15. What is reverberation? Answer: Reverberation means the prolonged refraction of sound from the walls, floor and ceiling of room. It is also defined as the persistence of audible sound after the source has stopped to emit sound. The time of reverberation is also defined as the time taken for the sound of fall below the minimum audibility measured from the instant when the source stopped sounding. The Sabine defined the standard reverberation time as the time taken by sound to fall to one millionth of the intensity just before the source is cut off. 16. Derive the expression for Sabine formula for reverberation time.



Answer: The derivation for the formula is based on assumption, that there is a uniform distribution of sound energy inside the room. Now, first, we shall calculate the rate at which the energy is incident upon the walls and other surfaces and hence the rate at which it is being absorbed.

Fig.1.

Consider the reception of sound energy by small element ds of a plane wall AB as shown in fig.1. from the centre of ds and with redii r and r+dr , draw two circles such that they lie in the plane containing the normal to the element ds .Now consider the area between the circles and also lying between the angles +d which the two redii make with the direction of normal at the surface ds . This area is shown shaded in fig.1. the arc of this area is rd and the radial length is dr . Hence its area is rd.dr. If the whole fig. is rotated about the normal through an angle d, the distance travelled by this area will be the circumferential distance rsind. Volume traced out by this area element=rd.dr(rsind) Suppose E is the value of sound energy density, the energy contained in the above volume is given by=EdV This sound energy is travelling through the element equally in all directions. Energy travelling per unit

solid angle any direction

4

EdV

But the solid angle subtended by ds at the element of volume considered 2r

cosds

the energy in the element of volume that is travelling towards ds is given by

)1(d.d.cossindr.4

Eds

r

cosds.

4

d.dr.d.sinEr

r

cosds.

4

EdV

2

2

2

In order to find out the value of the energy reaching ds in unit time, the above expression should be integrated from r=0 to r=C is velocity of sound, and it is the distance within the volume is lying. Integrating eqn. (1), first with respect to , we get

)2(dcossin.dr.ds.2

E2.dcossin.dr.ds.

4

E

2dddcossin.dr.ds.4

Edcossin.dr.ds.

4

E2

0

2

0

Now, integrating with respect to , we have energy received at d



)3(dr.ds.2

E1.dr.ds.

2

E

1d2sind2sindr.ds.2

E

dcos.sin2dr.ds.2

Ed

2

cos.sin2dr.ds.

2

Edcos.sindr.ds.

2

E

2

0

2

0

2

0

2

0

2

0

Now integrating again with respect to r whose values varies between 0 to C, we get

)4(C.ds2

Edrds

2

EC

0

If a be the absorption coefficient of the wall AB whose value ds is a part, then Energy absorbed by ds per unit time

ds.a.EC4

1

Hence total absorption all the surfaces of the wall where the sound is falling

)5(4

ECAds.aC.E

4

1

where a.ds=A, the total absorption on all the surfaces on which sound falls. (i). Rate of Growth of Sound Energy in a Room:- Let P be the power output i.e. rate of emission of energy from the source and V, the total volume of the room. Then the total energy in the room at the instant when energy density is E will be EV. Rate of growth of energy

dt

dEVEV

dt

d

But at any instant, rate of growth of energy in space= Rate of supply of energy from the surface- Rate of absorption by the entire surface

)6(4

ECAP

dt

dEV

when steady state is attained dE/dt=0, and if the steady state energy density is denoted by Em, then its is given by

4

CAEP

4

CAEP0 mm

)7(CA

P4Em

From equ.(6) A.V4

EC

V

P

dt

dE

CA

P4E

dt

dEE

CA

P4

dt

dE ; where

CA

4

V

1and

V4

CA



Multiplying both sides by te , we have tttt eCA

P4Ee

dt

de

CA

P4eE

dt

dE

Integrating the above equation, we get )8...(..........KeCA

P4Ee tt

where K is a constant of integration. Initial condition is that at t=0, E=0 .Applying this condition to above equation, we get

