Physics for You March 2015

physics for you | march 15 7

Training for research starts in schools

When should students start thinking, is a difficult question to answer. It is like asking when should children start learning music. children start learning music when they are in primary school stage, at home. By the time they are grown up children, it is amazing to see the advance they have made, when they have not yet reached 14 or 15 years of age.

research is only an attitude of mind which drives a person to think deeper and deeper. But to avoid the mistakes of repeating what others have done, a lot reading is also advised. Before starting something new, one should know what others have done earlier and what the great scientists are thinking about the problem. In modern books, written for graduate levels, one finds first a short history of the work and the thinking of the great scientists in about half a page. We are happy that in the NcErT books, particularly for high schools, the system of historic introduction and the thinking of the great scientists are also given. One may not gain extra marks for learning the history of science, but this gives extra inputs for the development of mind. reading the biography of scientists is as interesting as reading a novel.

To keep the attention of the students, after every heavy derivation, one should give a short digression on the scientists. as the editorial has a wider readership, every teacher would also like to know about the methods of research and also teach their students, how to succeed in research. For success in ones career, one must learn simultaneously how to concentrate on microproblems as well as the art of increasing a wide vision.

Anil Ahlawat Editor

Vol. XXIII No. 3 March 2015

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contents

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Physics Musing (Problem Set-20) 8

AIPMT Special 12 Practice Paper 2015

Thought Provoking Problems 23

Core Concept 26

You Asked We Answered 30

JEE Foundation Series 31

Brain Map 50

CBSE Board 58 Practice Paper 2015

JEE Main 67 Practice Paper 2015

JEE Advanced 75 Practice Paper 2015

Physics Musing (Solution Set-19) 91

25 Must Know Facts 92

Crossword 93

8 physics for you | march 15

single option correct type

1. A point object O is placed at a distance of 20 cm in front of a equiconvex lens (amg = 1.5) of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in the figure. Image will be formed at a distance h from lens. The value of h is(a) 5 cm (b) 10 cm(c) 20 cm (d) 40 cm

2. For a certain reflecting surface, the unit vector along the incident ray is i^ and that along the

outward normal of the surface is

i j

^^

23

2.

The unit vector along the reflected ray will be

(a) i j^

^2

32

+ (b) i j^

^2

32

(c) +32

i j^ ^ (d) 32

12

i j^ ^

3. A ray of light moving along the vector ( )^ ^i j 2undergoes refraction at an interface of two media, which is x-z plane. The refractive index

O

for y > 0 is 2 while for y < 0 it is 52

. The unit vector along the refracted ray is

(a) 3 5

34

i j^ ^ (b)

( )^ ^4 55

i j

(c) 3 4

5i j^ ^

(d) 4 3

5i j^ ^

4. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If s is the surface tension of soap solution, the energy spent in doing so is(a) 3psr2 (b) 6psr2(c) 12psr2 (d) 24psr2

5. Find the minimum vertical force required to pull a thin wire ring up as shown in figure, if it is initially resting on a horizontal water surface. The circumference of the ring is 20 cm and its weight is 0.1 N.

The surface tension of water is 75 dyne cm1. (a) 0.125 N (b) 0.225 N(c) 0.115 N (d) 0.130 N

6. A point mass m is welded to a ring of mass m and radius R as shown in the figure. Assuming that the ring does not slip and initially the

Physics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You.The readers who have solved five or more problems may send their solutions. The names of those who send atleast five correct solutions will be published in the next issue.We hope that our readers will enrich their problem solving skills through "Physics Musing" and stand in better stead while facing the competitive exams.

MUSINGMUSINGPHYSICSPHYSICS

By : Akhil Tewari


system is released from rest. What would be the speed of the point mass as seen from the ground after the ring has turned through an angle of 90 ?

m

Rm

(a) gR (b) gR2

(c) 2gR (d) gR3

7. Two infinitely long conducting parallel rails are connected through a capacitor of capacitance C as shown in the figure. A perfect conductor of length l is moved with constant speed v0. Which of the following graph truly depicts the variation of current through the conductor with time ?

B

l v0

(a) (b)

(c) (d)

8. An organ pipe of cross-sectional area 100 cm2 resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is

(Take velocity of wave = 320 m s1)(a) 800 cm3 (b) 1200 cm3(c) 1600 cm3 (d) 2000 cm3

9. Pulley and strings as shown in figure are massless. The force acting on the block of mass m is(a) 2F(b) F

(c) F2

(d) 4F

10. A particle of mass m moves along a circle of radius R. The modulus of average value of force acting on particle over the distance equal to a quarter of circle, if the particle moves uniformly with velocity v is

(a) 22mv

rp (b) 2 2

2

2mvrp

(c) 2 22mv

rp (d) mv

r

2

pnn

solution of february 2015 crossword

Winner (February 2015) riya Kataria

Solution Senders (January 2015)

Divya Acharya

Shubhneet Bhatia

Akash Kashyap


1. A 15 g ball is shot from a spring gun whose spring has a force constant 600 N m1. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression (Take g = 10 m s2) (a) 6 m (b) 8 m(c) 10 m (d) 12 m

2. Two weights w1 and w2 are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration g, the tension in the string will be

(a) 4 1 2

1 2

w ww w+

(b) 2 1 2

1 2

w ww w+

(c) w ww w

1 2

1 2

+

(d) w w

w w1 2

1 22( )+

3. A gas bubble from an explosion under water oscillates with a time period T, depends upon static pressure p, density of water r and the total energy of explosion E. Find the expression for the time period T. k is a dimensionless constant.(a) T = kp5/6r1/2E1/3 (b) T = kp4/7r1/2E1/3

(c) T = kp5/6r1/2E1/2 (d) T = kp4/7r1/3E1/2

4. A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after time t is

(a) E q t

m

2 2 2

2(b)

2 2 2E tqm (c)

Eqmt2

(d) Eq m

t

2

225. The intensity of magnetic field at a point X on

the axis of a small magnet is equal to the field intensity at another point Y on its equatorial

axis. The ratio of distances of X and Y from the centre of the magnet will be(a) (2)3 (b) (2)1/3 (c) 23 (d) 21/3

6. 5 mole of an ideal gas with g = 7/5 initially at STP are compressed adiabatically so that its temperature becomes 400C. The increase in the internal energy of gas in kJ is(a) 21.55 (b) 41.55 (c) 65.55 (d) 50.55

7. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 m s1 relative to the ground. Time taken by the man to complete one revolution with respect to disc is

(a) p s (b) 32p s (c) 2p s (d)

p2

s

8. A vessel contains oil (density = 0.8 g cm3 over mercury (density = 13.6 g cm3). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in g cm3 is(a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8

