physics (cbse 2011) - · pdf fileloop connected across a capacitor ... lower plate is negative...

1. Define electric dipole moment. Write its S.I. unit. [1]

SolutionAn electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space. By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole.

Mathematically, dipole moment is defined as

p qa= 2

Therefore the SI unit of the electric dipole is Cm.

2. Where on the surface of Earth is the angle of dip 90°? [1]

SolutionAt poles the angle of dip of a magnetic compass is 90°.

3. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere? [1]

SolutionThe net electric field inside the metal shell is zero. Therefore the work done to move the charge from the surface to the centre is zero. Potential at a point is the work done to move a unit positive charge from infinity. Therefore, the potential at the centre of the metal shell = potential at the surface = 10 V

4. How are radio waves produced? [1]

SolutionRadio waves are produced by the accelerated motion of charges in conducting wires. In practice electronic

oscillators which can generate waves of constant frequency is used to generate the radio waves.

5. Write any two characteristics properties of nuclear force. [1]

Solution(a) short range force.(b) Independent of charge

6. Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. [1]

NS NS

C

Solution

NS NS

C+-

While the two magnets approach each other the magnetic flux inside the loop increases, therefore reducing the flux change, induced currents are generated in the loop, direction of the external magnetic field is from north poles to south pole, and the direction of induced magnetic field is opposite to that. And from Ampere’s circuital law, the direction of current is as shown in figures. Therefore the lower plate is negative and the upper plate positive.

7. What happens to the width of depletion layer of a p-n junction when it is (a) forward biased, (b) reverse biased? [1]

PHYSICS (CBSE 2011)Time: 3 hours Max. Marks: 70

General Instructions

1. All questions are compulsory.2. There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27

carry three marks each and questions 28 to 30 carry five marks each.3. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks

and all three questions of five marks each. You have to attempt only one of the given choices in such questions.4. Use of calculators is not permitted.5. You may use the following values of physical constant wherever necessary: c = 3 × 108 ms–1 h = 6.626 × 10–34 Js e = 1.602 × 10–19 C μ0 = 4π × 10–7 T m A–1 ε0 = 8.85 × 10–12 C2N–1m–2

14

9 109 2

πε0

2N m C= × -

Mass of electron me = 9.1 × 10–31 kg

Physics Special market Book 2_2011 CBSE .indd 1 12/21/2011 4:50:20 PM

Physics (CBSE 2011)2

SolutionIf the p-n junction diode is forward biased the applied voltage opposes the barrier potential and therefore the width of the depletion layer decreases. When a p-n junction is forward biased the majority carriers moves away from the depletion region and therefore the width of the depletion layer increases.

8. Define the term stopping potential in relation to photoelectric effect. [1]

Solution The negative voltage applied to reduce the velocity of the most energetic photoelectron to zero is known as stopping potential. Alternatively the kinetic energy of the most energetic electron is eV where V is the stopping potential and e is charge of electron.

9. A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [2]

SolutionThe electric field of infinitely long wire

Er

r

= 12 0πε

λ

Therefore, the flux through the surface of cylinder

E da

rrl

li i= =1

22

0 0πελ π λ

ε

10. Plot a graph showing the variation of coulomb force (F)

versus 12r

, where r is the distance between the two

charges of each pair of charges : (1 μC, 2 μC) and (2 μC, – 3 μC). Intercept the graphs obtained. [2]

Solution The force between the 1µC and 2 µC charge is

F kq q

rk

r11 2

2

12

2

2 10= = × -

The force between the 2 µC and –3 µC is

Fk

r2

12

2

6 10= - × -

The graph of force (F ) versus 12r

, is a straight line.

F1

F2

From the graph we can see that the force Fr

∝

12 . The

force increase with increase in (1/r2) .

The slope of the curve gives the proportionality constant

in the first case i.e for F1 it is 2 10 12kNm

× - and for F2 it is

- × -6 10 12kNm

.

11. Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity

v in a

magnetic field B . Show that no work is done by this force

on the charged particle. [2]

OR

A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2) in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the expression for this force. [2]

Solution AThe magnetic Lorentz force or the magnetic force acting on a moving charge is given by the expression

F q v B= ×( )

The work done is defined as

W F dl= i

Therefore the work done by the magnetic force

W F dlmag mag=

i

= ( )×q dlv B

i

= ( ) = ⇒ =

×q v dtdldt

v dl v dtv B

i

W q v B vmag = =

0

(If one element is repeated in a box product, then the box product is zero)

OR

Solution B

a

b

d

Fba

BaIb

L

Ia

Let two straight parallel conductors a and b be separated by a distance d.


Physics (CBSE 2011) 3

According to Ampere circuital law the magnetic field experienced by the conductor b due to the current in a is

BIdaa= µ

π0

2

Along the direction perpendicular to wire b.Therefore, the magnetic force acting on the wire at a wire segment of length L

F q v Bba = ×( )

F I LB

I I Ldba b a

a b= = µπ

0

2

12. What are eddy currents? Write any two applications of eddy currents. [2]

Solution When a rapidly varying magnetic field is applied across a conductor small circulating current loops are formed throughout the volume of the conductor. These are known as eddy currents.

Applications

(a) Electromagnetic braking: The eddy current brakes are used in electric trains and roller coasters. A constant magnetic field is applied to a metallic drum attached to the wheels. The eddy currents generated in the drum slows down the trains.

(b) Induction heating: If eddy currents are generated in a metal with high resistivity, the metal heats instantly. Therefore the eddy currents are used in induction cookers and induction furnaces.

13. What is sky wave communication? Why is this mode of propagation restricted to the frequencies only upto few MHz? [2]

Solution In the frequency range from a few MHz up to 30 to 40 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation and is used by short wave broadcast services. Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape. Therefore the sky wave propagation is restricted to a few megahertz.

14. In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]

R12 Ω

2 A1 A D

R

A 1 V 2 AC

2 VB

Solution

R12 Ω

2 A1 A D

R

A 1 V 2 AC

2 VB

By applying Kirchhoff’s current law at junction 1

3 A = 1 A + current through 2 Ω.

Therefore the current through 2 Ω is 2 A, from D to C. Applying Kirchhoff voltage law in loop DCB

4 = 4 – 2 V or the resistance R1 is zero.

Therefore the potential at B and C is equal. Applying Kirchoff’s voltage law in loop ACBDA 4 V and 2 V are connected in series with B as a point between the two batteries. Therefore potential at B is 2 V.

15. A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. [2]

Solution Unlike a d.c. current an alternating current can pass through a capacitor. As the current moves from one plate to the other, there should be a magnetic field associated with that as explained by Ampere is circuital law. But in a capacitor the current transfer does not happen by transfer of charges but still the magnetic field exists. For explaining this, Maxwell modified the Ampere’s theorem with the help of continuity equation. According to Ampere’s circuital theorem

B dl ii =∫ µ0

Maxwell added an extra term to the right of the Ampere’s circuital theorem.

B dl i

ddt

Ei = +∫ µ ε0 0

Φ

The extra term has the same dimension of current and known as displacement current. According to Maxwell the displacement currents generates the magnetic field inside the capacitor plate. The displacement current is zero elsewhere except inside the capacitor because the electric flux is zero. Inside the capacitor the physical current is zero but the displacement current is present.

16. Net capacitance of three identical capacitors in series is 1 μF. What will be their net capacitance if connected in parallel? [2]

Solution The net capacitance of a series connection of capacitors is

1 1 1 1

1 2 3C C C C= + +



Therefore the effective capacitance of three identical cap-acitors in series is

CC

s = =3

1µF

Capacitance of one capacitor is

C = 3µF

If capacitors are connected in parallel the effective cap-acitance will be

C C C Cp = + +1 2 3

C CP = =3 9µF

17. Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release in energy in the processes of nuclear fission and nuclear fusion can be explained. [2]

Solution

9

8

7

6

Ave

rage

bin

ding

ene

rgy

per

nucl

eon

(MeV

)

5

4

3

2

1

00 30 50 90 120

Mass number of nucleons in nucleus150 180 210 240 270

U235Fe56

O16

C12

Li7

Li6

H3

He3

H2

H1

U238

He4

From the curve it is clearly visible that the binding energy of the atoms with lower atomic mass is low. So if two atoms with low mass number combined together to form an atom with higher mass number the binding energy increases which result in subsequent release of energy. This is called nuclear fusion. The Iron(Fe) has the highest binding energy. The binding energy of the atoms higher mass number than iron is lesser and the reduction binding energy increase with increase in the mass number. So if an atom with large mass number split to form two or more middle class atoms the binding energy per nucleon increases. This process also results in release of energy.

