C.B.S.E.12th Class

PHYSICSMOCK TEST PAPER

withSOLUTIONS

Presents

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Page 2 of 14

Page 3 of 14

Time : 3 Hours Maximum Marks : 70

GENERAL INSTRUCTIONSI. All questions are compulsory.

II. There are 30 questions in total. Questions 1 to 8 are very short answer type questions and carry one mark

each.

III. 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.

IV. 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 choice in such questions.

V. Use of calculators is not permitted. However, you may use log tables if necessary.

VI. You may use the following values of physical constants wherever necessary.

83 10 /c m s= × , 34

6.63 10h Js−= × , 191.6 10e C−= × ,

7 1

0 4 10 TmAµ π − −= × ,

29

2

0

19 10

4

Nm

Cπε= ×

1. Why is electric field intensity inside a charged conductor zero ?

2. A circular loop carrying current I is placed at right angles to a magnetic field. What is the net torque on the

loop ?

3. What is one henry ?

4. Give the formula for velocity of electromagnetic waves in free space.

5. Determine refractive index of a substance if critical angle is 45o .

6. In photo-electric effect, for a metal the K.E. of the photo-electron depends upon which factor ?

7. What is the function of transmitter antenna ?

8. What should be the length of a dipole antenna for a carrier wave having frequency 83 10 Hz× ?

9. Define the term electric dipole moment. Is it scalar or vector ?

10. Equipotential surfaces of some electric field are shown in figure. Given 1 2V V> .

V 2

V 2V 1 >

Draw the approximate distribution of the lines of force of the field and indicate their

direction. Determine in which region is the field intensity larger ?

11. Three identical resistors, each of resistance R , when connected in series with a d.c. source, dissipate

power X . If the resistors are connected in parallel to the same d.c. source, how much power will be

dissipated ?

12. A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0cm as shown in

figure (a). What is the direction and magnitude of Br

at the centre of the arc ? Would your answer be

different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown

in figure (b) ?

(a ) (b )

13. A straight wire of length L and carrying a current I stays suspended horizontally in mid air in a region

where there is a uniform magnetic field Br

. The linear mass density of the wire is λ . Obtain the magnitude

and direction of this magnetic field.

Page 4 of 14

14. An ordinary moving coil ammeter used for d.c. cannot be used to measure an alternating current even if its

frequency is low. Explain why.

15. A coil is wound on an iron core and looped back on itself so that the core has two sets of closely cound

wires in series carrying current in opposite senses. How is it self inductance affected ?

16. What are coherent sources of light ? Why no interference pattern is observed when two coherent sources

are (i) too close (ii) very far apart ?

OR

Mark the statement ture or false :

(a) In Young’s double slit experiment performed with a source of white light, only black and white fringes

are observed.

(b) Two slits in Young’s double slit experiment are illuminated by two different sodium lamps emitting light

of same wavelength. No interferenc pattern will be obtained.

17. A photodiode is fabricated from a semiconductor with a band gap of 2.8eV . Can it emit the wavelength of

6000 nm ? justify.

18. A silicon wafer is doped with phosphorus of concentration 13 310 /atoms cm . If all the donor atoms are

active, what is its resistivity at 20o . The electron mobility is 21200 / .seccm volt .

19. What is a Van-de-Graff generator ? Write the principle of Van-de-Graaf generator.

20. A long solenoid with closely wound turns has n turns per unit of its length. A steady current I flows

through this solenoid. Use Ampere’s circuital law to obtain an expression for the magnetic field at a point on

its axis and close to its mid-point.

OR

Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a toroidal sole-

noid.

21. What is meant by the term ‘interference of light’ ? Write any two conditions necessary for obtaining well

defined and sustained interference pattern of light.

22. Explain with the help of lens maker’s formula. Why does a convex lens behave as -

(i) Converging when immersed in water ( 1.33µ = ) and

(ii) A diverging lens when immersed in 2CS solution ( 1.6µ = ).

