1. Why does a metallic piece become very hot when it is surrounded by a coil carrying high frequency alternating current?

Ans. The eddy currents circulate on the metallic piece and dissipate heat.

Explanation : The metallic piece being a solid conductor has no definite current paths. The induced currents set up are eddy currents because of conductor being enclosed in a.c. producing change in magnetic flux.

2. How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers?

Ans. The total charge on an n-type semiconductor is the sum of charge of negative charge carriers i.e. electrons and the charge of positive ions of the lattice which donated these electrons. Both being equal and opposite charges making the crystal neutral.

Explanation : From the energy band diagram we observe that the electrons from the impurity levels IL cross over to the conduction band CB but remain in the crystal. So the overall charge is zero.

3. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance of separation between the slits and the screen is doubled?

Ans. Angular separation remains constant.

Explanation : Angular separation is .dλθ =

4. Give expression for the average value of the a.c. voltage

V = V0 sin ωt over the time interval t = 0 and t = πω

Ans. 02Vπ

Explanation : t = πω

= (2 / )T

ππ

= 2T

Vaverage over half cycle = 1( /2)T

/2

0

0

2sinT

V t dtTπ∫

= /2

00

2 2cos2

TT V tT T

π − π = 0 02

(2)V V

=π π

5. How is the band gap, Eg, of a photo diode related to the maximum wavelength, λm, that can be detected by it?

Ans. gm

hc E>λ

Explanation : The photon of frequency ν must have energy greater than the band gap so that the photon excites an electron from the valence band to the conduction band.

6. Keeping the voltage of the the charging source constant, what would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by 10%?

Ans. The expression for the capacitance C of a parallel plate capacitor is

C = 0Ad

ε

When d is decreased by 10% it is d′ = 10dd − = 9 .

10d

The new capacitance is C′ = 0 (10)9

Ad

ε

The electrostatic energy stored = 201 .( )2

AV

dε

′

The electrostatic energy stored becomes = 20 .101 ( )2 9

AV

dε

The change in energy = 201 10 12 9

AV

dε −

= 2

01 1.2 9

AVd

ε

The energy is increased by 100 %9

ie ≈11.1%

7. How is the resolving power of a microscope affected when,

(i) the wavelength of illuminating radiations is decreased?

(ii) the diameter of the objective lens is decreased?

Justify your answer.

Ans. The expression for the resolving power of a microscope is

R.P. = 2 sinn iλ

where n is the refractive index of the medium between the object and the lower surface of the objective lens and i is the angle subtended by a radius of the objective on one of the points of the object.

According to the relation for R.P.;

(i) R.P. is increased.

(ii) R.P. is decreased because i is decreased.

8. The following table gives the values of work function for a few photo sensitive metals :

S. No. metal Work Function (eV)

1. Na 1.92

2. K 2.15

3. mo 4.17

if each of these metals is exposed to radiations of wavelength 300 nm, which of them will not emit photo electrons and why?

Or

By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from 4 × 1015 Hz to 8 × 1015 Hz?

Given h = 6.4 × 10–34 J-s, e = 1.6 × 10–19 C and c = 3 × 108 ms–1

Ans. For photoelectric effect to occur hν ≥ φ0 or 0hc ≥ φλ

We calculate roughly hcλ

≈ 12375eVÅ

3000Å = 4.125 eV

Hence Mo will not emit photoelectrons when 300 nm light is shined on it.

Or

Using

hν – hν0 = eVS

and hν′ – hν0 = eVS′ We have

′ν − ν( )he

= (VS′ – VS)

Putting values VS′ – VS = 34

196.4 10 Js1.6 10 C

−

−××

. (8 × 1015 – 4 × 1015)s–1 = 16 V

9. The following data was recorded for values of object distance and the corresponding values of image distance in the experiment on study of real image formation by a convex lens of power +5D. One of these observation is incorrect. indentify this observation and give reason for your choice:

S.No. 1 2 3 4 5 6

Object distance (cm) 25 30 35 45 50 55

image distance (cm) 97 61 37 35 32 30

Ans. The given convex lens has a power +5D. Therefore its focal length f = 1( )mp

= 1 100 cm5

× = 20 cm.

