APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY

SECOND SEMESTER B TECH DEGREE EXAMINATION, APRIL 2018

PH 100: ENGINEERING PHYSICS

Max. Mark: 100 Duration: 3 hours

ANSWER KEY

PART A Answer all questions. Each question carries 2 marks.

1. Fundamental frequency m

T

l2

1=

m

T

lx

m

T

l

m

T

l

2

14'

4

2/2

1'

4

2/2

1'

=

=

=

𝝂′ = 4 ν

ie, the frequency will be four times the original frequency.

2. Sharpness of resonance means the rate of fall in amplitude with the change of forcing

frequency on each side of resonance frequency.

3. The loss of intensity due to reflection can be reduced by coating the reflecting surface with

a suitable transparent dielectric material such as calcium flouride. The refractive index of

such material must be in between that of air and glass. Such a film is called non reflecting

film.

4. Width of central maximum, 2x = 𝟐𝒇𝝀

𝒂

f- focal length of lens

x – distance of 1st secondary minimum from the centre of the principal

maximum (half width of central maximum)

a – slit width

since x α 1/a ,when the slit width increases, the width of central maximum decreases.

5. When a beam of ordinary light enters certain crystals like calcite or quartz it will split up in

to two refracted rays (Ordinary ray and Extra ordinary ray).This phenomenon is called as

Double refraction or Birefringence. 6. The property of expulsion of magnetic flux density from the interior of a superconducting

material during the transition from normal state to superconducting state is known as

Meissner effect.

Flux density, B = µo(H + M)

When B=0

H=-M

Magnetic susceptibility χ =M/H =-1

Magnetic susceptibility is -1 this shows that superconductors are perfect diamagnets.

7. After normalization the amplitude of the wave function has a fixed constant value. So

normalization is important to give Schrodinger wave equation a constant fixed amplitude.

Normalisation means the probability to find the particle in a particular state. If the particle

is ensured to be found in a particular state, means that its probability should be one. So,

by taking the normalisation of the wave function of a particular state, we are ensuring that

probability to find the particle in that state is 100%.

8. Consider particle of mass m incident on the barrier from left. Classically,If E <V0 all the

particle will be reflected at the barrier and If E>V0 all will be transmitted in to the region

III .Quantum mechanically, If E<V0 there will be a finite probability for the particle to be

transmitted in to the region III .This phenomenon of tunnelling through the barrier is

known as quantum mechanical tunnelling or barrier penetration .

9. Threshold hearing intensity is 10-12 Wm-2.

10. When certain crystals like quartz, tourmaline etc. are subjected to stress or pressure along

certain axis, a P.D is developed across the perpendicular axis. This is called Piezo electric

effect

11.

HOLOGRAPHY PHOTOGRAPHY

3D image of 3D object 2D image of 3D object

Intensity and phase variation recorded. Hologram provides full information of the object

Intensity variation only recorded. This provides only partial information.

Recorded film is called hologram Recorded film is called photograph

Each point of the film receives light from all parts of the object Point to point recording of the intensity of light

Each piece gives full information of object When cut into pieces, each piece gives partial information only

Lenses not needed Lenses are needed

Holograms have no resemblance to the object recorded.

There is resemblance between object and image(photograph)

Several images can be recorded on a film Only single image can be recorded on a film

12. The photovoltaic effect is the creation of voltage and electric current in a material upon exposure to light .

PART B

Answer any 10 questions. Each question carries 4 marks.