18. In the meter bridge experiment, balanced point was observed at J with AJ = l.

(a) The values of R and X were doubled and then interchanged. What would be the new position of balance point?

(b) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected? [2]

SolutionR

B

G

A J C

X

The balancing condition is

R XAJ AJ

=-1

Rl

Xl

=-1

If the R is doubled, then R1 = 2R; X1 = 2X

RX

RX

RX

ll

1

1

22 1

= = =-

The balancing point for the R and X do not change even if R and X is doubled. Therfore If the position of R and X is interchanged then the new balancing point is

AJ = -1 l

19. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (a) a medium of refractive index 1.65, (b) a medium of refractive index 1.33.

(a) Will it behave as a converging or a diverging lens in the two cases?

(b) How will its focal length change in the two media? [3]

Solution Given that Refractive index of the lens ng=1.5 Refractive index of first medium n1 = 1.5 Refractive index of the second medium n2 = 1.33

A convex lens behaves as a diverging lens if the surrounding medium is optically denser than the material of the lens. Therefore in the first medium the refractive index n1 =1.5 > ng=1.5. Therefore the lens behaves as a diverging lens if dipped in the first medium. In the second medium the refractive index ng=1.5 > n2 = 1.33. Therefore the lens behaves as converging lens in the second medium. The focal length of the lens is given by

11

1 1

1 2fn

R R= -( ) -

(1)

1

1 5 11 1

1 2f R R= -( ) -

. (2)

If the material is dipped in medium 1 then the new focal length

11

1 1

1 1 1 2f

n

n R Rg= -

-



1 1 51 6

11 1

1 1 2f R R= -

-

.

. (3)

By dividing Equation (1) by (3)ff1 0 5

0 625= - .

.Therefore, the new focal length is

f f1 80= -

The new focal length is the 80 times of the original focal length and diverging.In the second medium

1 1 51 33

11 1

2 1 2f R R= -

-

..

(4)

By dividing Equation (1) by (4)

ff2 0 5

0 13= .

.

or

f f2 3 84= .

Therefore, the focal length of the lens increases 3.84 times and it remains as converging lens.

20. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, n1 > n2, of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer. [3]

Solution The stopping potential increases with increase in frequency. Therefore n1 has the higher stopping potential compared to n2. We have the kinetic energy of the electron

K E h. = -ν Φ

Where Φ is the work function of the metal.But K.E = eV where e is the charge of the electron and V is the stopping potential. Therefore

Vh

e= -ν Φ

Therefore the stopping potential increases with increase in frequency.

Photoelectriccurrent

n2< n1

–V01 –V02 0 Collector plate potential

Saturation current

Retarding potential

n2

n1

21. Write briefly any two factors which demonstrate the need for modulating a signal. Draw a suitable diagram to show amplitude modulation using a sinusoidal signals as the modulating signal. [3]

Solution • For transmitting a signal, we need an antenna or an

aerial. This antenna should have a size comparable to the wavelength of the signal (at least l/4 in dimension) so that the antenna properly senses the time variation of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength l is 15 km. Obviously, such a long antenna is not possible to construct and operate.

• The power radiated by an antenna of length l is proportional to (l/l)2. This implies that for the same antenna length, the power radiated increases with decreasing l, i.e., increasing frequency. Hence, the effective power radiated by a long wavelength baseband signal would be small.

• If we send the baseband signal directly the signals from different transmitter get mixed up and the information will be lost.

Because of these reason we use modulation for transmitting information for long distance effectively.