CA

P4K

Putting this value in equation (8),we get

)9(e1Eme1CA

P4Ee

CA

P4

CA

P4E

CA

P4e

CA

P4Ee ttttt

The equation shows the growth of energy with time t. (ii). Decay of energy Density in a room:- Let the source is cutoff when E has reached the maximum value of Em, so that P=0 and t=0 when

CA

P4EE max

From equation (9) mmax EEK

0P,EEe mt

Or tmeEE .(10)

Equation (10) shows the decay of the energy density with time after the source is cutoff. (iii). Standard Reverberation Time (T):- Reverberation time T is defined as the time taken by sound to fall to one millionth of its value before the cutoff. Hence, to calculate T.

we put Ttand10EE 6m in eq.(10)

6TT6T

max

10ee10eE

E

Taking log, we have 63026.210log6T e

A

V.

340

63026.24T63026.2T.

V4

CA

(Taking the velocity of sound in air C=340 m/sec at room temperature)

Or VaS

165.0V

A

165.0T

..(11)

This equation is in good agreement with the experimental values obtained by Sabine. 17. Define Absorption Coefficient. Derive the expression for Absorption Coefficient of a Sound. Answer: The coefficient of absorption of a material is defined as the ratio of the sound energy absorbed by the surface to that of the surface to that of the total incident sound energy on the surface i.e.



Hence, the

absorption coefficient of a material is defined as the rate of the sound energy absorbed by a certain area of the surface to that of an open window of same area. The absorption coefficient of a surface is also defined as the reciprocal of its area which absorbs the same sound energy as absorbed at a unit area of an open windows. Measurement of Absorption Coefficient: The determination of standard times of reverberation in the room without and with a standard large sample of the material sample of the material inside the chamber. If the reverberation times are T1 and T2 respectively, then by applying Sabines formula, we have

V165.0

aS

V165.0

A

T

1

1

and V165.0

SaaS

T

1 11

2

where a1is the absorption coefficient of the area S1. From the above equations, we have

12

11

T

1

T

1

V165.0

Sa

Hence, knowing the terms on the right hand side of this equation a can be calculated. 18. Obtain an expression for fringe width in wedge shaped thin film. Answer: A wedge shaped film is one whose surfaces are inclined at certain small angle. Where- reflective index of wedge shaped film= . The path difference between the rays reflected at the upper

and lower surfaces= t2 . The additional path difference= 2

Fig.1.

The effective path difference between the two rays= 2

t2

The condition for the maximum intensity (bright fringes) is

n2

t2

Or 2

1n2t2

(1)

surfacetheonincidentenergysoundTotal

surfacethebyabsorbedySoundenergacoefficentAbsorption



The condition for minimum intensity (dark fringes) is

2

1n22

t2

Or nt2 . (2)

For the nth dark fringes, we have nt2 .(3)

Let this fringes be obtained at a distance nx from the edge

tanxt n .(4)

As is small

tan (5) nxt (6)

From eqn. (2) and (6),

nx2 n (7)

Similarly, if (n+1)th dark fringes is obtained at a distance, then

)1n(x2 1n (8)

From eqn. (7) and (8), we get

n)1n(x2x2 n1n

n1n xx2

Fringe width

2

xxX n1n ..(9)

Where is measured in radian. Similarly, we can calculate fringe width for bright fringes. 19. Give the theory of formation of Newtons Ring by reflected light. Explain why central ring is dark and higher order rings come closer to each other. Answer: When a plano- convex lens of large radians of curvature is placed with its convex surface in constant with a plane glass plate, an air film of gradually increasing thickness from the point of contact is formed between the upper surface of the plate and lower surface of the lens. If monochromatic light is allowed to fall normally on this film, then alternate bright and dark concentric rings with their centre dark are formed. These rings are known as Newtons rings. Experimental Arrangement: G= glass plate inclined at angle of 45 to the incident beam. L= plano-convex lens P= plane glass plate, S= monochromatic light source.

Fig.1. Experimental Arrangement



Theory:

Fig.2.