9. Maxwells velocity distribution curve is given for the same quantity two different temperatures. For the given curves(a) T1 > T2 (b) T1 < T2(c) T1 T2 (d) T1 = T2

N

v

T1T2

10. Two capacitors of 25 mF and 100 mF are connected in series to a source of 120 V. Keeping

PRACTICE PAPER 2015Exam on3rd May


their charges unchanged, they are separated and connected in parallel to each other. Find out energy loss in the process.(a) 5.2 J (b) 52 J (c) 50.2 J (d) 0.052 J

11. The steady state current in a 2 W resistor when the internal resistance of the battery is negligible and the capacitance of the condenser is 0.1 mF is

0.1 F

6 V

3

2 A B

4

2.8

(a) 0.6 A (b) 0.9 A (c) 1.5 A (d) 0.3 A12. In an experiment, a magnet with its magnetic

moment along the axis of a circular coil and directed towards the coil, is withdrawn away from the coil and parallel to itself. The current in the coil, as seen by the withdrawing magnet, is(a) zero (b) clockwise(c) anticlockwise (d) first (a) then (b)

13. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order to have an inverted image of the object coincident with it?(a) 12 cm (b) 30 cm (c) 50 cm (d) 60 cm

14. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 107 metre. The interference fringes are observed on a screen placed one metre from the slits. The distance between the third dark fringe and fifth bright fringe on the same side of centre is equal to(a) 0.65 mm (b) 1.63 mm(c) 3.25 mm (d) 4.8 mm

15. An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the heat and light energy produced by the bulb in 20 min?(a) 10 kJ (b) 15 kJ (c) 20 kJ (d) 30 kJ

16. A body hanging from a spring stretches it by 1 cm at the earths surface. How much will the same body stretch the spring at a place 16400 km above the earths surface?

(Radius of the earth = 6400 km)(a) 1.28 cm (b) 0.64 cm(c) 3.6 cm (d) 0.12 cm

17. A resistor R and 2 mF capacitor in series are connected through a 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Find the value of R to make the bulb light up 5 s after the switch has been closed.

(Take log10 2.5 = 0.4)(a) 1.7 105 W (b) 2.7 106 W(c) 3.3 107 W (d) 1.3 104 W

18. A coil of resistance 400 W is placed in a magnetic field. If the magnetic flux f (Wb) linked with the coil varies with time t (s) as f = 50t2 + 4. The current in the coil at t = 2 s is(a) 0.5 A (b) 0.1 A (c) 2 A (d) 1 A

19. An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with permittivity er = 4, then(a) the wavelength and frequency both remain

unchanged.(b) the wavelength is doubled and the frequency

remains unchanged.(c) the wavelength is doubled and the frequency

becomes half(d) the wavelength is halved and the frequency

remains unchanged.20. A polyster fibre rope of diameter 3 cm has a

breaking strength of 150 kN. If it is required to have 600 kN breaking strength. What should be the diameter of similar rope?(a) 12 cm (b) 6 cm (c) 3 cm (d) 1.5 cm

21. A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel forces of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is

(a) mal

ma F+ (b) 2( )F ma l

ma+

(c) l l Fma

+

(d) ( )F ma lma+2


22. The amount of work done in stretching a spring from a stretched length of 10 cm to a stretched length of 20 cm is(a) equal to the work done in stretching it from

20 cm to 30 cm(b) less than the work done in stretching it

from 20 cm to 30 cm(c) more than the work done in stretching it

from 20 cm to 30 cm(d) equal to the work done in stretching it from

0 to 10 cm.23. The rms value of the electric field of the light

coming from the sun is 720 N C1. The average total energy density of the electromagnetic wave is(a) 3.3 103 J m3 (b) 4.58 106 J m3

(c) 6.37 109 J m3 (d) 81.35 1012 J m3

24. In Youngs double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If Im be the maximum intensity, the resultant intensity when they interfere at phase difference f is given by

(a) Im3

1 22

2+

cos f (b) Im5

1 42

2+

cos f

(c) Im9

1 82

2+

cos f (d) Im9

82

2+

cos f

25. A compound microscope has an eye piece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective, so that the final image is formed at the least distance of distinct vision 20 cm.(a) 12 (b) 11 (c) 10 (d) 13

26. In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance S connected to the galvanometer is

(a) 19G (b) G19

(c) 20G (d) G20

27. The power factor of the circuit as shown in figure is

R1 = 40 220 V20 Hz

XC = 40

XL = 100 R2 = 40

(a) 0.2 (b) 0.4 (c) 0.8 (d) 0.628. At ordinary temperature, the molecules of an

ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this at higher temperatures(CV = molar heat capacity at constant volume)

(a) C RV =32

for a monoatomic gas

(b) C RV >32

for a monoatomic gas

(c) C RV >52

for a diatomic gas

(d) C RV =52

for a diatomic gas

29. A body is projected vertically upwards with a velocity of 10 m s1. It reaches the maximum height h in time t. In time t/2, the height covered is

(a) h2

(b) 25

h (c) 34

h (d) 58

h

30. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first 2 s, it rotates through an angle q1, in the next 2 s, it rotates through an

angle q2. The ratio of qq

2

1 is

(a) 1 (b) 2 (c) 3 (d) 531. A uniform chain of mass m and length l is

lying on a table with l4

of its length hanging

freely from the edge of the table. The amount of work done in dragging the chain on the table completely is

(a) mgl4

(b) mgl8

(c) mgl32

(d) mgl16


32. The diode used in the circuit shown in the figure has a constant voltage drop at 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with diode, for obtaining maximum current?

1.5 V

R 0.5 V

(a) 6.76 W (b) 20 W (c) 5 W (d) 5.6 W33. The half-life of a radioactive isotope X is

50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be(a) 100 years (b) 150 years(c) 200 years (d) 250 years

34. On shining light of wavelength 6.2 106 m on a metal surface photo-electrons are emitted. The work function of the metal is 0.1 eV. Find the kinetic energy of a photo-electron (in eV)(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4

35. A mass of 0.2 kg is attached to the lower end of a massless spring of force constant 200 N m1, the upper end of which is fixed to a rigid support. Which of the following statement is not true?(a) The frequency of oscillation will be nearly

5 Hz.(b) In equilibrium, the spring will be stretched

by 2 cm.(c) If the mass is raised till the spring is

unstretched state and then released, it will go down by 2 cm before moving upward

(d) If the system is taken to a planet, the frequency of oscillation will be the same as on the earth.