23. Calculate the Q − value in the fission reaction of an atom of mass no. 240 fragmenting into 2 atoms of

120A = .

24. Show that Bohr’s second postulate “The electron revolves around the nucleus only in certain fixed orbits

without radiating energy” can be explained on the basis of de-Broglie hypothesis of wave nature of elec-

tron.

25. A good quality mirror reflects 75% of light incident on it. How can we determine if the remaining 25% of

photons have been unreflected or energy of photons has been reduced by 25% ?

26. The energy gap of silicon is 1.14eV . Find the maximum wavelength at which Si starts energy absorption?

27. What are T.V. signals ? How can they be received.

28. State the working principle of a potentiometer. Explain, with the help of a circuit diagram, how the emf of

two primary cells are compared by using a potentiometer.

Page 5 of 14

In a potentiometer arrangement, a cell of 1.20volt gives a balance point at 30cm length of the wire. This

cell is now replaced by another cell of unknown emf. If the ratio of emfs of the two cells is 1.5 , calculate the

difference in the balancing length of the potentiometer wire in the two cases.

OR

Deduce the condition for balance in a Wheatstone’s bridge. Using the principle of Wheatstone bridge,

describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit dia-

gram and write the formula used. Write any two important precautions you would observe while performing

the experiment.

29. State the uses of electromagnetic radiation.

OR

What do you understand by electromagnetic wave ? Give its properties.

30. Name common optical defects of eye. How are they removed ?

OR

What are coherent sources of light ? Why are coherent sources required to obtain sustained interference

pattern ?

SOLU

TIONS

Page 7 of 14

SOLUTION1. This is because there are no electric lines of

force inside the body of a conductor.

2. Zero, because angle θ between area vector Ar

and magnetic field Br

is zero and hence torque

sin 0nIABτ θ= =

3. One henry is the self inductance of a coil in which

a current change at the rate of one ampere/sec.

induces an emf of one volt in the coil.

4.0 0

1v

µ ε=

⋅

5. Here, 45 , ?oC µ= =

1 1 12

1sin sin 45

2

oCµ = = = =

6. Frequency of incident light.

7. To radiate power fed from transmitter into free

space.

8. Length of dipole antenna

8

8

3 100.5

2 2 2 3 10

cHz m

f

λ ×= = = =

× ×

9. Electric dipole moment ( )p is the product of

either charge ( )q± and the distance (2 )a

between the charges, i.e., (2 )p q a=It is a vector quantity, directed from q− to q+ .

10. As the lines of force are perpendicular to

equipotential surfaces. Therefore, they are

shown in figure by dotted lines. Theirdirection is

from higher potential to lower potential.

Field intensity is larger on the left, where density

of equipotential surfaces is larger.

V 1 > V 2

V 2

11. Let the supply voltage be V . When three

resistors are connected in series, current flowing

through each of them is 1 ( / 3 )I V R= and

power of each

2 22

1 29 9

V VI R R

R R= = ⋅ =

∴ Total power of three resistors taken together

2 2

39 2

V VX

R R= × = =

When the same resistors are connected in

parallel, current flowing through each

2 ( / )I V R=

Hence, power of each

2 22

2 2

V VI R R

R R= = ⋅ =

∴ Total power of three resistors in parallel

arrangement

2

3 9V

XR

=

12.(a) Here 12I A= and 22.0 2 10R cm cm−= = ×∴ Magnetic field due to semicircular arc at the

centre point O will be

7

0

2

1 4 10 12

2 2 2 2 2 10

IB

R

µ π −

−

× × = = × × × 41.88 10 T−= ×

As per right hand rule, the magnetic field is

directed perpendicular to plane of paper pointing

inward,

O

(b) In this case magnitude of magnetic field will

remain as 41.88 10 T−× but it is directed

perpendicular to the plane of paper pointing

outward.

O

13. Magnetic field Br

should be such that force

acting on current carrying conductor due to it

acts vertically upward and just balances the

weight of the conductor so that conductor

remains freely suspended in mid air.