13. A pure LC circuit is an electrical analogue of simple pendulum.

Comparison between Mechanical and Electrical Oscillator

Mechanical Oscillator Electrical Oscillator

1. Displacement x Charge q 2. Velocity dx/dt Current dq/dt 3. Mass m Inductance L 4. Damping coefficient γ Resistance R 5. Force amplitude f0 Voltage amplitude v0 6. Driving frequency ωf Oscillator frequency ω

14. Path difference for dark band by reflection , 2µt cosr =nλ

For smallest thickness n = 1

i =300

Snell’s law, µ = 𝒔𝒊𝒏 𝒊

𝒔𝒊𝒏 𝒓

r = sin -1 (𝒔𝒊𝒏 𝒊

µ )

= sin -1 (𝒔𝒊𝒏 𝟑𝟎

𝟏.𝟓 )

= 19028’

t = 𝐧𝛌

𝟐µ 𝐜𝐨𝐬𝐫

= 𝟏×𝟔𝟎𝟎×𝟏𝟎−𝟗

𝟐×𝟏.𝟓× 𝐜𝐨𝐬𝟏𝟗𝟎𝟐𝟖′

= 2.12 x 10-7 m

Electrical Oscillator →

15. Force components acting on a Damped harmonic oscillator are :

Resistive force, F1 = -kx

K is the proportionality constant

Damping force, F2 = b

‘b’ is called the damping constant.

Force components acting on a forced harmonic oscillator are :

Resistive force, F1 = -kx

Damping force, F2 = b

External applied force ,F3 = tsin F f0

F0 – Force constant

16. Sinθ=Nnλ

λ1=500nm, λ2=400nm

Sinθ=Nnλ1

Sinθ=N(n+1)λ2

ie Nnλ1 = N(n+1)λ2

n = 𝝀𝟐

(𝝀𝟏 −𝝀𝟐 ) =

𝟒𝟎𝟎

𝟏𝟎𝟎 = 4

Sinθ=Nnλ1

N = 𝒔𝒊𝒏𝜽

𝒏𝝀𝟏 =

𝒔𝒊𝒏 𝟒𝟓

𝟒×𝟓𝟎𝟎∗×𝟏𝟎−𝟗

= 353553 lines/m

17. Here a QWP is used.

dt

dx

oscillatorharmonicforcedforeqnaldifferentitheisThis

mwhere

xdt

dx

dt

xd

m

kandcoefftdampingtheiswhere

m

bLet

mx

m

k

dt

dx

m

b

dt

xd

kxdt

dxb

dt

xdm

bygivenismotionofEquationThe

00

f0

2

2

2

2

f0

2

2

f02

2

Ff

)2(tsin f2

2

)1(tsin F

tsin F

=

−−−−−−−−=++

==

−−−−−−−−=++

=++

dt

dx

t = 𝝀

𝟒(µ𝒐~µ𝒆)

λ = t × 𝟒(µ𝒐~µ𝒆) = 2.5 x10-5 x 4 x 0.005 = 5 x 10-7 m

18. Superconductors are divided into 2 types depending on the magnetic properties. They

are

1. Type I or Soft superconductor

2. Type II or Hard superconductor

Type I or soft superconductors

When a superconductor is placed in a magnetic field H, the intensity of magnetization M is induced in it. When H increases, M also increases. The variation of M with H is shown in fig.

Up to the critical field Hc, magnetization increases proportional to the applied field and it abruptly drops to zero at the critical field. Above Hc it behaves as normal conductor. Below Hc, it is in the superconducting state. UptoHc, it behaves like a diamagnet and strictly obeys Meissner effect.

Type of superconductors for which M becomes zero abruptly when H=Hc are called Type 1superconductors. The transition from superconducting state to normal state is very abrupt. The critical field Hc is very small in the order of 0 .1 or 0.2 Tesla. Therefore it is easy to change type I superconductor in to a normal conductor. Hence they are also known as soft superconductors. The transition is reversible.

Eg: Al, Pb,Indium and almost all pure metals.

Type 2 superconductors

In this type there are two critical field,Lower critical field Hc1and upper critical fields Hc2. Up to Hc1 magnetism is proportional to the applied field and it behaves like Type 1 superconductors. It strictly obeys Meissner effect up to Hc1. In between Hc1 and Hc2 magnetization decreases gradually. At Hc2, the magnetisation vanishes completely and external field penetrates ,completely destroying the superconductivity .Above Hc2material behaves as normal conductor. Between Hc1 and Hc2material is in a mixed state (it behaves partially conducting as well as partially superconducting) known as vortex state. It does not strictly obey meissner effect in this region .