36. The equation of a wave is represented by

Y t x=

10 10010

5 sin ,m then the velocity of

wave will be (a) 100 m s1 (b) 4 m s1

(c) 1000 m s1 (d) zero

37. Force on a 1 kg mass on earth of radius R is 10 N. Then the force on a satellite revolving around the earth in the mean orbital radius 3R/2 will be (mass of satellite is 100 kg)(a) 4.44 102 N (b) 3.33 102 N(c) 500 N (d) 6.66 102 N

38. The far point of a near sighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 m high. How high is the image formed by the contacts?(a) 1.0 m (b) 1.5 m(c) 0.75 m (d) 0.50 m

39. You drive a car at a speed of 70 km h1 in a straight road for 8.4 km and then the car runs out of petrol. You walk for 30 min to reach a petrol pump at a distance of 2 km. The average velocity from the beginning of your drive till you reach the petrol pump is(a) 16.8 km h1 (b) 35 km h1

(c) 64 km h1 (d) 18.6 km h1

40. A fork A has frequency 2% more than the standard fork and B has a frequency 3% less than the frequency of same standard fork. The forks A and B when sounded together produced 6 beats s1. The frequency of fork A is(a) 116.4 Hz (b) 120 Hz(c) 122.4 Hz (d) 238.8 Hz

41. When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is 0.01 103. The Poissons ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m2, its Youngs modulus is(a) 1.6 108 N m2 (b) 2.5 1010 N m2

(c) 1.25 1011 N m2 (d) 16 109 N m2

42. In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is 2 107 m and stopping potential is 2.5 V. The threshold frequency of the metal is approximately

(Charge of electron e = 1.6 1019 C, Plancks constant h = 6.6 1034 J s)(a) 12 1015 Hz (b) 9 1015 Hz(c) 9 1014 Hz (d) 12 1013 Hz


43. Two cells of emf e1 and e2 (e1 > e2) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio is e1 : e2 is 1 2A

BC

(a) 3 : 1 (b) 1 : 3 (c) 2 : 3 (d) 3 : 244. A body is thrown horizontally from the top of a

tower of 5 m height. It touches the ground at a distance of 10 m from the foot of the tower. The initial velocity of the body is (Take g = 10 m s2)(a) 2.5 m s1 (b) 5 m s1

(c) 10 m s1 (d) 20 m s1

45. Two bodies of 6 kg and 4 kg masses have their velocity 5 2 10i j k + and 10 2 5i j k + respectively. Then the velocity of their centre of mass is(a) 5 2 8i j k + (b) 7 2 8i j k +

(c) 7 2 8i j k + (d) 5 2 8i j k +

solutions

1. (c) : Here, R u

gmu

mgmax= =

221

22

But 12

12

2 2mu kx=

\ = =R kxmg

kxmgmax

12

22 2

=

=600 0 050 015 10

102( . )

.m

2. (a) : For solving the problem, we assume that observer is situated in the frame of pulley (non-inertial reference frame).

m1g = w1m2g = w2

T

w2 m2a0

a

T

w1m1a0

am2 m1

From force diagram, T m2a0 w2 = m2aor T m2g w2 = m2a ( a0 = g)or T 2w2 = m2a ... (i)From force diagram, m1a0 + w1 T = m1aor m1g + w1 T = m1a ( a0 = g)or 2w1 T = m1a ... (ii) From eqs. (i) and (ii), we get

Tw w

w w=

+4 1 2

1 2

3. (a) : Time period, T parbEc

or T = kparbEc

k is a dimensionless constant.According to homogeneity of dimensions,LHS = RHS\ [T] = [ML1T2]a[ML3]b[ML2T2]c

[T] = [Ma+b+c][La3b+2c][T2a2c]Comparing the powers, we obtain a + b + c = 0 a 3b + 2c = 0 2a 2c = 1On solving, we get

a b c= = =56

12

13

, ,

4. (a) : Here, u a Fm

qEm

= = =0,

v u at qEm

t= + = +0

KE = = =12 2 2

22 2 2

2

2 2 2mv mq E t

m

E q tm

5. (d) : If d1 is the distance of point X on axial line and d2 is distance of point Y on equatorial line, then

B M

dB M

d1

0

13 2

0

234

24

= =mp

mp

,

As B1 = B2

\ =mp

mp

0

13

0

234

24

Md

Md

d d13

232= ;

dd

1

2

1 32= /


6. (b) : Here, n = 5, g = 7/5, T1 = 0C, T2 = 400C

dUnRdT

=g 1

=

=5 8 31 400 07 5 1

41550. ( )( / )

J = 41.55 kJ

7. (c) : Using angular momentum conservation,Li = 0, Lf = mvR Iw, so, mvR = I w

w= = =mvRI

50 1 2200

12

For one complete revolution,( )v R t R+ =w p2

1 12

2 2 2+

= t p

t = 2p s.

8. (c) : Weight = Buoyant force

V g V g V gmr r r= +2 2Hg oil

rr r

m =+

= +

= =

Hg oil

213 6 0 8

214 4

27 2

. .

. .

Oil

Mercury

9. (b) : Higher is the temperature, greater is the most probable velocity.

10. (d) : 1 1 1 125

1100

5100

1201 2C C Cs

= + = + = =

Cs = 20 mF = 20 106 F

U C Vs12 6 21

212

20 10 120= = ( )( ) = 144 103 JCharge on each capacitor,q1 = q2 = Cs V = 20 120 =2400 mCIn parallel, Cp = C1 + C2= 25 + 100 = 125 mF = 125 106 F

\ = = +

U Q

Cp2

2 6 2

622400 2400 10

2 125 10[( ) ]

= 92.16 103 JLoss of energy = U1 U2 = (144 92.16) 103 J = 51.84 103 J = 0.052 J

11. (b) : Capacitor will work as open key. Therefore no current flows through resistance

4 W. The total resistance of circuit

= + +

= + =2 8 2 32 3

2 8 1 2 4. . . W

\ Main current, I = =64

32

A

Potential difference across A and B

= =3

21 2 1 8. . V

\ Current through 2 W = =1 82

0 9. . A

12. (b) : As magnet is withdrawn from the coil, field into the coil decreases. To increase this field, current induced in the coil must be clockwise as seen by the withdrawing magnet.

13. (c) : For the lens, 1 1 1 120

130

160v f u

= + = =

v = 60 cm. Therefore, to have an inverted image of the object, coincident with it, image should tend to form at centre of curvature of convex mirror. Therefore, distance of convex mirror from the lens = 60 10 = 50 cm.

14. (b) : For dark fringes,

Y n D

dn= ( )2 1

2l

Y D

d352

= l

For bright fringes,

Y n D

dn= l

Y D

d55= l

Y = Y5 Y3

Y Dd

= =

52

5 6 5 10 12 10

7

3l .