O

(a)

m(t)

t

(b)

O

c(t)

t

(c)

O

AAc

t

(Ac + Am sin wmt), top envelope

(Ac + Am sin wmt), bottom envelope

Am

Am

Amin

Amax

22. Use the mirror equation to show that(a) an object placed between f and 2f of a concave mirror

produces a real image beyond 2f.(b) a convex mirror always produces a virtual image

independent of the location of the object.(c) an object placed between the pole and focus of a concave

mirror produces a virtual and enlarged image. [3]

Solution The equation of a mirror is

1 1 1v u f

+ =

v

fuu f

=-

The focal length of the concave mirror is negative and u is also negative. So if |2f| > u > |f|



f

uC

F2F

Then fu u f> - <0 0and

Therefore,

vVeve

ve= +-

= - Therefore the image formed is real.

Also vfu

u ff u f=

-> >2 ( )

(b) For a convex mirror

vfu

u f=

-

f

uC

F2F

f u> <0 0;

Therefore

vveve

ve= -+

= -

Therefore, the image is always virtual irrespective of the postion of the object.

(c) The image distance

vfu

u f=

-

f

uC

F2F

If an object placed between the pole and focus of a concave mirror then |f |>|u|

vveve

ve= ++

= +

Therefore, the image is virtual. Also |fu|>|u-f | therefore the v > u.The magnification

mvu

= - >1

The image is enlarged.

23. Draw a labeled diagram of a full wave rectifier circuit. State its working principle. Show the input-output waveforms. [3]

Centre-taptransformer

Diode 2(D2)

Diode 1(D1)

Output

Centre

Tap BA X

RL

Y

Solution For the positive half cyle of the a.c. current the diode D1 will be forward biased and the diode D2 will be reverse biased. So, the current pass through X to Y of the load resistance RL During the negative half cycle the diode D2

is forward biased and D1 is reverse biased. Therefore the current passes through the load resistance from X to Y. In both the cycle the direction of the current in the load resistance is the same. Therefore the output of the given circuit is d.c.

Wav

efor

m a

t AW

avef

orm

at B

Out

put w

avef

orm

(acr

oss

R L

)

t

t

t

(b)

(a)

(c)

(i)

(ii)

Due toD2

Due toD1

Due toD2

Due toD1

24. (a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.

(b) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state? [3]



Solution According to de Broglie’s hypothesis the quantization of the orbits can be explained by the wave particle duality. For having a standing wave pattern the length of the orbit of the electron should be an integral multiple of the wavelength of the electron. The wavelength

λ = =hp

hmv

And the perimeter of the orbit

2π λr n=

or nhp

r= 2π

or rp Lnh= =2π

The Bohr’s second postulate is that the angular momentum

of the electron in the orbit is an integral multiple of h

2π.

25. You are given a circuit below. Write its truth table. hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. [3]

A

Z

Y

X

B

Solution

A B X Y Z

0 0 1 1 0

1 0 0 1 0

0 1 1 0 0

1 1 0 0 1

Therefore, the given gate is an AND gate

O

B

A

26. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm form the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope. [3]

OR

A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.

If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m. [3]

Solution

D

B

A

A

h

Bb

b

B

Objective

O E

Eyepiece

u fo feA

h

The distance to the primary image formed by the objective lens O is

1 1 1

0 0 0v f u= +

We have f0 is 4 cm and u0 is -6 cm. then

1 14

16

120

0vor v= - = cm

The magnification of the microscope

m m mvu

Dfe

e

= = +

0

0

0

1

=

-

+

= -126

12510

7

The length of the microscope is v0 + u also u = ue is the object distance of the eyepiece.

1 1 1

0u D fe

= -

Therefore, ue = -7 14. cm

Therefore, the length of the microscope

v ue0 + = 19.14 cm

OR

Magnifying power

A

B

fe

fo

EO

Objective Eyepiece

α α

βh

Angular magnification

= - +

= - +

= -f

ffDe

e0 11500

11

125

1560



Let d is the diameter of the moon and r is the diameter of the image formed. Then the angular magnification tan α =α = d/r = d’/f0 where f0 is the diameter of the image. Therefore,

d fdr

′ =

=

×( )×

=0

8

10

1 3 42 10

3 8 100 135

.