Transition from superconducting state to normal state is very slow and gradual. The field Hc2 value is very high in the order of 10T to 20T. A strong magnetic field is required to convert it into a normal conductor. They are known as hard superconductors. The transition is irreversible. Very strong magnetic field can be produced using type II superconductors.

Eg: Niobium, Germanium and all alloys

19. Wave length λ = 𝒉

√𝟐𝒎𝑬

ie, λ α 𝟏

√𝒎

me = 9.11 x 10-31

mp = 1.673 x 10-27

mp = 1836 me

Since mass of proton is higher than electron proton will be having shorter wavelength.

20. Let the phase space be divided into tiny cells whose sides are ∆x, ∆y, ∆z, ∆Px, ∆Py, ∆PzThe

volume of each cell is called phase volume given by dԏ.

dԏ= ∆x .∆y .∆z .∆Px .∆Py .∆Pz. From Heisenberg’s Uncertainty Principle, ∆x . ∆Px = h

Therefore, dԏ= ∆x . ∆y .∆z .∆Px .∆Py . ∆Pz = h3 Volume of each cell (Phase volume) = h3 Thus a point in phase space is not actually a point, but a cell of minimum volume of the

order of h3 and particle is some where within the cell.

21. Frequency, ν= 𝑽

𝝀

Velocity, v =

Y

For fundamental frequency

λ = 2l

KHz

x

xx

Y

l

73.1330

3000

105.8

1022

1

2

1

10

3

=

=

=

−

22.

2652

2

8000163.0

163.0

163.0

Sabinem

x

T

VA

A

VT

=

=

=

=

23.

SPONTANEOUS EMISSION STIMULATED EMISSION

An excited atom spontaneously jumps from higher energy level to the lower energy level without the help of any external agency

An external photon is incident on the excited atom and stimulate it to jump from the higher energy level to the lower energy level

There is no photon multiplication There is photon multiplication

The emerging beam is not coherent The emerging beam is highly coherent

The beam is not monochromatic The beam is monochromatic

The rate of emission depends on the no of atoms or molecules present in the excited state. So the beam is not intense

The rate of emission depends on a) the no of atoms or molecules present in the excited state b) the no of incident photons. Hence beam is highly intense.

The emission is not due to population inversion

Emission is due to population inversion and hence it become monochromatic

This is an uncontrolled random process This is a controlled regular process

Travelling in different direction as white light

Travelling in a particular direction as a narrow beam with definite frequency

The spectrum is broad and wide The spectrum is very narrow and sharp

24. Advantages of optical fibre over conventional transmission lines

• Extremely wide Bandwidth

• Very high information carrying capacity

• Smaller diameter, lighter weight cable

• Lack of cross talk between the parallel fibers.

• Immunity to electromagnetic interference

• Electric isolation

• Low cost

• Low transmission loss

• Signal security

• Longer life span

• Temperature resistant

PART C

Answer any three questions. Each question carries 6 marks.

25. DAMPED HARMONIC OSCILLATION

A harmonic oscillator in which the motion is damped by the action of an additional

force is called a damped harmonic oscillator. Damped oscillations are oscillations under

the action of resistive forces.

Damping force is α to its velocity.

ie Damping force = b

‘b’ is called the damping constant. Including the damping force, the differential eq.

for a damped harmonic oscillator is

dt

dx

oscillatorharmonic

dampedforeqnaldifferentitheisThis

xdt

dx

dt

xd

coefftdampingtheiswherem

bLet

xm

k

dt

dx

m

b

dt

xd

dt

dxbkx

dt

xdm

dt

dxbkx

dt

xdm

)2(02

.2

)1(0

0

2

2

2

2

2

2

2

2

2

−−−−−−−−=++

=

−−−−−−−−=++

=++

−−=

√𝑘

𝑚= 𝜔𝑜 = 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑏𝑠𝑒𝑛𝑐𝑒 𝑜𝑓 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒.