Real image is located

at f < v < 2f, m < 1

at v > 2f, m > 1

But m = v/u

Therefore readings 1 and 2 being v > 2f are consistent

reading 3 has both u and v < 2f is not consistent

readings 4, 5, 6 being v < 2f are consistent

Therefore reading 3 is incorrect.

10. What does the term lOS communication mean? Name the types of waves that are used for this communication. Which of the two-height of transmitting antenna and height of receiving antenna can affect the range over which this mode of communication remains effective?

Ans. LOS means line of sight communication. The two types of waves used for LOS communication are (i) space waves (ii) waves from transmitting stations near the ground to flying aircrafts.

The maximum distance or range upto which signals can be received on the earth is given by

d = 2 2T RRh Rh+

Hence both hT and hR affect the range over which the signal be received. where h is the height of the transmitting antenna.

11. (a) What is the basic difference between the atom or molecule of a diamagnetic and a paramagnetic material? Why are elements with even atomic number more likely to be diamagnetic?

(b) represent diagramatically an electron moving in a circular orbit of radius ‘r’ with speed v. Give expression for equivalent current and magnetic dipole moment. mark the direction of the dipole moment in the diagram.

Ans. (a) The atoms of a paramagnetic material have permanent magnetic moment whereas atoms of a diamagnetic material have no permanent magnetic moment.

When there are even number of electrons e.g. in helium there are two electrons which revolve around the nucleas with same velocity in like orbits but in opposite directions. There is no permanent magnetic moment.

When such an atom is put in a magnetic field B→

, the velocity of electron 1 is increased whereas the velocity of electron 2 is reduced. The magnetic moment appears. The net effect is an

increase of moment in a direction opposite to B→

, as shown below.

This is a diamagnetic behaviour

(b) For the electron in its orbit I = eT

= 2 / 2

e evr v r

=π π

pm = IA = 2

2 2ev evrr

r⋅ π =

π 12. Name the characteristics of electromagnetic waves that (i) increases (ii) remains constant in

the electromagnetic spectrum as one moves from radiowave region towards ultraviolet region.

Why are infrared radiations referred to as heat waves also? Name the radiations which are next to these radiations in electromagnetic specturm having

(i) Shorter wavelength. (ii) longer wavelength.

Ans. (i) Frequency

(ii) Velocity

Explanation : Velocity of em waves in vaccum is constant however the ultraviolet region has higher frequency than radio-waves.

Infrared are referred as heat waves because water molecules present in most materials readily absorb these waves and after absorption, their thermal motion speeds up i.e. they are heated

(i) Red wavelength in the visible part is next to infrared, it has a shorter wavelength than infrared

(ii) Microwaves are the next to the infrared having higher wavelength.

13. list any two conservation laws in radioactive decay. prove that the instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of its half life.

Ans. (i) Conservation of nucleon number

(ii) Conservation of linear momentum.

dNdt

i.e. number of nuclei which decay in a unit time (with minus sign) is called the activity A of a

radioactive substance. It is proportional to the no. of radioactive nuclei present in a sample

The law is dNdt

= – λN …(i)

where λ is decay constant, λ = ½

0.693T

Eqn(i) also gives N = N0e–λt

\ A = dNdt

= 0½

0.693 tN eT

−λ×

The rate of change of activity is

dAdt

= 2

2d Ndt

= ( )d dNdt dt

= 12

0.693T

No (–λ)e–λt

= –No1 12 2

–.693 .693 teT T

λ

= 2

–12

.693– . toN e

Tλ

= 12

20.693– NT

or dAdt

∝ ( )2

1/2

1

T

14. The following data was obtained for the dependence of the magnitude of electric field, with distance, from a reference point O, within the charge distribution in the shaded region.

Field point A B C A′ B′ C′magnitude of electric field E E/8 E/27 E/2 E/16 E/54

(i) identify the charge distribution and justify your answer.

(ii) if the potential due to this charge distribution, has a value V at the point A, what is its value at the points. A’, B’, C’.