.. m

27. Two heating elements of resistances R1 and R2 when operated at a constant supply of voltage, V, consume powers P1 and P2 respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in (a) series and (b) parallel across the same voltage supply. [3]

Solution

The PVR

and PVR1

2

12

2

2

= =

When the heating elements are connected in series the

PVR

VR Req

eq

= =+

2 2

1 2

PV

VP

VR

P PP Peq =

+=

+

2

2

1

2

2

1 2

1 2

If the heating elements are connected in parallel then

PVReq

eq

=2

But

1 1 1

1 2R R Req

= +

Therefore,

PVR

VReq = +

2

1

2

2

P P Peq = +1 2

28. (a) State the principle of the working of a moving coil galvanometer, giving its labeled diagram.

(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement.

(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. [5]

OR

(a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.

(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.

(c) How is the magnetic field inside a given solenoid made strong? [5]

Solution Moving Coil GalvanometerScale

Sp

Coil

Pointer

Pivot

Soft-ironcore

PermanentmagnetUniform radial

magnetic field

N S

If a current carrying loop is placed in a uniform magnetic field, a torque acts on the loop which is proportional to the current in the loop. This is the working principle of a moving coil galvanometer.

The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis, in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it. This torque is given by

τ = NIAB

Where N is the number of turns of the coil, A area of the coil, I current passing through the coil and B is the magnetic field. The torque tends to rotate the coil. A spring Sp provides a counter torque kF that balances the magnetic torque NIAB; resulting in a steady angular deflection Φ. In equilibrium

k NIABΦ =

Φ =

NABk

I

The quantity in the bracket is called the galvanometer constant. The sensitivity of the galvanometer depends upon the number of turns of the coil, area of the galvanometer coil, pole strength of the permanent magnet and the spring constant. The voltage sensitivity of the instrument is

ΦV

NABk

IV

NABk R

=

=

1

For converting a galvanometer into ammeter a shunt resistance which can bypass most of the current is connected to the galvanometer.



S

Ammeter

(I - Ig) (I - Ig)

IgIg II BAG

The value of the resistance is given by the equation

SI

I IGg

g

=-

Where Ig is the maximum galvanometer current and G is the resistance of the galvanometer coils.

ORMagnetic field due to solenoid: Consider a rectangular Amperian loop PQRS near the middle of solenoid shown in figure where PQ = l

a

P

d c

ω

b

h

Q

B

Let the magnetic field along the path ab be B and zero along cd. As the paths bc and da are perpendicular to the axis of solenoid, the magnetic field component along these paths is zero. Therefore, the path bc and da will not contribute to the line integral of magnetic field B.

The total number of turns in the length l =Nl. The line integral of magnetic field induction on B over the closed path PQRS is

B dl B dl B dl B dl B dla

b

b

c

c

d

d

a

i i i i i= + + +∫ ∫ ∫ ∫∫

B l nIi = µ0

(b) The toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a solenoid which has been bent into a circular shape to close on itself.

The magnetic field inside a torrid is not extended to space. And inside the toroid the magnetic field is constant.

Q

P

Magnetic field of Toroid

3

Q

S

2

r2

r3

r1P1

O

(c) The magnetic field of a solenoid can be made strong by 1) inserting a ferromagnetic core in the solenoid 2) by increase the number of the turns of the coil 3) by increase the current inside in the coil.

29. State the working of a.c. generator with the help of a labeled diagram. The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity w. Deduce the expression for the alternating e.m.f. generated in the coil.What is the source of energy generation in this device? [5]

OR

(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by p/2 in phase.

(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.

(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.

(ii) What is the direction of the e.m.f. ? (iii) Which end of the wire is at the higher potential? [5]

Solution

A.C. Generator

Slipring

Carbonbrushes

Alternating emf

Coil Axle

sN



The device used to generate a.c. voltage (or current) using the principle of electromagnetic induction is called a.c. generator. An a.c. generator consists of a metallic coil called armature rotating between the poles of the field magnet as shown in the figure. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The ends of the armature coil are welded to two metallic rings called slip rings which rotate along with the armature. Two carbon brushes rub against the slip rings and the wires which carries the power to the external circuit is connected to the carbon brushes. As the armature rotates in the magnetic field, The flux linked to the coil changes and hence induced currents generated in the coil. When the coil rotated with constant angular velocity w, then the angle θ between the magnetic field vector and the area vector is

θ ω θ= = =t t( )Assuming at0 0

Therefore the flux at time t is

Φ( ) cos cost BA BA t= =θ ω

Therefore the emf induced in the coil

ε ω ω= - =Nddt

NBA tΦ

sin

The maximum emf induced

ε ω0 = NBA

Therefore

ε ε ω= 0sin t

An ac generator is an instrument which converts mechanical energy into electrical energy. The energy is generated in is given by the driving agency usally stem or running water.