= 1.625 103 m = 1.63 mm15. (d) : Here, power of the bulb, P = 100 W

Supply voltage, e = 230 VLet R be the resistance of the bulb.

As PR

= e2

= = =R Pe2 2230

100529( ) W

Changed supply voltage, e = 115 VHeat and light energy produced by the bulb in 20 min.

=

= e 2 2115 20 60529

tR

= 30,000 J = 30 kJ


16. (b) : In equilibrium, weight of the suspended body = stretching force.\ At the earths surface, mg = k x At a height h, mg = k x

gg

xx

RR h

e

e

= =+

=+

2

2

2

26400

6400 1600( )( )

( )

=

=6400

80001625

2

x x = = 1625

1625

1 cm = 0.64 cm

17. (b) : As VC = e(1 et/RC)

or 1 et/RC = = =VCe

120200

35

or et/RC = 2.5 or log et/RC = loge 2.5

or tRC

= =2 3026 2 5 0 9210. log . .

or R tC

= =

= 0 92

50 92 2 10

2 7 106

6. .

. W

18. (a) : e f= = + = ddt

ddt

t t( )50 4 1002

When t = 2 s, |e| = 200 VInduced current at t = 2 s,

IR

= = =| | .e200400

0 5 A

19. (d) : Frequency remains unchanged with change of medium.

m (refractive index) = cv r r= =

1

10 0/

/

e m

eme m

Since, mr is very close to 1, m e= = =r 4 2

Thus, lmedium =

lm

l=2

20. (b) : Y FA

LL

=

The breaking strength F A.

\ = = =FF

AA

D

D

D

D2

1

2

1

22

12

22

12

4

4

p

p

( / )

( / )

or cmD DFF2 1

2

1

1 2 1 23 600

1506=

=

=/ /

21. (b) : Let L be the length of the rod of mass m, with centre of mass at C. Suppose F1 is the magnitude of other force. Let F1 > F.\ F1 F = ma or F1 = F + ma

L CF1 l

F

As the rod moves translationally and there is no rotation, therefore, net torque about C must be zero.

\

=

= +

F L F L l F ma L l2 2 21

( )

F L F L Fl ma L mal

2 2 2

=

+

ma L l F ma

2= +( )

\ L F ma l

ma= +2( )

22. (b) : W K x x= ( )12 22 12

= =1

220 10 1502 2K( ) kJ

which is less than work done in stretching it from 20 cm to 30 cm.

viz. kJ12

30 20 2502 2K( ) =

23. (b) : Total average energy density of electro-magnetic wave is

< > = +u E B12

120

2

0

2emrms rms

= +

=

12

120

2

0

2

2e

mE

E

cB

Ecrms

rmsrms

rms

= +12

120

2

0

20 0e me mE Erms rms

= + =12

120

20

20

2e e eE E Erms rms rms

= (8.85 1012) (720)2 = 4.58 106 J m3

24. (c) : Here, A2 = 2A1 Intensity (Amplitude)2

\ =

=

=

II

AA

AA

2

1

2

1

21

1

22

4


I2 = 4I1

Maximum intensity, I I Im = +( )1 2 2

= +( ) = ( ) = =I I I I I Im1 1 2 1 2 1 14 3 9 9or ...(i)

Resultant intensity, I I I I I= + +1 2 1 22 cosf

= + +I I I I1 1 1 14 2 4( ) cosf = 5I1 + 4I1cosf = I1 + 4I1 + 4I1cosf = I1 + 4I1(1 + cosf)

= +I I1 1

282

cos f

1 2

22+ =

cos cosf f

= +

I121 8

2cos f

Putting the value of I1 from eq. (i), we get

I

Im= +

9

1 82

2cos f

25. (a) : Here, uo = 5 cm, fo = 4 cmfe = 10 cm, D = 20 cmAccording to lens formula,

1 14

15

14

15

120vo

= +

= =

vo = 20 cm

Magnification, Mvu

Df

o

o e= +

1

= +

=20

51 20

1012

26. (b) : As shunt is a small resistance S in parallel with a galvanometer (of resistance G) as shown in figure.(I IG)S = IGG

SI G

I IG

G=

( )

S( )I IG

G

IG

I I

Here, I IG =5

100

\ =

=SIG

I I

G5

1005

10019

27. (c) : Resistance of the circuit,R = R1 + R2 = 40 W + 40 W = 80 WImpedance of the circuit,

Z R X XL C= + = + 2 2 2 280 100 40( ) ( ) ( )

= + =( ) ( )80 60 1002 2 W

Power factor, cos .f = = =RZ

80100

0 8

28. (c) : Monoatomic gas C RV =32

This value is same for high temperature also.In case of diatomic gas

C RV =52

(at low temperature)

Also, C RV >52

(at high temperature due to

vibrational kinetic energy)29. (c) : As v2 v0

2 = 2gh, 0 (10)2 = 2(10) hor h = 5 mAlso, v = v0 + at, 0 = 10 + (10) tor t = 1 sHeight covered in time t/2, i.e., (1/2 s),

= + =

h v t g t02

212

10 12

12

10 12

( )

= 3.75 m = (3/4) h

30. (c) : As q w a a a1 02 21

20 1

22 2= + = + =t t ( )

( ) ( )q q w a a a1 2 02 21

20 1

24 8+ = + = + =t t

Thus, q2 = 6a or qq

2

13=

31. (c) : Mass of l4

length of the chain = m4

The weight of this part of the chain acts as its

CG which is at a distance l8

from the edge of the table.

Work done = m g l mgl4 8 32

=

32. (c) : As RVPD

D

D= = =

2 20 50 1

2 5( . )

..

VW

W

IVRD

D

D= = 0 2. A

Total resistance required in the circuit,


R VID

eq = = =1 50 2

7 5..

. W

Resistance of the series resistor, R = Req RD= 7.5 2.5 = 5 W

33. (c) : As NN

N N NXY

X Y X= + =1

1516,

Thus, N

N NX

X Y+= 1

16

or N N N N NX X Y X Y= + = +1

161

24( ) ( )

Age of the rock = number of half-lives of isotope X passed = 4 = 4 50 years = 200 years

34. (a) : Here, l = 6.2 106 m, f0 = 0.1 eV

Energy of the incident photon, E h hc= =ul

or JE =

( . )( )

.6 6 10 3 10

6 2 10

34 8

6

=

=

6 6 3 10

6 2 10 1 6 100 2

26

6 19.

( . )( . ).eV eV

As E K K E= + = f f0 0,

= 0.2 eV 0.1 eV = 0.1 eV

35. (b) : up p

= = =12

12

2000 2

5km .