Ans. (i) The charge distribution for whih the results are given is a small electric dipole for which

E = 30

21 .4

p

rπε along point on its axis along ABC and

E = 30

1 .4

p

rπε on the equatorial plane along A’B’C’

At A EA = 3

2.

pk

r = E

At B EB = 3

2.(2 )

pk

r = E/8

EC = 3

2.(3 )

pk

r = E/27

EA’ = 3.p

kr

= E/2

EB’ = 3.(2 )

pk

r = E/16

EC’ = 3.(3 )

pk

r = E/54

(ii) The potentials at points A′, B′, C′ on the equatorial line are zero.

15. A charge Q located at a point →r is in equilibrium under the combined electric field of three

charges q1, q2, q3. if the charges q1, q2 are located at points 1r→

and 2r→

respectively, find the

direction of the force on Q, due to q3 in terms of q1, q2, 1r→

, 2r→

and r→

.

Ans. Since charge Q is in equilibrium the forces →

1F , 2F→

and 3F→

will

form a balanced system. i.e. 1 2 3F F F→ → →

+ + = 0.

The direction of 1F→

is along 1r r→ →

− , of 2F→

is along 2r r→ →

− .

The direction of 3F→

must comply with 1 2 3F F F→ → →

+ + = 0

or 3[ ]F→

= 1 2[ ]F F→ →

− +

\ The direction is – 1 2[( ) ( )]r r r r→ → → →

− + − = – 1 22 r r r→ → →

+ + The direction of electrostaticforce is independent of themagnitude of the charges.

It is to be noted that stable equilibrium is not possible in electrostatics for charges which are unequal and similar. As an example if q3 is disturbed from its straight line formation with q1 and q2 in equilibrium, there will always exist a force on q3 which does not allow it to return to the straight line formation.

16. 12 cells, each of emf 1.5 V and internal resistance 0.5 Ω, are arranged in m rows each containing n cells connected in series, as shown. Calculate the values of n and m for which this combination would send maximum current through an external resistance of 1.5 Ω.

Or

For the circuit shown here, calculate the potential difference between points B and D.

Ans. Let there be m rows in parallel, each row with n cells each of internal resistance r.

Then, p.d across the circuit = nE

Internal resistance of each row = nr

Internal resistance of m such rows in parallel = nrm

or Internal resistance = nrm

Therefore, total resistance = nr Rm

+

Current, I, through R = nEnr Rm

+ = mnE

nr mR+

For maximum current the denominator nr + mR be minimum.

nr + mR = 2( ) 2nr mR nrmR− +

R

Since nrmR cannot be zero nr = mR

or R = nrm

From the data 1.5 = 0.5nm

× ⇒ nm

= 1.50.5

= 3

Also, nm = 12 (Total no. of cells)

Therefore nnmm

× = 36

or n = 6

and m = 2

Or

Applying junction rule at C, i = i1 + i2 …(1)

Applying Kirchhof’s laws to the mesh CABC

2 × i + 1 × i + 2 × i1 = 2 – 1 = 1

3i + 2i1 = 1 …(2)

Applying Kirchhof’s law to mesh DCBD

3 ×i2 + 1 × i2 – 2 × i1 = 3 – 1

4i2 – 2i1 = 2 …(3)

Solving i1 = – 1 A13

, i2 = 6 A13

and i = 5 A13

Sign of i1 shows that B is at a higher potential than C.

VB – VC = V1 = i1 × 2 = 2 V13

.

VB – VD = – 1 × i2 – 1 = 6 191=13 13

V− −

AlTErNATiVElY

VC – VD = 3 i2 –3 = 18 213 =13 13

V− −

VB – VD = ( ) ( ) 2 21 1913 13 13B C C DV V V V V− + − = − = −

Note: The student may practise calculating for VAB, VAC, VCB.