OR

L

ε

According to Kirchhoff’s Law net voltage around the loop is zero.

∑ =ε i 0

or vmsinωtLdidt

- = 0

or didt

v tL

= msinω

By integrating both sides

i

vL

tvL

tm m= - = -

ω

ωω

ω πcos sin

2

(b) (i) The emf induced in the coil

ε = Blv

Due to the earth’s magnetic field electrons in the rod move to the west thus making the east end of the rod positive. As more and more electrons reaches the end the rod become equilibrium. At equilibrium the voltage between the ends

ε = Blv

(ii) The east end of the positively charged and the west end negatively.

(iii) The east end of the wire is at higher potential because the east end is positively charged and the west end negatively.

30. State the importance of coherent sources in the pheno-menon of interference.

In Young’s double slit experiment to produce interference pattern, obtain the conditions for construc-tive and destructive interference. Hence deduce the expression for the fringe width. How does the fringe width get affected, if the entire experimental apparatus of Young’s is immersed in water? [5]

OR(a) State Huygens principle. Using this principle explain

how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe.

(c) If a monochromatic source of light is replaced by white light, what change would you observed in the diffraction pattern? [5]

SolutionTwo sources are said to be coherent if they have a con-stant phase relation among themselves. The interference patters are visible only when two coherent sources of light are superpostioning. Let two slits are separated by a sistance d. Screeen is placed from at a distance D. The slits S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2. Thus, the two sources S1 and S2 will be locked in phase; The constructive interference happens when the wavelets emitting from both sources are in phasei.e. S2P – S1P = nλ; n = 0, 1, 2

Then

S P D xd

Dx

d

D22

212

2

21

12

2= + +

≈ +

+

Similarly

S P Dx

d

D1

2

112

2= +-



S1

d

G

P

OZ

G

S2 D

x x

zy

Therefore path difference

S P S P

xd

xd

DxdD2 1

2 2

2 22

- =+

- -

=

Therefore the point P will be maximum intensity if

xdD

n= λ

or

xn D

d=

λ

The condition for destructive interference is

Path difference = +

n12

λ

Therefore

xdD

n= +( )12

λ

or

xn D

d=

+( )2 12

λ

The fringe width is the separation between the consecutive bright or dark fringes. Taking bright fringes

β λ λ λ= - = + - =+x x

n Dd

n Dd

Ddn n1

1( )

If the whole apparatus is immersed in water then the path length D increase by a factor of n where n is refractive index of water. Therefore the new fringe width

β λ’ = n D

d

Therefore the fringe width increases by a factor of n.

OR(a) According to Huygens principle:

1. Each and every point on the given wavefront, called “primary wavefront,” acts as a source of new disturbances, called “secondary wavelets,” that travel in all directions with the velocity of light in the medium.

2. A surface touching these secondary wavelets tangentially in the forward direction at any instant gives a new wavefront at that instant, which is known as the secondary wavefront.

P1

D

P0

r1

r2 a/2

a/2

B

Viewingscreen

C

Central axis

Incidentwave

θ

Totally destructiveinterference

b

θ

r1

r2a/2 b

θ

Path lengthdifference

θ

(b) The time taken to travel one wave front to other is same along any ray.

a O

l/a

2l/a

sin–1q

–l/a

–2l/a

L

q

The width of the first interference fringe Δθ = l/d. And the width of the second (diffraction) fringe

Δθ = l/2d. (c) While doing diffraction with white light the central

fringe will be white but the subsequent bright fringes will be colored .

The fringe width and angular fringe width will not be constant.


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