BLI mg Lgλ∴ = = [ m Lλ= =Q mass of wire]

gB

I

λ⇒ = Keeping in mind Fleming’s left hand

rule, we can say that Br

must act horizontally in

a direction perpendicular to the wire carrying

current.

Page 8 of 14

14. An ordinary moving coil ammeter measures

average value of current. The average value of

alternating current over a full cycle, even when

its frequency is low is zero.Hence the ordinary

moving coil ammeter cannot measure such an

a.c.

15. As two sets of closely wound sires carry currents

in opposite senses, therefore, their induced

effects cancel. The self inductance reduces. In

a special case we may have

. 1 2 2 2 0equivL L L M L L L= + − = + − =

16. The sources of light which emit light waves of

same wavelength, same frequency and in same

phase or having a constant phase difference are

called coherent sources.

(i) When these sources are too close, distance ( d )

between them tends to zero. Fringe width

( / )D dβ λ= tends to infinity. Hence no

interference pattern will be observed.

(ii) When distance ( d ) between between them

becomes to large, fringe width β becomes too

small and may no longer be in the visible region.

Hence no interference pattern will be observed.

OR

(a) False. With a source of white light central

maximum is white. The fringes are coloured i.e.

bright bands are coloured and dark bands are

dark. The centre of pattern is white.

(b) True. Two different sodium lamps cannot be

coherent even when they emit light of same

wavelength. Therefore, no interference pattern

will be obtained.

17.

34 8

6

6.6 10 3 100.2

6 10g

hcE eV

λ

−

−

× × ×= = =

×

As 2.8gE eV< hence it cannot emit the

wavelength 6000 nm .

18. As qnσ µ= , where, µ = electron mobility,

q = charge, n = concentration of donor atoms

19 131200 1.6 10 10σ −= × × ×

419.2 10 / cm−= × Ω

19. A Van-de-Graaff generator is a device used for

building up high potential differences of the order

of a few million volts which are used to accelerate

charged particles like electrons, protons, ions

etc in various experiments of nuclear physics.

This generator is based on

(i) The action of sharp points, i.e., the phenomenon

of corona discharge.

(ii) The property that charge given to a hollow

conductor is transferred to outer surface and is

distributed uniformly over it.

20. Consider a long solenoid having n turns per unit

length as shown in figure. The upper view of

dots in the figure is like a uniform current sheet

coming out of the plane of the paper. From the

right hand rule, the field due to this is to the left

at point Q (above) and to the right at point

P (below) . The lower row of crosses in the figure

is like a uniform current sheet going into the

plane of the paper. The field at any point above

it ( P as well as Q ) is to the right. The two fields

reinforce each other at P and exactly cancel at

Q .

Thus, a uniform magnetic field Br

is present

along the axis of solenoid at any point inside the

solenoid and is zero at any point outside the

solenoid. Consider a rectangular Amperian loop

abcd . Along cd the magnetic field is zero as

explained just above. Moreover along bc and

da the field Br

is at right angle to bc or da .

dQ

c

a b

PB

b c d a

dcbaB dl B dl B dl B dl B dl⋅ = ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫

r r r r rr r r r r

Ñ Ñ Ñ Ñ Ñ

b b b

a a aB dl B dl B dl B l⋅ = ⋅ = = ⋅ ∆∫ ∫ ∫

rr

Ñ Ñ Ñ

[where ab l= ∆ (say)]

According to Ampere’s circuital law

0B dl µ⋅ =∫

rr

Ñ (current enclosed in length l∆ )

0 ( )n l Iµ= ∆ [Q Number turns n l= ∆ ]

Hence, we have

0B l n l Iµ∆ = ∆ 0B n Iµ⇒ =

The direction of the field is given by the right

hand rule.If the solenoid has a total length l and

total number of turns N , then 0( / )B NI lµ=

OR

Page 9 of 14

Consider a toroidal solenoid consisting of an

anchor ring of mean radius R , over which a

large number of turns (say N ) of an insulated

metallic wire is wound. When a current I is

passed through the toroid, the magnetic field

produced will be same at all points on the central

axis of the ring (shown by dotted curve in the

figure) and directed along tangent to the ring.