17. A beam of light of wavelength 400 nm is incident normally on a right angled prism as shown. it is observed that the light just grazes along the surface AC after falling on it. Given that the refractive index of the material of the prism varies with the wavelength λ as per the relation µA = 1.2 + b/λ2

Calculate the value of b and the refractive index of the prism material for a wavelength λ = 5000 Å. [Given θ = Sin–1 (0.625)]

Ans. The R.I. for 4000Å is higher and this beam emerges grazing the surface i.e. Snell’s law at AC is µ sinθ = sinr.

µ4000Åsinθ = sin 90°

21.20 0.625(4000)

b + × = 1

or b = ( )2 1 1.20 0.625(4000)0.625

− ×

= 6 0.2516 100.625

× × Å2 = 16 × 106 × 0.4 Å2 = 6.4 × 106 Å2

µ = 6

26.4 101.2 1.456(5000)

×+ =

18. Three students X, Y, and Z performed an experiment for studying the variation of alternating currents with angular frequency in a series lCr circuit and obtained the graphs shown as. They all used a.c. sources of the same r. m.s. voltage and inductances of the same value.

What can we (qualitatively) conclude about the

(i) capacitance value

(ii) resistance values

used by them? in which case will the quality factor be maximum?

What can we conclude about nature of the inpendance of the set up at frequency ω0?

Ans. (i) The ciruits have the same values of L for the students X, Y and Z. The resonance peak 01LC

ω =

is the same. Therefore we conclude that the three circuits have the same value of capacitance C.

(ii) For smaller values of R, the band width 2 1( ) RL

ω − ω = is smaller and the resonance curve is

sharper. Therefore the curve X has the smallest resistance. For ω1 and ω2, one higher than ω0

and the other smaller than ω0, the rms current is I0/ 2 .

The quality factor Q = 0LR

ω is maximum for curve X which has the smallest resistance.

19. An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image conicides with the needle itself. The distance of the needle from the lens is measured to be ‘a’. On removing the liquid layer and repeating the experiment the distance is found to be ‘b’.

Given that two values of distance measured represent the focal length values in the two cases, obtain a formula for the refractive index of the liquid.

Ans. The distance of the needle in the first case gives the focal length ‘F’ of the the system which is combination of equiconvex lens and planoconcave liquid lens of focal length f1 and in the second case focal length ‘f1’ of the lens alone. Thus the focal length f2 of the planoconvex lens of liquid

is given by 1 2

1 1 1f f F

+ =

1F

= 1 2

1 1f f

+

we have 2

1f

= 1 1a b

−

⇒ f2 = abb a−

…(1)

Using lens maker’s formula for planoconvex lens

2

1f

= 1 2

1 1( 1)R R

µ − −

Where µ is the RI of the liquid, using sign convention R = – r and R 2 = ∞

\ 2

1f

= 1( 1)r

µ − −

Putting the value of f2 from ‘1’ we have

b aab− = 1( 1)

r µ − −

or b arab− = (1 )− µ

and µ = 1 – r. b aab−

20. A circular coil having 20 turns, each of radius 8 cm, is rotating about its vertical diameter with an angular speed of 50 radian s–1 in a uniform horizontal magnetic field of magnitude 30 mT. Obtain the maximum, average and r.m.s. values of the emf indued in the coil.

if the coil forms a closed loop of resistance 10Ω, how much power is dissipated as heat in it?

Ans. At an instant when the normal to the coil makes an angle θ with the field, the magnetic flux passing through the coil is

φ = BA cos ωt

where A is the area of the coil. If ω is the angular velocity θ = ωt.

We have taken at t = 0, θ = 0

Thus φ = BA cos θ

Since ε = d ddt dt

− φ −= (BA cos ωt) = BA ω sin ωt

For N turns ε = BAωN sin ωt

(i) Maximum value is BAωN = 30 × 10–3 T × π (8 × 10–2)2m2 × 50 rad/s × 20

≈ 0.60 V

(ii) Average value of emf in one cycle is zero.

(iii) RMS value is 0.60 / 2 volt.

The power dissipated is VRMS IRMS = 2

RMSVR

=

20.6

210

= 0.36

20 = 0.018 joule/sec.