.Hz

In equilibrium, kx = mg

or x mgk

= = =0 2 10200

0 01. . m

When mass is raised till the spring is unstretched,

the work = =12

2kx mgx

When the mass is released from the unstretched position of spring, then total work done

mgx mgx kx mgx = + =( ) 1

222

or x = 2x = 2 0.1 = 0.02 mAs u of spring is independent of g so that the frequency of oscillation will be the same as that on the earth.

36. (c) : Here, Y t x=

10 10010

5 sin

Comparing it with, standard equation of wave motion

Y rT

t x=

sin 2 2p pl

2 100 2100 50

p p pT

T= = =, s ; 2 1

1020p

ll p= =,

Velocity, vT

= = = l pp20

501000 1

/m s

37. (a) : On the surface of earth, the force on a mass of 1 kg is

F GMmR

GMR

= = =2 2

1 10 ... (i)

When the radius of the satellite, r = 3R/2, the force on the satellite is

F GMmr

GMR

==

2 2 2100

3 2( / )

= = 10 4 1009

4 44 102. N (Using (i))

38. (d) : The far point of 6.0 m tell us that the focal length of the lens is f = 6.0 m, u = 18 m and h = 2 m

Using, 1 1 1f v u=

= + =

1 1 1 16 0

118 0v f u . .

v = 4.5 m\ The image size,

h h vu

=

=

=2 4 5

18 00 50.

.. m

39. (a) : Here, displacement = 8.4 + 2 = 10.4 km

Total time taken = 12

8 470

0 62+ =. . h

Average velocity = DisplacementTotal time taken

= = 10 40 62

16 8 1..

.kmh

km h

40. (c) : Let u be frequency of standard fork. The

frequency of A, u u uA = +2

100

and the frequency of B, u u uB = 3

100According to question, uA uB = 6


\ +

=u u u u2

1003

1006

or or Hz5100

6 6005

120u u= = =

The frequency of A

u u uA = +

= + =2

100120 2

100120 122 4. Hz

41. (a) : Poissons ratio = Lateral strain

Longitudinal strain

Longitudinal strain = Lateral strain

Poissons ratio

= 0 01 100 4

3.. ... (i)

Youngs modulus, Y = Normal stress

Longitudinal strain

Y F

A

=

0 01 100 4

3..

(Using (i))

=

100 0 4

0 025 0 01 10 32.

. .N m = 1.6 108 N m2

42. (c) : Energy of incident photon, E hc=l

E =

6 6 10 3 10

2 10

34 8

7.

= 9.9 1019 J = 9 9 101 6 10

6 219

19..

.

=

eV eV

Kmax = eVs = e 2.5 V = 2.5 eVAccording to Einsteins photoelectric equation

K hcmax = lf0

where the symbols have their usual meanings.

or fl0

= hc Kmax = 6.2 eV 2.5 eV = 3.7 eV

Threshold frequency, uf

00=

h

u019

343 7 1 6 10

6 6 10=

. .

.= 0.9 1015 Hz = 9 1014 Hz

43. (d) : When potentiometer is connected between A and B, then it measures only e1 and when connected between A and C, then it measures e1 e2.

\

=

=e

e ee ee

1

1 2

1

2

1 2

1

2

1

ll

ll

,

or or1 100300

1 13

2

1

2

1 = = ee

ee

or oree

ee

2

1

1

2

23

32

= =

44. (c) :

Ground

5 m

Tower

u

10 m Let t be time taken by the body to reach the

ground.

\ = = = =H gt t H

g12

2 2 510

12 or s

R = ut or 10 = u 1 or u = 10 m s1

45. (c)nn

Form IV1. Place of Publication : New Delhi2. Periodicity of its publication : Monthly3. Printers and Publishers Name : Mahabir Singh Nationality : Indian Address : Physics for You, 406, Taj Apartment, New Delhi - 110029.4. Editors Name : Anil Ahlawat Nationality : Indian Address : Physics for You, 19, National Media Centre, Gurgaon Haryana - 122002 5. Name and address of : Mahabir Singh individuals who own the 406, Taj Apartment newspapers and partners or New Delhi shareholders holding more than one percent of the total capitalI, Mahabir Singh, hereby declare that particulars given above are true to the best of my knowledge and belief.Mahabir SinghPublisher

PHYSICS FOR YOU | March 15 23

1. A sample of hydrogen gas in its ground state is irradiated with photons of 10.2 eV energies. The radiation from the above sample is used to irradiate two other samples of excited ionised He+ and excited ionised Li2+ respectively. Both the ionised samples absorb the incident radiation.(i) How many lines are obtained in the He+

and Li2+ emission spectra?(ii) What are the smallest and biggest

wavelengths in their spectra?

2. The half-value thickness of an absorber is defined as the thickness that will reduce the intensity of a beam of particles by a factor of 2. Calculate the half-value thickness for lead, assuming an X-ray beam of wavelength 20 pm. Total linear attenuation coefficient, m = 55 cm1 for X-rays in lead at wavelength l = 20 pm.

3. When an electron beam interacts with atoms on the surface of a solid, by studying the angular distribution of the diffracted electrons, one can indirectly measure the geometrical arrangement of atoms. Assume that the electrons strike perpendicular to the surface of a solid as shown in figure, and that their energy is low, K = 100 eV, so that they interact only with the surface layer of atoms. If the smallest angle at which a diffraction maximum occurs is at 24, what is the separation d between the atoms on the surface?

4. Initial activity of a b emitter isotope 90Sr is 10 mCi. How many decays per second will be taking place after 84 years. The half life of 90Sr is 28 years.

5. Nuclei of a radioactive element A are being produced at a constant rate a. The element has a decay constant l. At time t = 0, there are N0 nuclei of the element.(a) Calculate the number N of nuclei of A at

time t.(b) If a = 2lN0, calculate the number of nuclei

of A after one half life of A, and also the limiting value of N as t .

6. Find the decay constant and mean life time of 55Co radio nuclide if its activity is known to decrease by 4% per hour. The decay product is non-radioactive.

SOLUTIONS

1. After absorbing photons of energy 10.2 eV, hydrogen atom would reach the first excited state of 3.4 eV, since energy difference

By : Prof. Rajinder Singh Randhawa*

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699

24 PHYSICS FOR YOU | March 15

corresponding to n = 1 and n = 2 is 10.2 eV. When this excited hydrogen atom deexcites, it would release 10.2 eV, which is absorbed by He+ and Li2+.

Energy of nth state of a hydrogen like atom with atomic number Z is given by,

E Z

nn= 13 6

2

2. eV

After absorbing 10.2 eV, He+ electron moves from n = 2 to n = 4 and Li2+ electron moves from n = 3 to n = 6.