The condition of maximum intensity of bright rings 1airfor is given by

n2

t2

Or

2122

nt (1)

Where ...........3,2,1n

The condition of minimum intensity of dark rings 1airfor is given by

212

22

nt

Or nt2 (2) Where .....,.........3,2,1n

Determination of Diameters of Bright & Dark Rings: Let r be the radius of Newtons ring. R be the radius of curvature of the circle. t be the thickness of wedge shaped film. From the geometrical property of a circle, DEODDBAD ODR2ODrr tR2tr2 22 tRt2r .(3)

Since t is very small compared to R , hence 2t can be neglected. Hence

Rt2r2

Or R

rt

2

2 .. (4)

From eqn.(1) and (4), we have

2

1n2R

r2

Radius of thn bright ring

2

R1n2rn

(5)

The diameters of thn bright ring nn r2D ,



2

R1n22Dn

R21n2R21n22

R1n24D n

Or 1n2Dn .(6) Thus, the diameter of bright rings are proportional to the square roots of the odd natural numbers.

From eqn. (2) and (4), radius of thn dark ring, we have

nR

rt2

2

n

Rnrn .(7)

The diameters of thn dark ring is given by

RnRnRnrD nn 4422

Or nDn (8)

Thus, the diameter of dark rings are proportional to the square roots of natural numbers. Determination of wavelength of sodium light using Newtons rings: From eqn.(8), we have

Rn4D2n (9)

Similarly, the diameter of thpn ring is given by R)pn(4D2 pn (10)

From eqn. (9) and (10), we have

Rp4Rn4R)pn(4DD 2n2

pn

pR4

DD 2n2

pn

(11)

20. Explain the formation of interference fringes by means of Fresnel biprism using monochromatic source of light. Answer: The biprism is a device to obtain two coherent sources to produce sustained interference. It is a combination of two prisms of very small refracting angles, placed base to base. Actually, it is constructed as a single prism with one of its angle about 179 and the other two about 03 each. Production of Fringes: Let, S= A narrow adjustable slit; P= biprism; S1 and S2= virtual image & coherence source

(a) (b)



Theory: Let, S= A narrow adjustable slit; P= biprism; S1 and S2= virtual image & coherence source

Hence 2

dOROQ;xPO;DRSQS;dSS 2121

From PQS1 and PRS2 , According to Pythagoras theorem

4

dxdxD

2

dxDOQPODQPQSPS

2

22

2

22222

1

2

1

(1)

And

4

dxdxD

2

dxDORPODPRRSPS

222

2

22222

2

2

2

..(2)

From eqn. (1) and (2), we have

xd24

dxdxD

4

dxdxDPSPS

2

2

2

22

1

2

2

xd2PSPSPSPS 1212

PSPS

xd2PSPS

12

12

..(3)

But DPSPS 12 (approximately)

Path difference = D

xd

D2

xd2PSPS 12 .(4)

For P be the centre of bright fringes; nD

xdPSPS 12 where n= 0,1,2,3,.

Or nd

Dx .(5)

For the P to be the centre of a dark fringes; 2

1n2D

xdPSPS 12

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

Or 2

1n2d

Dx

(6)

Let nx and 1nx denote thn and th1n bright fringes. Then the distance between th1n and thn bright

fringes is given by

d

Dn1n

d

Dn

d

D1n

d

Dxx n1n

Or d

DxxX n1n

.(7)

And let nx and 1nx denote thn and th1n dark fringes. Then the distance between th1n and thn

dark fringes is given by



2

1n23n2d

D

21n211n2

d

D

21n2

d

D

211n2

d

Dxx n1n

Or d

DxxX n1n

. (8)

Where X is known as width of fringes. Experimental Arrangement and Adjustment: The apparatus used for this purpose are optical bench with four uprights, biprism sodium lamp, a convex lens, plumb line, bench error rod and reading lamp. Following procedure is adopted for obtaining sharp fringes:-