21. The nucleus of an atom of 23592 Y , initially at rest, decays by emitting an a-particle as per the

equation 231235 49092 2He +EnergyY X→ +

it is given that the binding energies per nucleon of the parent and the daughter nuclei are 7.8 meV and 7.835 meV respectively and that of α-particle is 7.07 meV/nucleon. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, calculate the speed of the emitted α-particle. Take mass of α-particle to be 6.68 × 10–27 kg.

Ans. 231235 49092 2+ He+EnergyY X→

Y has 92 protons and 143 neutrons. The rest mass energy of this nucleus would be

E = (mp × 92 + mn × 143)c2 – 7.8 × 235 MeV

X has 90 protons and 141 neutrons. The rest mass energy will be

E1 = (mp × 90 + mn × 141)c2 – 7.835 × 231 MeV

The α particle has 2 protons and 2 neutrons.

The rest mass energy will be

E2 = (mp × 2 + mn × 2)c2 – 7.07 × 4 MeV

The energy released due to breaking is

E – (E1 + E2) = [(7.835 × 231 + 7.07 × 4) – 7.8 × 235] MeV

= [(1809.9 + 28.28) – 1833] MeV

= [1838.18 – 1833] MeV

= 5.18 MeV = 5.18 × 1.6 × 10–13 joule.

This energy is taken up by the α particle

212

mv = Ek or v = 13

27

2 5.18 1.6 102

6.68 10kE

m

−

−× × ×

=×

≈ 1.6 × 107 ms–1

22. Define the term ‘modulation index’ for an Am wave. What would be the modulation index for an Am wave for which the maximum amplitude is ‘a’ while the minimum amplitude is ‘b’?

Ans. The process of changing amplitude of the high frequency carrier wave in accordance with the changes in the intelligence (code, voice or music) to be transmitted, keeping the frequency and the phase of the carrier wave unaltered is called amplitude modulation.

Modulation index m = amplitude change of carrier wave

amplitude of unmodulated carrier wavem

c

VV

=

From the graph 2Vm = Vmax – Vmin

Or Vm = max min

2V V−

…(1)

and Vc = Vmax – Vm = Vmax – max min

2V V−

= max min

2V V+

…(2)

Dividing (1) by (2) m = max min

max min

V VV V

−+

= a ba b

−+

23. Subhashish, a grade 12 student visits his uncle Sachin’s place. His uncle has just received his electricity bill from the electric supply company and is complaining about the huge charges he will have to pay. Subhashish quietly inspects the electrical appliances and observes that his uncle is using tungsten filament lamps for lighting. He talks to his uncle about the bill and suggests

immediate replacement of five 100 W electric lamps in the house by 40 W fluorescent tubes; three 60 W lamp by 15 W CFls and 20 W night lamp by a 2 to 4 watt diode lamp. He also explains the effect of use of high power devices on the environment and the simple means by which the energy can be saved in the kitchen as well. His uncle agrees to try out the changes and is happy to see a marked reduction in energy charges in the subsequent bill. His uncle agrees to talk to his friends also about the need for such changes in their homes.

The 100 W lamps are used in the house for an average 6 hours daily; 60 W lamps for 8 hours each day and the night lamp for 10 hours a day. The electricity charges with taxes are about rs 5 per kWhr.

(i) Calculate the cost of lighting for Sachin’s household for a month.

(ii) What would be the cost if the lamps are replaced as suggested by Subhashish?

(iii) in your opinion, what are the two qualities Subshashih posesses?

Ans. (i) Energy consumed by lamps in kWhr in one month

= (watts)1000

P × t (hours/day) × 30

= 5 100 3 60 206 30 8 30 10 301000 1000 1000× ×

× × + × × + × ×

= (90 + 43.2 + 6) kWhr

= 129.6 kWhr

Cost = 129.2 × 5 = ` 642.

(ii) Energy consumed for lighting by the new devices in one month

= 5 40 6 30 3 15 8 30 4 10 30

1000 1000 1000× × × × × × × ×

+ +

= 36 + 10.8 + 1.2

= 48 kWhr

Reduced cost = 48 × 5 = Rs 240. (iii) (a) Concern for environment (b) Desire to apply the knowledge acquired for welfare of the society. 24. Two circular coils X and Y having radii R and R/2 respectively are placed in horizontal plane

with their centres coinciding with each other. Coil X has a current I flowing through it in the clockwise sense. What must be the current in coil Y to make the total magnetic field at the common centre of the two coils, zero?