Hence,

(2 )B dl B dl B dl B Rπ⋅ = ⋅ = =∫ ∫ ∫rr

Ñ Ñ Ñ

RO

P

I

and according to Ampere’s circuital law

0B dl µ⋅ =∫

rr

Ñ (total current enclosed)

0( )NIµ=

∴ 02B R NIπ µ⋅ = 00

2

NIB nI

R

µµ

π⇒ = =

where ( / 2 )n N Rπ= = Number of turns per

unit length of toroid.

21. Interference of light is the phenomenon of

redistribution of light energy in a medium on

account of superposition of light waves from two

coherent sources. Following are some of the

important conditions for obtaining sustained

interference of light :

(i) The two sources of light must be coherent i.e.,

they should emit continuous light waves of same

wavelength or frequency, which have either the

same phase or a constant phase difference.

(ii) The two sources should be strong with least

background.

(iii) The amplitudes of waves from two sources

should preferably be equal.

(iv) The two sources should preferably be

monochromatic.

22. As 2

1 1 2

1 1 11

f R R

µµ

= − −

for lens material 2 1.5µ =

(i) If the convex lens is immersed in water

( 1.33µ = ) its focal length will be positive hence

it behaves as converging lens.

(ii) If the convex lens is immersed 2CS solution

( 1.6µ = ) its focal length will be negative hence

it behaves as diverging lens.

23. For 240, 7.6bnA E MeV= ;

For 120, 8.6bnA E MeV= ;

∴ Gain in binding energy for nucleon is about

0.9 MeV . Hence total gain in binding energy is

240 0.9×216 MeV=

24. When an electron of mass m is confined to

move on a line of length l with velocity v , the

de-Broglie wavelength λ associated with

electron is

h h

mv pλ = = or

2 / 2

h h nhp

l n lλ= = =

When electron revolves in a circular orbit of

radius r ; then 2 2l rπ=

2

nhp

rπ∴ = or

2

nhp r

π× =

i.e., angular momentum ( )p r× of electron is

integral multiple of / 2h π , This is Bohr’s

quantisation condition of angular momentum.

25. If energy of incident light is reduced by 25%upon reflection, then the wavelength of reflected

light would increase and the colour of reflected

light would change. This mean that 25%photons have not been reflected at all.

26.hc

Eλ =

34 86.62 10 3 10

1.14eV

× × ×=

34 8

19

6.62 10 3 1010888

1.14 1.6 10Å

−

−

× × ×= =

× ×

27. T. V. signals are those signals which have

frequency range 80 to 200 MHz

Since these waves neither follow curvature of

earth, nor get reflected by ionosphere their

reception is possible by using tall antennas for

direct transmission/reception or by geostationary

satellite communication.

Page 10 of 14

28. Working principle of a potentiometer can be

explained by a circuit diagram shown in Fig.

G

+

+

A B

R

J

If r be the resistance of potentiometer wire and

L be its total length (i.e., AB L= ) and an

external resistance R is joined along with a

driver cell of emf 0ε , then current flowing through

the potentiometer wire.

( )0I

R r

ε=

+Fall in potential along the potentiometer wire.

( )0rIr

R r

ε= =

+and the fall in potential per unit length (i.e., the

potential gradient) of the potentiometer wire

( )0 r

kR r L

ε ⋅=

+ ⋅

If an auxiliary cell of emf ε is connected as

shown and galvanometer gives null point when

AJ l= , then klε = .

Comparison of emfs of two cells :

A labelled circuit diagram arrangment to

compare the emfs of two primary cells using

potentioneter is shown in Fig.