1. The optical bench is made horizontal with the help of sprit level. 2. Widen the slit and focus the eyepiece on the cross wires. One of the cross wires is set vertical by

means of a plumb-line. 3. Rotate the slit with the help of tangent screw provided in the plane of the slit so that its image

coincided with the vertical cross-wires. Hence the slit becomes vertical. 4. Using the tangent screw provided with the biprism mount and the slit will be nearly parallel. 5. The line joining slit and the edge of the biprism is made parallel to the length of the optical bench

by removing the lateral shift. The following measurements are now made:-

1. Measurements of Fringe Width X :-Adjust the vertical cross wire of the eyepiece on a bright fringe. Take the reading. Then the eyepiece is moved laterally so that the vertical cross-wire coincides with successive bright fringes and the corresponding readings are noted. From these

readings the fringe-width X is determined. 2. Measurement of D :- Takes the reading of position of the slit and the eyepiece on the optical

bench. The difference these readings gives D . 3. Measurement of d the distance between Virtual Sources in Biprism:- the distance between

the two virtual sources in biprism is determined by any of the following two methods:- I. Displacement Method and

II. Deviation Method 15. Discuss the Fraunhofer diffraction at a single slit. Derive the necessary conditions for minima and maxima produced. Answer: This phenomenon of bending of light corners and spreading the region of geometrical shadow of an object is called Diffraction. There are two types:-

I. Fresnel diffraction and II. Fraunhofer diffraction

Fraunhofer Diffraction at a Single Slit: Let wavelength of parallel beam of monochromatic light, e width of narrow slit AB, L Convex lens, angle of diffraction



Fig.1.

Let the disturbance caused at P by the wavelet from width of the slit at C be

tcosAy0 . (1)

Then the wavelet from with dx at C, when it reaches P has the amplitude Adx and phase

sin2

t .

Let this small disturbance be dy, we have

sinx2tcosAdxdy .(2)

For the total disturbance at the point of observation at an angle , we get

2

e

2

e

dxsinx2

tcosAy

But we know that Bsin.AsinBcos.AcosBAcos

2

e

2

e

2

e

2

e

2

e

2

e

dxsinx2

sintsinAdxsinx2

costcosA

dxsinx2

sin.tsinsinx2

cos.tcosAy

2

e

2

e

2

e

2

e

sin2

sinx2cos

tsinAsin2

sinx2sin

tcosAy

sine2

sine2cos

sine2cos

tsinAsine2

sine2sin

sine2sin

tcosAy

We know that coscosandsinsin



sine2

sine2cos

sine2cos

tsinAsine2

sine2sin

sine2sin

tcosAy

sine2

sine2sin2

tcos.Ae0sine2

sine2sin2

tcosAy

Let

sine2and 0AAe , therefore we get tcos

sinAy 0

.(3)

Resultant amplitude

sinAR 0

The resultant intensity at P is given by 2

2

02

22

0

2 sinIsin

ARI

(4)

Where 200 AI represent the intensity at 0 .

21. Discuss the Fraunhofer diffraction at a N- slit. Derive the necessary conditions for minima and maxima produced. Answer: A diffraction grating is an arrangement equivalent to N number of parallel slits of equal widths and separated from one another by equal opaque space. Diffraction grating is made by ruling large number of fine, equidistant and parallel lines on an optically plane glass plate with a diamond point makes diffraction grating. The ruled portion scatters the light while unruled portion transmits light.

Fig.1.

Let amplitude of each N-waves

sinAA 0 ; Phase difference

sinde2; Grating element

de ; Width of each silt d The resultant disturbance y is given by

termsN..............2tcostcostcosAy

We know that 2

eecos

ii ; therefore



termsN............2

ee

2

ee

2

eeAy

2ti2tititititi

termsN.............2

e

2

e

2

e

2

e

2

e

2

eAy

2iti2itiitiitititi

termsN.......2

e

2

e

2

e

2

e

2

e

2

e

2

e

2

e

2

e

2

eAy

i2tii2tiitiitititi

termsN.......ee12

eAtermsN.......ee1

2

eAy i2i

ti

i2i

ti

Using only real part of above equation, therefore

termsN.......ee1eAy i2iti The above equation is GP (Geometric Progression) series, so we get

i

iNti

e1

e1eAy (1)

Intensity yyI (2)

where y is the complex conjugate of y.