In the spectrum of He+ there would be 4C2 = 6 lines.

lmax.n n

hcE= =

= =

=4 3

1242

13 6 49

416

470

nm

lmin.

.n n

hcE= =

= =

=4 1

1242

13 6 4 416

24 4

nm

Similarly, in spectrum of Li2+ there will be 6C2 = 15 lines.

lmax.

.n n= =

=

=6 5

1242

13 6 925

936

830 2 nm

lmin.

.n n= =

=

=6 1

1242

13 6 9 936

10 4 nm

2. The intensity varies with distance travelled in the medium, according to relation

I x I eI

I ex x( ) = \ = 00

02m m

e ex x = =m m12

2or

Taking log both sides, mx = ln 2

x x= = =

= =

ln ln .

. .

2 255

0 69355

1 26 10 0 1262m

or

mmcmHence, we conclude that lead is a very good absorber for X-rays.

3. The path difference is dsinq.For constructive interference, dsinq = nl.For the smallest value of q, dsinq = 1 lThe kinetic energy is

K

pm

hm

p h

e e= = =

2 2

22 2 l l

\ =

=

=

l hm Ke2

6 63 10

2 9 1 10 100 1 6 100 123

34

31 19

.

( . ) .. nm

\ On the surface, interatomic spacing of atoms is

d = = =l

qsin.

sin.0 123

240 30 nm

4. As T1/2 = 28 year, there will be three half lives in 84 years.

So the activity will be only 18

12

12

12 8

as = 1 .

So the activity of source will be,

A = = = 10 18

1 25 1 25 10 3. .mCi Ci

(Q 1 Ci = 3.70 1010 decays/sec)\ =

=

A ( . ) ( . )

.

1 25 10 3 70 10

4 63 10

3 10

7 decays per second

PHYSICS FOR YOU | March 15 25

5. (a) The rate of production of A is a while its rate of decay is (lN). Thus the net rate of change of the nuclei of A is given by

dNdt

N dNN

dt=

=( )a la l

or

Integrating, we get

dNN

dt N tN

N t

e NN

( )log ( )

a l la l

= =

000

1or

or ( ) ( )a l a l l = N N e t0which gives the number of nuclei at time t,

N N e t= 1 0la a l l[ ( ) ] (i)

(b) For a = 2lN0, eqn. (i) becomes N N N e t= 2 0 0

l (ii)

To obtain the number of nuclei after one

half life of A, put t T= =1 20 693

/.l

in eqn. (ii)

N N N e N N e= =

2 20 0

0 693

0 00 693l l

..

\ = ={ }N N N e2 2 120 0 0 693 . =

32 0

N

The limiting value of N as t isN N N e N N e N

tt= = =

lim 2 2 20 0 0 0 0l

6. Initial activity, A0 = lN0Activity of nuclei at time t, A = lN = lN0elt = A0elt (i)Since activity decreases at h = 4% per hour, so activity of 55Co radio nuclide at t = 1 h,A = A0 hA0 = A0 (1 h) (ii)Taking natural log of equation (i), we get

lnA

At

0= l

ln( )AA

0

0

1 =

hl (using (ii))

or l = ln(1 h) (h)\ l = hPut the value, we get l 1.1 105 s1

Mean life time, t = 1l

= 9 104 s

nn

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26 physics for you | march15

The apparent changes in frequency detected due to relative motion between source and observer is termed as Dopplers effect. Now, the question is how does one layman understand whether there has been any change in frequency or not? The answer is very easy. If the sound appears to be more or less shrill, its frequency has either increased or decreased. Shrillness of sound is directly related to frequency. This is a common phenomenon experienced in day to day life. Supposedly, a bike is stationary and it blows its horn. The sound of the horn will be more shrill if the bike starts approaching us.Now, let us explore the details of the how and why of Dopplers effect.Case-I : Source(S) moving, observer(O) stationary

vw = wave velocity f0 = frequency of source towards observer vS = velocity of source towards observerIn this case, the waves, once emitted, the propagation speed is only medium dependent, since the observer is stationary too.But what about wavelength?Wavelength is defined as the shortest distance between two points oscillating in same phase. Had the source been stationary, this distance surely would not have changed. Let me explain.If f0 is the frequency of the wave emitted, its time period,

T f= 1

0Let, a wave be emitted at t = 0, hence, the wave would travel a distance l0 = vwT in a time t = T. But the source itself has moved towards observer by a distance vST as shown in figure.

Clearly, the waves appear to have been compressed, hence wavelength decreases for the observer.

\ = = lapp v T v T v v fw S w S( ) 1

0

= vf

v vf

ww S

app( ) 1

0

=

fv

v vfw

w Sapp 0

where, fapp = apparent frequency detected by observer.Clearly, fapp > f0, where source approaches observer. Similarly, had the source been moving away (receding) from the observer, the waves would have expanded, i.e., wavelength increased, hence frequency decreased.In general, due to motion of source towards/away from observer, the apparent frequency detected would be

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

28 physics for you | march15

f

vv v

fww s

app =

0

Adjust the or + sign using the simple logic, that when separation between S and O decreases, fapp should be more than f0, hence obviously sign has to be used here.Case-II : Source(S) stationary, observer(O) moving

vO = velocity of observer away from source.In this case, as the source is stationary, there is no scope of wavelength appearing to have changed. But here, as the observer is moving away, the wave appears to be coming at a slower rate, hence the wave velocity changes in this case.\ vapp = vw vO = apparent wave velocity for observer

\ = =v v f fvfw Ow

app appl00

\ =

fv v

vfw O

wapp 0

Clearly, the apparent frequency is lesser than the frequency emitted by source.A more generalised result when the observer moves towards/away from stationary source, would be

f

v vv

fw Ow

app =

0

Again, the same rule for using + or sign, i.e., separation decreases then frequency should increase and vice versa.Case-III : Source(S) as well as observer(O) moving

This case taken is just case-I and II combined, i.e. due to the motion of source, wavelength changes whereas due to motion of observer, wave velocity changes, and due to cumulative effect of both, frequency of wave appears to have changed. A more

generalised expression for finding out the apparent frequency would be

f

v vv v

fw Ow S

app =

0

Let us learn, how to apply this formula for the above case shown where the source move towards observer and observer moves away from source. Due to motion of source, the separation decreases, hence frequency should increase (therefore, negative sign in denominator) whereas due to the motion of observer, the separation increases, hence frequency should decrease (hence negative sign in numerator too).