Plug is applied in key K and by adjusting the

rheostat Rh a constant current I is allowed to

flow through potentiometer wire due to the driver

cell Ba . Put the plug in key 1K so as to connect

cell 1E and slide the pencil jockey J till a

balance (null) point is obtained. Let the length of

potentiometer wire AJ in this position be 1l ,

then

1 1E kl= where k is the potential gradient along

the wire.

D riv e r c e ll ke yRheostat

Potentiom eter w ire

G a lv an om e te r

P en cil J oc k ey

R hBa A

1E

2E2K

1K

++

+

+

A

O L

B

K

G

J

Now plug is removed from key 1K and, in turn,

plug is applied in key 2K so as to bring 2E in

the electric circuit. Again slide the pencil jockey

and obtain the position of jockey for null

deflection in galvanometer. Let the length now

be 2l , then

2 2E kl=

1 1

2 2

E l

E l∴ =

We know that in potentiometer arrangement

1 1

2 2

l

l

εε

=

In present problem 1

2

1.5εε

= and 1 30l cm=

2

3020

1.5l cm= =

OR

C

QG

D

1I

I I

1K

S

1I

( )1I I- ( )1

I I-

R

P

B

A2K

Circuit arrangement for a Wheatstone bridge is

being shown in the adjoining Fig. In balance

condition of bridge no current flows through

galvanometer and it gives no deflection. Currents

in various branches of network as per Kirchhoff’s

first law are, thus, represented in figure.

Applying Kirchhoff’s second law to mesh

ABDA , we have

( )1 1. 0P I R I I= + − = .......(i)

or ( )1 1PI R I I= −

Page 11 of 14

Again for mesh BCDB , we have

( )1 1 0QI S I I= + − =

or ( )1 1QI S I I= − .............(ii)

Dividing (i) by (ii), we get

P R

Q S=

which is the balance condition of Wheatstone’s

bridge.

Specific resistance or resistivity of the material

of a wire can be calculated provided that its

resistance length and diameter are known. A

meter bridge is a paractical form of Wheatstone

bridge and the circuit diagram is shown in Fig.

R .B .R X

D

AB

C

λ cm (100 - ) cmλ

G alvanom eter

Resistance boxU nknow n resistance

C u s trips

Jockey B ridge w ire

C e ll E K key

G

A current is allowed toflow from the cell E and

a suitable resistance R is inserted from the

resistance box R. B. Now the pencil jockey is

gently slided over the bridge wire and its position

is adjusted at point B at a length l from end

A of bridge wire till galvanometer gives no

deflection i.e., a balance point is obtained.

According to Wheatstone bridge balance

condition.

R esistance of o f b ridge w ire

R esistance of leng th of bridge w ire

R A B

X B C=

( ) ( ).

100 100

l l

l l

ρρ

= =− −

where ρ is the resistance per unit length of the

bridge wire.

⇒ Unknown resistance ( )100R l

Xl

⋅ −=

Knowing the resistance X of given wire and

by measuring its length l by a metre rod and

diameter D by using a screw gauge, the

resistivity ρ can be calculated as

2

4

X D

l

πρ =

Precautions :

(i) Null point should be obtained in the middle

of bridge wire.

(ii) Cell circuit is to be completed only when

reading is to be taken.

29. 1. Radiowaves :

(i) The electromagnetic waves of frequency

range for 530kHz to 1710kHz form

amplitude modulated (AM) band It is used in

ground wave propagation.

(ii) The electromagnetic waves of frequency

range 1710kHz to 54MHz are used for short

wave bands. It is used in sky were propagation.

2. Microwaves :

(i) Microwaves are used in Radar dsystems for

air craft navigation.

(ii) A radar using microwave can help in

detecting the speed of tennis ball, cricket ball,

automobile while in motion.

3. Infrared waves :

(i) Infrared waves are used in physical therapy,

i.e., to treat muscular strain.

(ii) Infrared waves are used to provide electrical

energy to satellite by using solar cells.