i

iN

ti

i

iN

ti

e1

e1eA

e1

e1eAI

iN

i

iN

i

iNtiti2

i

iN

i

iNtiti e

e1

e11

e1

e1eA

e1

e1

e1

e1eeAAI

1ee1

1ee1A

eee1

eee1eA

eeee1

eeee1eAI

ii

iNiN2

0ii

0iNiN02

iiii

iNiNiNiN02

cos1

Ncos1A

2

ee1

2

ee1

Aee2

ee2A

ee2

ee2

4

1AI 2

ii

iNiN

2

ii

iNiN2

ii

iNiN2

....(3)

We know that 2sin212cos , therefore we get

sindesin

sindeNsin

Asinde2

2

1sin

sinde2N

2

1sin

A

2sin

2

Nsin

A

2sin2

2

Nsin2

AI2

2

2

2

2

2

2

2

2

2

2

2

(4)

Since

sinAA 0 ;

Therefore

2

2

2

22

0sin

Nsin.

sinAI .(5)

where

sin)de(



Hence the intensity distribution is product of two terms. The first term 2

22

0

sinA

represent the

diffraction pattern due to a single slit and second term

2

2

sin

Nsinrepresent the interference pattern due

to N-slits.

For N=1 2

22

02

2

2

22

0

sinA

sin

sinsinAI

(single slit diffraction)

For N=2

2

22

2

2

2

02

2

2

2

2

02

2

2

2

2

0

sin

cossin4sinA

sin

cossin2sinA

sin

2sinsinAI

Or

2

2

22

0 cossin

A4I (double slit diffraction)

Hence for N-slit

2

2

2

22

0cos

NcossinAI (generalized expression)

Principal Maxima Intensity would be maximum when

0sin Or when n ; where n=0,1,2,3,

Also 0Nsin

Thus 0

0

sin

Nsin

i.e. indeterminate.

To solve this, we will use L Hospitals rule i.e.

N1.Nsin

N

Nsin

Lim.Nsin

N

Nsin

NLimsin

Nsin

N

NLim

sin

NsinLim

nnnn

22

22

0

p NsinA

I

..(6)

Theses maxima are most intense and are called Principal maxima. Now as n

n

sinde

Or nsinde (7) where n=0,1,2,3.. This is known as grating equation. 22. What is ruby laser? Describe the construction and action of the ruby laser. Answer: A ruby laser is a solid-state laser that uses a synthetic ruby crystal as its gain medium. The first working laser was a ruby laser made by Theodore H. "Ted" Maiman at Hughes Research Laboratories on May 16, 1960.



Fig.1. Ruby Laser

Ruby lasers produce pulses of visible light at a wavelength of 694.3 nm, which is a deep red color. Typical ruby laser pulse lengths are on the order of a millisecond. Working and theory: A ruby laser most often consists of a ruby rod that must be pumped with very high energy, usually from a flashtube, to achieve a population inversion. The rod is often placed between two mirrors, forming an optical cavity, which oscillate the light produced by the ruby's fluorescence, causing stimulated emission. Ruby is one of the few solid-state lasers that produce light in the visible range of the spectrum, lasing at 694.3 nanometers, in a deep red color, with a very narrow line width of 0.53 nm. The ruby laser is a three level solid-state laser. The active laser medium is a synthetic ruby rod that is energized through optical pumping, typically by a xenon flashtube. Ruby has very broad and powerful absorption bands in the visual spectrum, at 400 and 550 nm, and a very long fluorescence lifetime of 3 milliseconds. This allows for very high energy pumping, since the pulse duration can be much longer than with other materials. While ruby has a very wide absorption profile, its conversion efficiency is much lower than other mediums.

