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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.