4. Visible light :

The visible light emitted or reflected from objects

around us provides the information about the

world surrounding us.

5. Ultraviolet rays :

(i) Ultraviolet rays are used for checking the

mineral samples through the property of

ultraviolet rays causing flourescence.

(ii) Ultraviolet rays are used in the study of

molecular structure and arrangement of

electrons in the external shell through ultraviolet

absorption spectra.

6. X-raysX-rays are used in surgery for the detection of

fractures, foreign bodies like bullets, diseased

organs and stones in the human body.

7. γ - rays :

(i) γ -rays are used in the treatment of cancer

and tumours.

OR

Farady in his experlmental study of

electromagnetic induction concluded that a

magnetic field changing with time produces an

electric field in that region. Maxwell in 1865 from

his theoretical study concluded that there is a

Page 12 of 14

great symmetry in nature ”, i.e. an electric field

changing with time in a region produces

magnetic field there. It means a change in either

field (electric or magnetic) with time produces

the other field. This idea leads Maxwell to

conclude that the variation in electric and

magnetic field vectors perpendicular to each

other leads to the production of electromagnetic

disturbances in space. These disturbances have

the properties of wave and can travel in space

even without any material medium. These waves

are called electromagnetic waves.

According to Maxwell,

The electromagnetic wave are those waves in

which there are sinsusoidal varidation of electric

and magnetic field vectors at right angles to each

other as well as at right angles to the direction

of wave propagation.

These fields vary with time and space and have

the same frequency. In Fig., the electric field

vector Er

and magnetic field vector Br

are

vibrating along Y and Z directions and

propagation of electromagnetic wave is shown

in X -direction.

E

B

O

Y

Z

X

B

B

E

E

E n velo pe o f E le ctric In te ns ity Ve c to r

E n velo pe o f M ag n e ticIn d u c tio n Ve c to r

Properties :

1. The electromagnetic wave are produced by

accelerated or oscillating charge.

2. These waves do not require any material

medium for propagation.

3. These waves travel in free space with a

speed 8 13 10 ms−× (i.e., speed of light) given by

the relation0 01/c µ= ∈

4. In electromagnetic waves the sinusoidal

variation in both electric and magnetic field

vectors ( Er

and Br

) occurs, simultaneously. As

a result, they attain the maximum and minimum

values at the same place and at the same time.

The amplitudes of the electric and magnetic

fields in free space are related by ; 0 0/E B c= .

5. The directions of variation of electric and

magnetic field vectors are perpendicular to each

other as well as perpendicular to the direction of

propagation of waves. Therefore,

electromagnetic waves are transverse in nature

like light waves.

6. The velocity of electromagnetic waves

depends entirely on the electric and magnetic

properties of the medium in which these waves

travels and is independent of the amplitude of

the field vectors.

30. Some of the common optical defects of the eye

are

1. Myopia or short sightedness.

2 Hypermetropia or long sightedness.

3. Presbyopia

4. Astigmatism.

We discuss here the first two defects in some

detail.

(a) Myopia or short sightedness

Myopia or short sightedness is that defect of

human eye by virtue of which, the eye can see

clearly the objects lying near it, but the far off

objects cannot be seen distinctly i.e. objects lying

beyond a particular distance cannot be seen

clearly by the eye. In other words, for a myopic

eye, the far point shifts towards the eye. It is not

longer at infinity. It is at F .

In Fig (a) parallel rays from infinity are focussed

on the retina by the normal eye. In Fig (b) the

myopic eye focuses the parallel rays from infinity

at P in front of the retina. However, the defective

eye focusses rays from a point F is the far point

of the myopic eye. It cannot see clearly beyond

F .

To correct a myopic eye of this defect, the person

has to use spectacles with a concave lens of

suitable focal length. In fact, parallel rays of light

from infinity, after refraction through the cancave

lens, should appear to come from F , the far

point of the defective eye. This is shown in Fig

(c).