Fig.2. Energy Level Diagram

Modern lasers often use rods with antireflection coatings, or with the ends cut and polished at Brewster's angle instead. This eliminates the reflections from the ends of the rod. External dielectric mirrors then are used to form the optical cavity. Curved mirrors are typically used to relax the alignment tolerances and to form a stable resonator, often compensating for thermal lensing of the rod. Ruby also absorbs some of the light at its lasing wavelength. To overcome this absorption, the entire length of the rod needs to be pumped, leaving no shaded areas near the mountings. The active part of the ruby is the dopant, which consists of chromium ions suspended in a sapphire crystal. The dopant often comprises around 0.05% of the crystal, and is responsible for all of the absorption and emission of radiation. Depending on the concentration of the dopant, synthetic ruby usually comes in either pink or red. 23. What is He-Ne laser? Describe the construction and action of the He-Ne laser.



Answer: A heliumneon laser or He-Ne laser is a type of gas laser whose gain medium consists of a mixture of helium and neon inside of a small-bore capillary tube, usually excited by a DC electrical discharge. Working and theory: The gain medium of the laser, as suggested by its name, is a mixture of helium and neon gases, in approximately a 10:1 ratio, contained at low pressure in a glass envelope. The gas mixture is mostly helium, so that helium atoms can be excited. The excited helium atoms collide with neon atoms, exciting some of them to the state that radiates 632.8 nm. Without helium, the neon atoms would be excited mostly to lower excited states responsible for non-laser lines. A neon laser with no helium can be constructed but it is much more difficult without this means of energy coupling. Therefore, a He-Ne laser that has lost enough of its helium will most likely not lase at all since the pumping efficiency will be too low. The energy or pump source of the laser is provided by a high voltage electrical discharge passed through the gas between electrodes within the tube. A DC current of 3 to 20 mA is typically required for CW operation. The optical cavity of the laser usually consists of two concave mirrors or one plane and one concave mirror, one having very high (typically 99.9%) reflectance and the output coupler mirror allowing approximately 1% transmission.

Fig.1. Schematic diagram of a heliumneon laser

Commercial He-Ne lasers are relatively small devices, among gas lasers, having cavity lengths usually ranging from 15cm to 50cm, and optical output power levels ranging from 0.5 to 50 mW. The red He-Ne laser wavelength of 633 nm has an actual vacuum wavelength of 632.991 nm, or about 632.816 nm in air. The wavelength of the lasing modes lie within about 0.001 nm above or below this value, and the wavelengths of those modes shift within this range due to thermal expansion and contraction of the cavity. Frequency-stabilized versions enable the wavelength of a single mode to be specified to within 1 part in 108 by the technique of comparing the powers of two longitudinal modes in opposite polarizations. Absolute stabilization of the laser's frequency (or wavelength) as fine as 2.5 parts in 1011 can be obtained through use of an iodine absorption cell.

Fig.2. Energy level diagram of a He-Ne laser

The mechanism producing population inversion and light amplification in a He-Ne laser plasma originates with inelastic collision of energetic electrons with ground state helium atoms in the gas mixture. As shown in the accompanying energy level diagram, these collisions excite helium atoms from the ground state to higher energy excited states. This excitation energy transfer process is given by the reaction equations:

He*(23S1) + Ne 1S0 He (1S0) + Ne* 2S2 + E and



He*(21S) + Ne 1S0 + E He (1S0) + Ne*3S2 where (*) represents an excited state, and E is the small energy difference between the energy states of the two atoms, of the order of 0.05 eV or 387 cm1, which is supplied by kinetic energy. The medium becomes capable of amplifying light in a narrow band at 1.15 m and in a narrow band at 632.8 nm. The remaining step in utilizing optical amplification to create an optical oscillator is to place highly reflecting mirrors at each end of the amplifying medium so that a wave in a particular spatial mode will reflect back upon itself, gaining more power in each pass than is lost due to transmission through the mirrors and diffraction. When these conditions are met for one or more longitudinal modes then radiation in those modes will rapidly build up until gain saturation occurs, resulting in a stable continuous laser beam output through the front (typically 99% reflecting) mirror. Spectrum of a helium neon laser illustrating its very high spectral purity. The 0.002 nm bandwidth of the lasing medium is well over 10,000 times narrower than the spectral width of a light-emitting diode, with the bandwidth of a single longitudinal mode being much narrower still. The visible output of the red He-Ne laser, long coherence length, and its excellent spatial quality, makes this laser a useful source for holography and as a wavelength reference for spectroscopy. Prior to the invention of cheap, abundant diode lasers, red He-Ne lasers were widely used in barcode scanners at supermarket checkout counters. Laser gyroscopes have employed He-Ne lasers operating at 0.633 m in a ring laser configuration. He-Ne lasers are generally present in educational and research optical laboratories. 24. Describe the construction and working of Nd:YAG laser. Answer: Nd:YAG (neodymium-doped yttrium aluminum garnet; Nd:Y3Al5O12) is a crystal that is used as a lasing medium for solid-state lasers. The dopant, triply ionized neodymium, Nd(III), typically replaces a small fraction of the yttrium ions in the host crystal structure of the yttrium aluminium garnet (YAG), since the two ions are of similar size. It is the neodymium ion which proves the leasing activity in the crystal, in the same fashion as red chromium ion in ruby lasers. Generally the crystalline YAG host is doped with around 1% neodymium by atomic percent. Laser operation of Nd:YAG was first demonstrated by J. E. Geusic et al. at Bell Laboratories in 1964.