Solutions to the Equation Assume the solution is of the form

( )2

442

02

02

),2(

22

22

22

22

2

2

−−=

=++

=++

==

=

=

solutionThe

ie

xxx

getweinvaluesthesePutting

xAedt

xd

Aedt

dx

Aex

t

t

t

( )

( ) ( )

( ) ( )

.mindet''

.

tan

)3(

21

21

21

2211

2222

2222

22

systemtheofbehaviourtheeser

velocityandpositionofvaluesinitialtheon

dependswhichtsconsareAandAWhere

eAeAx

xxxissolutiongeneralThe

eAxandeAx

aswrittenbecansolutionsThe

Aex

tt

tt

t

−−−−+=

+=

==

=

−−−

−+−

−−−

−+−

−−

Equation (10) shows that the motion is oscillatory. The amplitude x0 Is not a constant

but decreases with time. The motion is not periodic.

26. NEWTON’S RINGS

The arrangement consists of a plano convex lens of large radius of curvature placed on an optically

plane glass plate. A thin film of air of varying thickness is formed between the lens and the glass

plate . The thickness of thin film is zero at the point of contact and gradually increases towards the

edge of the lens.

A beam of monochromatic light is incident normally on the lens. A part of the light is reflected from

the top surface of the thin film another part of light is reflected from the top surface of the glass

plate. These two reflected beams interfere destructively or constructively and produce dark or

bright band.

If a point appears dark, all the points along a circle through this point are dark since the

thickness of air film is same along a circle. So we get a dark ring. If the point appears bright, we get

a bright ring. Thus alternate dark and bright rings of increasing radii are observed. As the radii

increase, the rings become thinner and closer. These are called newton’s rings.

RADIUS OF THE nthRING

Let C be the centre of curvature of the curved surface of the lens and R be its radius of curvature.

Let rn be the radius of dark ring , BH=rn

As n increases, the distance between the rings decreases.

REFRACTIVE INDEX OF LIQUID

The refractive index of the liquid, 𝝁 =(𝑫𝒏+𝒌𝟐−𝑫𝒏𝟐)

(𝒅𝒏+𝒌𝟐−𝒅𝒏𝟐)

27. POSITIVE CRYSTAL

In a positive crystal , the o-ray travels faster than the e-ray. When a point source of light

is placed in a positive crystal, it gives out wave fronts due to o-ray and e- ray. The

ordinary wave travels with the same velocity in all direction and so the corresponding

wave front will be spherical. Extra-ordinary waves have different velocities in different

directions, so the corresponding wave front will be elliptical. In positive crystal spherical

wave front lies outside the ellipsoidal wave front. Along the direction of optic axis Ve=Vo.

The two wave fronts touch each other along the direction of optic axis. The refractive

index of the crystal for o-ray is less than the refractive index for e- ray ie μo<μe.

Example : Quartz

NEGATIVE CRYSTAL In a negative crystal , the e-ray travels faster than the o-ray. The ordinary wave travels

with the same velocity in all direction and so the corresponding wave front will be

spherical. Extra-ordinary waves have different velocities in different directions, so the

corresponding wave front will be elliptical. In negative crystal spherical wave front lies

inside the ellipsoidal wave front. Along the direction of optic axis Ve=Vo. The two wave

fronts touch each other along the direction of optic axis. The refractive index of the

crystal for o-ray is greater than the refractive index for e- ray ie μe<μo.

Example : Calcite

QUARTER WAVE PLATE (QWP) When a plane polarized light is incident normally on a doubly refracting crystal in

a direction perpendicular to optic axis, the ray splits up into O-ray and E- ray. They travel

in the same direction, but with different speeds. Hence they come out with a path

difference depending on the thickness of the crystal.

If the thickness of the crystal is such that path difference between the o-ray and

e-ray is λ/4 (phase difference of π/2), the crystal is called a quarter wave plate.

let ‘t’ be the thickness of the crystal , μo refractive index of ordinary ray and μe refractive

index of extra-ordinary ray. The optical path difference between the two rays when they

come out is

μot - μet = t (μo - μe) For a quarter wave plate, t (μo - μe) = λ/4

t = 𝝀

𝟒(𝛍𝐨 ∼ 𝛍𝐞)

For a positive crystal, μo< μe For a negative crystal, μo> μe Thickness of a QWP depends on wavelength λ .QWP is used for the production and

detection of circularly polarized light.