\ =

fv vv v

fw Ow S

app 0

This is the frequency detected by observer for the case shown.Typical Examples(1) Source(S) and observer(O), both moving with

same velocity in same directionv v

S O In this case, since there is no relative motion

between source and observer, the observer would not detect any change in frequency but this does not mean that wavelength or wave velocity does not appear to have changed for the observer. Infact, the wavelength has decreased due to motion of source whereas the wave speed decreases due to the motion of observer and their cumulative effect is that there is no change in frequency.

f

v vv v

f fww

app =

=0 0

(2) If the direction of motion of source/observer does not match with the line joining them.

In this case, break the components of the velocity of the source and observer along the line joining them and use the same generalised formula.


\ =

+

fv vv v

fw O Ow S S

appcoscos

0

(3) One among the source or observer is stationary while the other moves perpendicular to the line joining them

Now, clearly there is no component of velocity along OS, there would not be any change in frequency.

\ fapp = f0(4) Frequency detected after reflection from a

rigid boundary (wall/building/cliff)

We have two observers here, A and B. A will get to hear two frequencies, one of the

wave which has been emitted from the source directly and the other after reflection from the wall. Hence he can also hear beats if the difference of frequencies is less than 10 Hz (due to limitation of resolution). The frequency received by the wall,

f

vv v

fww

rec =

0

,

since the wall becomes a stationary observer and the source is approaching the observer.

The received frequency will be equal to the reflected frequency (fref).

\ = =

f fv

v vfw

wref rec 0

Now the wall behaves as a stationary source of sound of frequency fref whereas A is the observer (moving with speed v towards wall).

\ =

+

fv v

vfw

wapp ref

=

+

v vv

vv v

fww

w

w0

f

v vv v

fww

app =+

0

where fapp is the apparent frequency detected by the observer A after reflection from wall.

\ Beat frequency detected by him (A) is fb = fref f0

=

+

v vv v

f fww

0 0

=

20

vv v

fw

Now, what about B? Since B is stationary, he will receive the same

frequency as received by wall, and the reflected frequency being equal to the received frequency, both the frequencies received by B, directly from the source as well as after reflection are identical and hence he would not hear any beats.

A shortcut can also be used to find the reflected frequency as detected by A, where we create a virtual source S of S by taking reflection on the wall as shown in figure.

\ =

+

fv vv v

fww

app 0

nn

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Q1. Why does a mobile phone blast while attending call during charging ? How can it be prevented ?

Sachin Vats (New Delhi)

Ans. It is a fact that mobile phones when answered while charging can sometimes lead to electrocution of the person and other fire and explosion hazards, but it is important to note that such accidents happen rarely and that to because of faulty mobile phone manufacturing, battery problem and low quality chargers. As a matter of safety, it is important to follow some precautions. Make sure the battery and charger are of the same brand as mobile phone. Avoid using phone on charging. Dont tamper with the battery or bring it to contact with other metal objects outside the phone. Plug off as soon as mobile is fully charged. Avoid over heating of battery. Most importantly, follow the instructions of manufacturer for battery usage, storage, and recharging.

Q2. What are auroras and how are they form? -Aditya Prabhakar Warke

Ans. The bright dancing lights of aurora are caused by collisions between fast-moving particles (electrons) from space and the oxygen and nitrogen gas in our atmosphere. These electrons originate in the magnetosphere, the region of space controlled by Earths magnetic field. As they enter into the atmosphere, the electrons impact energy to oxygen and nitrogen molecules, making them excited. When the molecules return to their normal state, they release photon, small bursts of energy in the form of light. When billions of these collisions occur and enough photons are released, the oxygen and nitrogen in the atmosphere emit enough light for the eye

to detect them. This ghostly glow can light up the night sky in a dance of colours. But since the aurora is much dimmer than sunlight, it cannot be seen from the ground in the daytime. The colour of the aurora depends on which gas is being excited by the electrons and on how much energy is being exchanged. Oxygen emits either a greenish-yellow light or a red light, nitrogen generally gives off a blue light. Auroras usually occur in ring-shaped areas centered around the magnetic poles of Earth. The brighter the colour, the more intense the aurora. The crescent of colour on the left is from sunlight scattered over the upper atmosphere.

Q3. Why magnetic field intensity at the end of long solenoid is half than at the centre of the solenoid? Suraj Gohel (Rajkot)

Ans. A solenoid is made out of a current carrying wire which is coiled into a series of turns. In a solenoid, a large field is produced parallel to the axis of the solenoid. Components of the magnetic field in other directions are cancelled by opposing fields from neighbouring coils. Outside the solenoid the field is also very weak due to this cancellation effect and for a solenoid which is long in comparison to its diameter, the field is very close to zero. Inside the solenoid the fields from individual coils add together to form a very strong field along the center of the solenoid. The magnetic field at any point in space can be computed by summing over the magnetic fields produced by each turn of wire in the solenoid. It turns out that for an infinitely long solenoid, with the same number of turns per unit length of the solenoid, the magnetic field is constant in strength everywhere inside. If solenoid has ends, then you can think of it as an infinitely long solenoid minus the end parts that stretch off to infinity. The magnetic field strength on the axis going right through the solenoid, in the place on the end of the solenoid is then the field of an infinitely long solenoid minus half of it because half is missing, and so the field strength is half as big on the ends (but right in the middle). The field strength in the middle of a long solenoid is almost exactly that of an infinitely long solenoid, or twice that on the ends.

nn

Do you have a question that you just cant get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough.The best questions and their solutions will be printed in this column each month.

Y U ASKEDWE ANSWERED


reflection of light

When a light ray strikes the surface between two media, a part of it get return back in the initial medium. It is known as reflection.

laws of reflectionThe incident ray, the reflected ray and the normal to the surface, all lie in the same plane.The angle of incidence is equal to the angle of reflection, i = r

reflection of light at plane surfaceIn case of a reflection from plane surface such as plane mirror, the image is always erect, virtual and of same size as the object. It is also at the same distance behind the mirror as the object in front of it.When two plane mirrors are inclined at an angle q and an object is placed between them, multiple images of the object are formed as a result of multiple successive reflections.

If 360q is an even integer, then the number of

images (n) is given by n =

360 1q

If 360q is an odd integer, then the number of

images (n) is decided according to the following two situations :

r

Normal

i

If the object lies symmetrically, then

n =

360 1q .

If the object lies unsymmetrically, then

n =

360q .

If 360q is a fraction, the number of images formed

will be equal to its integral part.

KEYPOINT

When two plane mirror are placed parallel to each other and an object is placed between them, the number of image formed will be infinite.

reflection of light at spherical surfaceA spherical mirror is a part of a spherical reflecting surfaces. They are of two types :

Concave mirror : If the reflection occurs from the inner surface of the spherical mirror, the mirror is called a concave mirror.Convex mirror : If the reflection occurs from the outer surface of the spherical mirror, the mirror is called a convex mirror.