Fig.1. Nd:YAG Laser

Working & Theory: Nd:YAG lasers are optically pumped using a flashtube or laser diodes. These are one of the most common types of laser, and are used for many different applications. Nd:YAG lasers typically emit light with a wavelength of 1064 nm, in the infrared. However, there are also transitions near 940, 1120, 1320, and 1440 nm. Nd:YAG lasers operate in both pulsed and continuous mode. The high-intensity pulses may be efficiently frequency doubled to generate laser light at 532 nm, or higher harmonics at 355 and 266 nm. Nd:YAG absorbs mostly in the bands between 730760 nm and 790820 nm. At low current densities krypton flashlamps have higher output in those bands than do the more common xenon lamps, which



produce more light at around 900 nm. The former are therefore more efficient for pumping Nd:YAG lasers.

Fig.2. Energy level diagram of Nd:YAG Laser

The amount of the neodymium dopant in the material varies according to its use. For continuous wave output, the doping is significantly lower than for pulsed lasers. Nd:YAG lasers and variants are pumped either by flashtubes, continuous gas discharge lamps, or near-infrared laser diodes (DPSS lasers). Prestabilized laser (PSL) types of Nd:YAG lasers have proved to be particularly useful in providing the main beams for gravitational wave interferometers such as LIGO, VIRGO, GEO600 and TAMA. 25. Give principle of propagation of light through optical fiber. Derive an expression for acceptance angle. Answer: Let us consider light propagation in an optical fibre. The end at which the light enters the fibre

is called launching end. Let the refracting index of the core be 1n and the refractive index of cladding be

)nn(n 122 . Let the outside medium from which the light is launched into the fibre have a refractive

index 0n . Let the refracted angle r , with the axis and strikes the core-cladding interface at an angle .

Fig.1. Light propagation in fibers

If c (critical angle), the ray undergoes total internal reflection at the interface. As long as the angle

c , the light remains within the fibre.

Applying Snells law to the fibre, we get

0

1i

n

n

sin

sin

r

..(1)

Now largest value of i occurs when c

From ABC we have

cos)90sin(sin r ..(2)

From eqn.(1).



cosn

nsin

n

nsin

0

1r

0

1i ..(3)

When maxic,

c0

1

max cosn

nsin ..(4)

But 2

1c

n

nsin

Therefore

0

2

2

2

1

cn

nncos

.(5)

Substituting the expression (5) into (4), we get

0

2

2

2

1

maxn

nnsin

(6)

If 202221 nnn , then for all values of i , total internal reflection will occur. Assuming 1n0 , the maximum value of isin for a ray to be guided is given by

222

1max nnsin .........(7)

22211max sin nn (8)

The angle m is called the acceptance angle of the fibre. Acceptance angle may be defined as the

maximum angle that a light ray can have relative to the axis of the fibers and propagates down the fibre. 26. Give the construction and theory of Huygens eyepiece. Why cannot across wire be used with it? Answer: An eye-piece is a combination of lenses designed to magnify the image already formed by the objective of a telescope and microscope. An eyepiece consists of two planoconvex lenses. F is called the field lens and E the eye lens (Fig. 1.). The field lens has large aperture to increase the field of view. The eye lens mainly magnifies the image. To reduce the spherical aberration, the lenses taken are plano-convex lenses. Further the focal lengths of the two lenses and their separation are selected in such a way as to minimize the chromatic and spherical aberrations. A combination of lenses is used in an eyepiece of a simple lens magnifier for the following reasons: (i) The field of view is enlarged by using two or more lenses. (ii) The aberrations can be minimized.