HALF WAVE PLATE (HWP) When a plane polarized light is incident normally on a doubly refracting crystal in

direction perpendicular to optic axis, the ray splits up into O-ray and E- ray. They travel in

the same direction but with different speeds. Hence they come out with a path difference

which depending on the thickness of the crystal

If the thickness of the wave plate is such that it can produce a path difference of λ/2 and a

phase difference of π between E-ray and O-ray, the crystal is called half wave plate.

let ‘t’ be the thickness of the crystal , μo refractive index of ordinary ray and μe refractive

index of extra-ordinary ray. The optical path difference between the two rays when they

come out is μot - μet = t (μo - μe)

For a Half wave plate, t (μo - μe) = λ/2

t = 𝝀

𝟐(𝛍𝐨 ∼ 𝛍𝐞)

For a positive crystal, μo < μe

For a negative crystal, μo> μe

Thickness of a HWP depends on wavelength λ

28. The Uncertainty Principle states that the simultaneous determination of the exact position

and momentum of a moving particle is impossible. The product of uncertainty in

position(Δx) and momentum (ΔPx) of the particle is of the order of Planks constant.

Δx. Δ Px =ħ (ħ= h/2π )

If Δx is small, Δ Px will be large and vice versa. If one quantity is measured accurately, the

other quantity become less accurate. If Δx=0 Δ Px=infinity; that is if we know the exact

position of a particle we shall never know its momentum with certainty

Position & momentum

Non-existence of electron in the nucleus

The nuclear diameter is of the order of 10-15 m . If an electron exists in the nucleus, the

maximum uncertainity, Δx,in position is of the order of the diameter .

2

hpx x =

So the uncertainly in the momentum,

The energy of electron is given by

E = pc = 1.055x10-19x3x108

=3.165x10-11 J

=3.165x10-11/1.6x10-19 eV

=1.98x108 eV

=198MeV

For an electron to exist in the nucleus, it should have an energy of this order. Energy of

an electron is of the order of only a few MeV . So it is not possible for electron to exists

inside nucleus.

PART D Answer any three questions. Each question carries 6 marks.

29. Numerical aperture NA of an optic fibre is the sine of acceptance angle θa.

NA = Sin θa

NA depends on the acceptance angle. If θa is larger, NA will also be higher.

Numerical aperture is also the light gathering power of the fibre and it measures the

amount of light accepted by the fibre. NA is between 0.13 abd 0.5. If NA is larger, fibre

can accept more light from the source. NA depends on the refractive indices of core and

cladding.

NA = 2

2

2

1 nn −

Consider a light ray AB incident at B at the edge of the core of an optic fibre from

air. It is incident at an angle θa with the axis of the core. This maximum angle is known as

acceptance angle. Since it travels from air to core, it is refracted along BC at an angle θr.

This refracted ray is now incident at C at the core-cladding interface with an angle slightly

greater than the critical angle θc. Hence the ray is undergoing total internal reflection and

it is travelling along CD. Thus the ray is propagated through the fibre by multiple total

internal reflections.

skgmx

xx

x

xp

/10055.1

1014.32

10626.6

19

15

34

−

−

−

=

=

=

At C , CN is drawn normal to the axis. The angle at C is taken as the limiting angle

θc , the critical angle. Let n0 be the refractive index of air, n1 that of core and n2 that of

cladding.