Let x be the distance of far point of the myopic

eye and f be the focal length of concave lens

to be used.

Now, for concave lens, ,u v x= ∞ = − .

As 1 1 1

f v u= − or

1 1 1

f x

−= −

∞

Hence focal length of concave lens put in front

of the eye should be equal to disance of far point

of the defective eye.

Page 13 of 14

R N orm al Eye (Sharp im age of ob ject a t in fin ity)

(a )

RM yopic Eye[B lurred im age o f ob jec t a t in fin ity and sh arp im age of o b je ct a t F(far po in t)]

(b )

x

P

R C orrec tedm yopic eye(Sha rp im a ge ofob jec t a t in fin ity)

(c)

x

F

(b) Hypermetropia or Long slghtedness :

Hypermetropia or long slghtedness is that defect

of human eye, by virtue of which the eye can

see clearly the far objects but nearby objects

cannot be seen clearly. In other words, for a

hypermetric eye the near point no longer at

25cm .

The hypermetropic eye focusses the rays from

N to 'N .

R N orm al E ye (Sha rp im a ge of o b je ct a t N (near po int)]

(a )

d

N

RH yperm etrop icEye (B lurred im age of ob je ct a t N(N ear po int)sharp im age of o b je ct a t N ')

(b )

x '

NN '

P '

RC orrec ted H yperm etrop icEye :(Sha rp im a geof o b je ct a t N )

(c)

NN '

Suppose 'x = the distance of near point 'N of

the defective eye,

d = the least distance of distinct vision (of

normal eye),

f = focal length of convex lens to be used.

For the correcting lens, , 'u d v x= − = − .

As 1 1 1

f v u= −

1 1 1 '

' '

d x

f x d x d

− +∴ = + =

−

or '

'

x df

x d=

−As 'x d> , f is + , the correcting lens must be

convex.

(c) Presbyopia (old sight)

With increasing age, the ciliary muscles holding

the eye lens weaken and the lens loses some of

its elasticity, therefore, power of accommodation

of the eye decreases with age. This defect is

called presbyopia. To remove this defect,

converaging spectacle lenses are employed (as

in the case of hypermetropia). The person needs

to wear these spectacles for reading or similar

close work.

(d) Astigmatism

To a normal eye, all the lines in Fig look equally

black. But an astimatic eye will find variation in

the intensity of different lines. This defect arises

when cornea has different curvature in different

directions.

Astigmatism can be corrected by using a

cylindrical lens of suitable radius of curvature,

and suitable axis.

OR

The sources of light, which emit continuous light

waves of the same wavelength, same frequency

and in same phase or having a constant phase

difference are called coherent sources.

The phase difference φ between light waves at

a given time and at a given position, is sum of

(i) Initial phase difference between two sources

A and B, and

Page 14 of 14

(ii) Phase difference on account of difference in

paths. The latter does not change with time (for

a given point P).

Hence, initial phase difference between two

sources must be constant.

Conditions for Obtaining two CoherentSources of Light

1. Coherent sources of light should be

obtained from a single source by some device.

In this case, any phase change in one is

simultaneously accompanied by the same phase

change in other. Due to it, the phase difference

between the two waves of light reaching at a

point from the two coherent sources remains

constant with time and a steady interference

pattern is obtained.

The coherent source can be obtained either by

(i) the source and its virtual image or (ii) the two

virtual images of the same source. (iii) two real

images of the same source.

2. The two sources should give monochromatic

light.

If the sources give white light, consisting of a

number of wavelengths (colours), then the light

of each wavelength gives its own set of

interference fringes. Since the fringe width is

different for different wavelengths, overlapping of

the fringes occurs, resulting in a few coloured

fringes only near the central white fringe. These

fringes are hazy, whereas those with

monochromatic light are quite sharp.

3. The path difference between light waves from

two sources should be small.

If the path difference between two interfering

waves is large, there is intermixing of the two at

every point, resulting in uniform illumination.