With the same (i.e. equal) currents flowing in the two coils in same direction, if the coil Y is now lifted vertically upwards through a distance R, what would be the net magnetic field at the centre of coil Y?

Or A straight thick long wire of uniform cross section of radius ‘a’ is carrying a steady current

I. Use Ampere’s circuital law to obtain a relation showing the variation of magnitude of the magnetic field (B) inside the wire with distance r, (r ≤ a) and (r > a) of the field point from the centre of its cross section. plot a graph showing the nature of variation of the field.

Calculate the ratio of magnetic field at a point 2a above the surface of the wire to that at a

point 2a below its surface. What is the maximum value of the field of this wire?

Ans. (i)

The magnetic field B at the centre of 0 2X

R= µ I The direction is as given by RH rule pointing

downward.

To make the net B field zero, the current in the coil Y be I’ in the anticlockwise direction given by

B = 0 2( /2)R′µ I

\ 0 :R

µ I = 0

2Rµ I

⇒ I′ = 2I

in anti-clock wise direction

(ii)

The magnetic field B at the centre of Y due to current I in X

= 2

0 2 2 3/22 ( )R

R Rµ

+I = 0

1.2 8R

µ I

The magnetic field B2 at the centre of Y due to current in Y itself

= 0 2( /2)Rµ I = 0

2.2 R

µ I

Net magnetic field B is the vector sum of B1 and B2.

Therefore, B→

= 1 2B B→ →

+ = 01 2

2 8 RR +µ +

I

Or

(i) r > a According to Ampere’s law

C

B dl→→

∫ = µ0I [For loop C as shown above]

B⋅ 2πr = µ0I

B = .0

2µ

πIr

…(i)

(ii) r = a. Result (i) is also valid for this case

B = 0 .2 aµ

πI

(iii) r < a. In this case the curent passing through the area bounded by the circular path is not I but it is given by I(πr2/πa2) i.e. Ir2/a2

B dl→→

∫ = 2

0 2ra

µ

I

B2πr = µ0 2

2raI

B = µ

π0

22I r

a

Ratio of Bext to Bint is

3 /2

/2

a

a

B

B =

0

02

2 (3 /2) 4/2 3

2 ( )

µaaa

π=µ

π

I

I

25. State the principle which helps us to determine the shape of the wavefront at a later time from its given shape at any time. Apply this principle to

(i) Show that a spherical/plane wavefront continues to propagate forward as a spherical/plane wavefront.

(ii) Derive Snell’s law of refraction by drawing the refracted wavefront corresponding to a plane wavefront incident on the boundary separating a rarer medium from a denser medium.

Or What do we understand by ‘polarization’ of a wave? How does this phenomenon help us to

decide whether a given wave is transverse or longitudinal in nature? light from an ordinary source (say a sodium lamp) is passed through a polaroid sheet P1. The

transmitted light is then made to pass through a second polaroid sheet P2 which can be rotated so that the angle (θ) between the two polaroid sheets varies from 0° to 90°. Show graphically the variation of the intensity of light, transmitted by P1 and P2, as a function of the angle θ. Take the incident beam intensity as I0.

Why does the light from a clear blue portion of the sky, show a rise and fall of intensity when viewed through a polaroid which is rotated?

Ans. Huygen’s principle (i) Once a wavefront is formed all points on a given wavefront act as new source of disturbance

called secondary wavelets. These wavelets send disturbance in all directions with the velocity of wave.

Secondarywavelet

Secondarywavefront

Primarywavefront

P1

S1

S2P2

The forward (tangential) envelope of all these wavelets at any time gives new wavefront as shown in figure (i) If the primary wavefront (P1 P2) in spherical; the secondary wavefront (the forward envelope S1S2) is also spherical.