By Snell’s law at B, 𝐒𝐢𝐧 𝛉𝐚

𝐒𝐢𝐧 𝛉𝐫=

0

1

n

n

n0 Sin θa = n1 Sin θr

n0 for air = 1 and NA = Sin θa

NA = n1 Sin θr ……….(1)

At the critical angle, 𝐒𝐢𝐧 𝛉𝐜

𝐒𝐢𝐧 𝟗𝟎=

1

2

n

n

Sin θc =1

2

n

n …………..(2)

But Sin θc = Cos (90 – θc) θc + θr = 900

Sin θc =Cos θr

From (2) Cos θr =1

2

n

n

But Cos2 θr + Sin2 θr = 1

Sin2 θr = 1 - Cos2 θr

Sin2 θr = 1 - 2

1

2

2

n

n

2

1

2

2

2

1

2

1

2

2

2

12

n

nnrSin

n

nnrSin

−=

−=

Substituting for Sin θr in eqn (1)

NA = n1 2

1

2

2

2

1

n

nn −

Numerical Aperture , NA = 2

2

2

1 nn −

Where,

n1 = Refractive index of core

n2 = Refractive index of cladding

APPLICATIONS INDUSTRY

Fiberoscope(direct inspection is impossible) Inspection of various parts of huge machines Used for signaling Electric railway system To Enlarge the pictures in a TV screen In computer in connection with CPU &memory

MILITARY AND DEFENCE

Used for secret communication(difficult to tap) Optic sensors mounted on the missiles are used for video picture transmission and monitoring correct destination of the target. Used in aircrafts ships and submarines

30. When a ferromagnetic rod is placed in an alternating M.F, with its length parallel to the

field, the length of the rod increases and decreases rapidly (or the rod vibrates) This

phenomena is known as Magnetostriction. This principle is used to produce ultrasonic

waves

MAGNETOSTRICTION OSCILLATOR

Magnetostriction oscillator is used to produce ultrasonic waves using

magnetostriction principle. When high frequency alternating magnetic field is applied

parallel to the length of a ferromagnetic rod, its dimension changes alternately and very

rapidly and hence the rod is set into vibrations. When the natural frequency of the rod

becomes equal to the applied frequency , resonance takes place and the rod begins to

vibrate with maximum amplitude. At this condition ultrasonic waves will be produced

from the end of the rod.

It consists of a thin uniform ferromagnetic rod of length about 10 cms clamped at

the middle O. The coil L1 is loosely wound over the rod at one end. A variable capacitor C1

is connected across this coil.The coil L1and the variable capacitor C1 constitute a tank

circuit. One end of tank circuit is connected to the collector of transistor and the other

end to the positive of battery. The coil L2 is also loosely wound over the rod at its other

end .One end of this is connected to base of transistor and the other end to negative of

the battery. When the key is closed a current passes through

the tank circuit andthe tank circuit set in to vibrations. Thus high frequency oscillations

are obtained from tank circuit.

The high frequency oscillations will produce an alternating emf in the coil L1 which in turn

induces an alternating magnetic field with in the coil. Then rod begin to vibrate with its

natural frequency based on magnetostriction principle.

Now the variable capacitor C1 is adjusted so that the applied frequency becomes equal

to the natural frequency of the rod . So resonance takes place and the rod begins to vibrate

with maximum amplitude.At this condition ultrasonic waves will be produced from the

end of the rod.

When the applied frequency and the natural frequency becomes same resonance

takes place and the rod vibrate with maximum amplitude. At this condition ultrasonic

waves will be produced from the end of the rod.

31. Sound produced in a hall suffers multiple reflections from the walls, ceiling, floor

,furniture etc. Hence a listener hears a series of reflected sound waves in additional to the original wave . So the sound appears to persists for a longer time even after the source has stopped the production of sound. The persistence of sound even after the source of sound is stopped is called reverberation

REVERBERATION TIME Reverberation time(T) is defined as the time taken for a sound to decrease in intensity to

10-6 of its original intensity, from the moment when the source of sound is cut off. The

time of reverberation is also defined as the time required for the intensity to decrease by

60 dB, from the moment when source sound cut off.

SIGNIFICANCE OF REVERBERBATION TIME

For good acoustics of a hall, the reverberation time should have an optimum value.In a lecture hall the reverberation time should be smaller because the speeches will be clear only if a sound note decays rapidly before the next note falls on the ear of listener. For

Y

lLC

Y

lcrystalofnoscillatiooffrequencythe

l

vFrequency

lv

l

v

YvsultrasonicofVelocity

LCcircuitTankoffrequencyThe

2

1

2

1

2

1

2

2.