Fig.1. Eyepiece



Construction: It consists of two plano- con vex lenses of focal lengths 3f (field lens) and f (eye lens), placed a distance 2f apart [Fig.2.]. They are arranged with their convex faces towards the incident rays. The eye-piece satisfies the following conditions of minimum spherical and chromatic aberrations. (i). The distance between the two lenses for minimum spherical aberration is given by a=f1-f2. In Huygen's eyepiece, a=f1-f2=3f-f=2f. Hence this eye-piece satisfies the condition of minimum spherical aberration.

(ii). For chromatic aberration to be minimum 2

ffa 21

. In Huygens' eyepiece, f2

2

ff3a

. Hence

this eyepiece satisfies the condition of minimum chromatic aberration.

Fig.2.

Working: An eye-piece forms the final image at infinity. Thus the field lens forms the image I2 the first Field Lens focal plane of eye lens, i.e., at a distance f to the left of eye-lens. Now the distance between the field lens and eye-lens is 2f. Therefore, the image I2 lies at a distance f to the right of field lens. The image I1 formed by the objective of microscope or telescope acts as the virtual object for the field lens. Thus we treat I1 as the virtual object for the field lens, and I2 as the image of I1 due to it (Fig.2.) or

?uandf3F,fv . We have

2

f3u

f3

f2u

u

1

f3

ff3

u

1

f3

1

f

1

f3

1

u

1

f

1

F

1

u

1

v

122

------- (1)

i.e. I1 should be formed at a distance 2/f3 from the field lens. Therefore, the rays coming from the objective which converge towards I1 are focused by the field lens at I2. The rays starting from I2 emerge from the eye-lens as a parallel beam.

Fig.3.

Cardinal Points of Huygens Eyepiece: The equivalent focal length F of this eyepiece is

2

f3F

f3

2

ff3

f2

f

1

f3

1

ff

a

f

1

f

1

F

1

2121

The second principal point is at a distance from the eye lens.



ff2

f2

f2ff3

f2f

aff

af 2

21

2

The first principal point is a distance from the field lens.

Fig.4.

The position of the principal points P1 and P2 and the principal focal point F1 and F2 are shown in Fig.4. Since the system is in air, the nodal points coincide with the principal points. 27. Obtain the expression for the focal length of combination of two thin lenses. Also find the positions of cardinal points for a lens combination. Answer: Consider two thin convergent lenses L1 and L2 of focal length f1 and f2 placed coaxially at a distance d apart. Let a ray AB parallel to the axis be incident at B on the lens L1 at a height C1B=h. It surfers deviation through an angle 1 at the first lens and proceeds towards the principal focus F1. It means the second lens at a point D at a height h2 from the axis on refraction it is further deviated through an angle 2 and meets finally the axis at F2. As the incident parallel ray PA after refraction through both the lenses meets the axis at F2, the point F2 must be the second principal focus of the combination. The lens of focal F placed at H2 is called the equivalent lens, which can be replace the combination.

Fig.1

Deviation produced by the equivalent lens F

h1

Deviation produced by the first lens 1

11

f

h

Deviation produced by the second lens 2

22

f

h

As both deviations are in the same direction, the total deviation is

f3f2

f6

f2ff3

f2f3

aff

af 2

21

1



2

2

1

1

21f

h

f

h

From fig.1 111112 dhtanBChCDhh

11tan

1

112

f

hdhh

2121212121211

11

21

11

ff

d

f

1

f

1

F

notes of applied physics

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