(ii)

Let AB be a plane wavefront incident on a refracting surface XY at an angle of incidence i. Let medium (1) be the rarer medium where the speed of light is c1, and medium 2 be the denser medium where the speed is c2 (c2 < c1).

First of all, disturbance from wavefront AB strikes at the point B. By the time 1

'AAtc

=

disturbance from A reaches A′, disturbance from B would have spread in the second medium

in the form of hemispherical wavelet of radius BB′ 2 21

AAc t cc

′= = ×

Tangent from A′ on this wavelet gives refracted wavefront A′B′. Wavefront A′B′ makes an angle r with refracting surface.

In ∆ BAA′, sin i = AABA

′′

In ∆ BB′A′, sin r = BBBA

′′

sinsin

ir

= //

AA BABB BA

′ ′′ ′

= AABB

′′

= 1

2

c tc t

= 1

2

cc

= constant

sinsin

ir

= constant = n21

= refractive index of second medium with respect to first medium. This is Snell’s law.

Or

In an unpolarised wave the displacements are randomly changing with time in all directions although always perpendicular to the direction of propagation. When the wave is plane polarised the oscillations are confined to a single direction.

Only transverse waves can be polarised. A given wave is transverse or longitudinal can be known from its polarisation. If it can be polarised it is a transverse wave.

When the unpolarised light of intensity I0 with electric vector E0 is passed through the sheet P1 its intensity is reduced to (I0/2). This is because vectors of electric oscillations can be resolved into two components (i) perpendicular to the pass direction of the polariser and (ii) parallel to the pass direction. One set of these components is blocked and the transmitted light has an intensity is (I0/2).

According to Malus, law, if the pass axis of P1 an P2 make an angle θ with each other the component E cos θ along the pass axis of P2 will pass through P2. The intensity of light will vary as I = (I0/2) cos2θ, when any of the two polaraoids is rotated.

intensity variation of blue light:

The incident sunlight is unpolarised. When it is scattered by molecules of atmosphere it gets polarised. The polarised light will have intensity variation (∝ I0 cos2 θ) when viewed through a polariser when it is rotated.

26. A student has to study the input and output characteristics of a n-p-n silicon transistor in the common emitter configuration. What kind of a circuit arrangement should she use for this purpose?

Draw the typical shape of input characteristics likely to be obtained by the student. What do we understand by the cut off, active and saturation states of the transistor? in which of these does the transistor not remain when being used as a switch?

Or

input signals A and B are applied to the input terminals of the ‘dotted box’ set-up shown here. let Y be the final output signal from the box.

Draw the wave forms of the signals labelled as C1 and C2 within the box, giving (in brief) the reasons for getting these wave forms. Hence draw the wave form of the final output signal Y. Give reasons for your choice.

What can we state (in words) as the relation between the final output signal Y and the input signals A and B?

Ans.

Circuit arrangement to study input and output characteristics in CE configuration.

(i) In active region IC depends on IE.

EB junction is forward biased and CB junction is reverse biased.

(ii) In saturation region, IC is independent of IB.

EB is forward biased and CB is reverse biased.

(iii) In cut off region there is no current in the transistor it is like an open switch because the base does not attract electrons from the emitter. Both EB and CB junctions are reverse biased or VBE < VBarrier. When a transitor is working as a switch, the cut off (off) and saturation (on) regions are stable regions of its operation, while active region is the unstable (i.e. transient) region of its operation. So the transistor should not be in active region.

Or

(i) Gate 1 is a NAND gate. Both terminals are connected together. Therefore, it acts as a NOT gate. A becomes A .

(ii) Gate 2 is also a NOT gate. B becomes B .

(iii) The output C1 is from AND gate. Therefore it is A .B.

(iv) The output C2 is from AND gate. Therefore it is A. B .

(v) Gate 3 is NOR gate. Therefore, Y is A.B + A.B

Y = A.B+ A.B

Interval A B A B C1 = .A B 2 .C A B= . .A B A B+0 – 1 1 0 0 1 0 1 0

1 – 2 1 1 0 0 0 0 1

2 – 3 1 1 0 0 0 0 1

3 – 4 0 0 1 1 0 0 1