2

2

1

=

=

=

=

=

=

=

=

a lecture hall the reverberation time is about 1.03 sec.For a music hall it should be a little longer so that there is continuity between the successivemusical notes. For a musical hall the reverberation time is slightly greater than 1.03 sec.

Reverberation time can be controlled by

1. Using heavy curtains

2. Providing proper doors, windows & ventilators

3. Covering the walls and ceilings with sound absorbing materials like felt, fibre board

etc.

4. Covering floors with carpets

5. Using sound absorbing tiles

6. Providing full capacity to the auditorium

SABINE’S FORMULA

Sabine derived an expression of the reverberation time T in terms of the volume of

the hall and the total energy absorbed by the materials in the hall.

Reverberation time,

T = sec

This is called Sabine’s formula

A = 11S the total energy absorbed

V = Volume of the hall

32. PHOTO DETECTORS

The functions of a photo detector is to detect the optical signal and are based on semiconductor pn junction devices . The device can be operated either in photo voltaic mode or photo conductive mode. In photo voltaic mode the junction is treated as an open circuit to measure the voltage across the junction. In photoconductive mode the pn junction is connected in the reverse bias and the current is made to flow externally to complete the circuit between the junction. When the pn junction is connected in reverse bias a large depletion region is created. When the light with energy hט>>Eg falls on the depletion layer of junction electrons are excited to the conduction band leaving a hole in valence band. Thus electron- hole pairs are produced and hence electric current. The number of electron-hole pairs depends on the amount of incident light.

Commonly used photo detectors are,

•Photodiode

•Avalanche photodiode

•Phototransistor

PHOTO DIODE

Photo diode is a silicon or germanium p-n junction working under reverse biased condition. Reverse current increases when light falls on the junction.

CONSTRUCTION

Photodiode consist of a p-n junction mounded on an insulating substrate and sealed inside a metal case.A glass window is provided at the top for allowing light to strike at the junction.Two terminals acts as anode and cathode.

When a small reverse voltage is applied a very small current flows through the diode. This is due to minority charge carriers.At the same time majority charge carriers are prevented from crossing the junction.The current flowing through the reverse biased.

photodiode when no light is incident on the junction is called ‘ dark current ‘ .The corresponding resistance is called ‘dark resistance’

Dark resistance=VR/Id

where VR is reverse voltage and Id the dark current. When light falls on the junction the energy of the photons produce more electron-holepairs.The number of new electron-hole pairs is proportional to the number of incident photons.

APPLICATIONS

•Used in CAT scanner •Climate control-sunlight detector

•Widely used in optical communication •For brightness control and barcode scanning

•In security systems

AVALANCHE PHOTO DIODE(APD)

An APD is a semiconductor based photo detector (photodiode) which is operated with a relatively high reverse voltage(100-200 volt) just below breakdown.Here electrons and holes excited by absorbed photons are strongly accelerated in the strong internal electric field.They can generate secondary carriers by avalanche process.This amplifies the photocurrent by a significant factor.The carrier multiplication mechanism is known as

impact ionization.

APD consist of p+ and n+ regions which are heavily doped semiconductors and have very low resistance.The π region is basically an intrinsic region which is slightly p- type.Light is allowed to fall through p+ layer which is very thin .Most of the light enters in to the π region.

At low reverse biased voltage, the depletion layer of p-n+ junction extends slightly in to the π region.The electric field in the π region is used to separate the generated charge carriers by the drifting phenomenon.The depletion region around p-n+ forms the avalanche region where carrier multiplication takes place by impact ionization mechanism.

The avalanche photodiode has an internal gain when compared with photodiode.

Under large reverse bias an avalanche carrier multiplication occurs and the photo generated current gets amplified .As a result a large current can be produced with a small amount of incident light .

USES 1.In long range optical communication system

2.Used in laser range finders 3.Particle detectors

4.For fibre testing 5.In Lidars