CF P

f

Concave Mirror

CFP

Convex Mirror

f

Optics and Modern Physics8


Here, P = pole of mirror, F = principal focus f = focal length, C = centre of curvatureNew cartesian sign conventions : All distances have to be measured from the pole of the mirror. Distances measured in the direction of incident light are taken as positive, while those measured in opposite direction are taken as negative. Heights measured upwards and normal to the principal axis of the mirror are taken as positive, while those measured downwards are taken as negative.

the Mirror equation

1 1 1u v f+ = where u is the distance of object from

the pole of the mirror and v is the distance of image from the pole of the mirror. f = R/2 where R is the radius of curvature of mirror.

KEYPOINT

f or R is negative for concave mirror and positive for convex mirror

Linear Magnification

m

IO

vu

ff u

f vf= = = =size of image( )

size of object( )m is positive for erect image and m is negative for inverted image.Axial Magnification

mvuax =

2

Areal Magnification

mar

area of imagearea of object=

Newtons formula is f 2 = xy, where x is distance of object from the focus and y is distance of image from the focus of the mirror.

refraction of lightRefraction of light is the change in the path of light due to change in velocity, when it goes from one medium to another.

laws of refractionThe incident ray, the refracted ray and the

normal to the interface at the point of incidence, all lie in the same plane.The ratio of sine of angle of incidence to the sine of angle of refraction for any two media is constant.

i.e. sinsin

ir =

12m

where 1m2 is the refractive index, of the medium 2 with respect to medium 1. This is also known as Snells law.If 1m2 > 1, r < i the refracted ray bends towards the normal. In such a case medium 2 is said to be optically denser in comparison to medium 1. If 1m2 < 1, r > i the refracted ray bends away from the normal. In such a case medium 2 is said to be optically rarer in comparison to medium 1.

Absolute refractive index : Refractive index of a medium with respect to vacuum (or in practice air) is known as absolute refractive index of the medium

m = =cvspeed of light in vacuumspeed of light in medium

General expression for Snells law

12

21

2

1

12

mmm= =

= =

cvcv

vv

ir

sinsin

where c is the speed of light in air, v1 and v2 be the speeds of light in medium 1 and medium 2 respectively.Principle of Reversibility :

12 2

1

1mm

=

lateral shiftWhen the medium is same on both sides of a glass slab, then the deviation of the emergent ray is zero. That is the emergent ray is parallel to the incident ray but it does suffer lateral shift with respect to the incident ray and is given by

r

iMedium 1Medium 2


Lateral shift,sin ( )

cosd ti r

r=

where t is the thickness of the slab.

real Depth and apparent DepthWhen one looks into a pool of water, it does not appear to be as deep as it really is. Also when one looks into a slab of glass, the material does not appear to be as thick as it really is. This all happens due to refraction of light.If a beaker is filled with water and a point lying at its bottom is observed by someone located in air, then the bottom point appears raised. The apparent depth is less than the real depth. It can be shown that

apparent depthreal depth

refractive index= ( )m

total internal reflectionIt is the total reflection of light from a boundary between two mediums. It occurs when the angle of incident is greater than the critical angle for the two surfaces involved. Critical angle is the angle of incidence for which the angle of refraction is 90. It exists only when light passes from a denser to a rarer medium.

Critical angle, sin iC RD

= 1m

If the rarer medium is air or vacuum, then

sin iC =

1m

When light travels from a higher refractive medium (water) into a lower refractive index medium (air), the refracted ray is bent away from the normal.

Air ( )2Refracted

rayi2

Water ( )1Reflected

rayIncidentray

i1 i1

When the angle of incidence is equal to the critical angle iC, the angle of refraction is 90.

2i2 = 90

iC iC1

If i1 is greater than iC, there is no refracted ray, and total internal reflection occurs.

iC Totalinternal

reflection

21

KEYPOINT

Critical angle depends on nature of media in contact and on the wavelength of light.

application of total internal reflectionThe brilliance of diamond : The critical angle for diamond air interface is 24.4. The diamond is cut suitably so that light entering the diamond from any face falls at an angle greater than 24.4, and suffers multiple total internal reflections which results in sparkling of diamond.Mirage : It is an optical illusion which occurs in hot, sunny days. The object such as a tree appears to be inverted, as if the tree is on the bank of a pond of water.Optical fiber : It is a thin tube of transparent material that allows light to pass through, without being refracted into the air. Light undergoes successive total internal reflections as it moves through an optical fibre.


refraction through a prismPrism is a homogeneous, transparent medium enclosed by two plane surfaces inclined at an angle. These surfaces are called the refracting surfaces and angle between them is known as the refracting angle or the angle of prism.The angle between the incident ray and the emergent ray is known as the angle of deviation.

For refraction through a prism it is found thatd = i + e A where A = r1 + r2 When A and i are small\ d = (m 1) AIn a position of minimum deviation d = dm, i = e, and r1 = r2 = r

\ =+

=i

Ar Am

d2 2and

The refractive index of the material of the prism is

m

d

=

+

( )sin

( )

sin

A

A

m2

2This is known as prism formula, where A is the angle of prism and dm is the angle of minimum deviation.

Dispersion of lightIt is the phenomenon of splitting of white light into its constituent colours on passing through a prism. This is because different colours have different wavelengths (lR > lV). According to Cauchys formula

ml l

= + +A B C2 4

where A, B, C are arbitrary constants. Therefore, m of material of prism for different colours is different (mV > mR). As d = (m 1) A, therefore different

colours turn through different angles on passing through the prism. This is the cause of dispersion.

RedYellow

Violetangular DispersionThe difference in deviation between any two colours is known as angular dispersion. Angular dispersion dV dR = (mV mR)A where mV and mR are the refractive index for violet and red rays.

Mean deviation dd d

=+V R2 .

Dispersive power,

wd dd=angular dispersion

mean deviation( )

( )V R

wm mm=

V R( ) ,1

where mean refractive indexmm m

=+

=V R2Dispersion without DeviationSuppose we combine two prisms of refracting angles A and A, and dispersive powers w and w respectively in such a way that their refracting angles are reversed with respect to each other.For no deviation, the condition is d + d = 0

(m 1) A + (m 1) A = 0 or = A

A( )( )mm

11

Under this condition, net angular dispersion produced by the combination

= ( ) + ( )d d d dV R V R

= ( ) + ( ) m m m mV R V RA A

Deviation without DispersionThe condition for no dispersion is

m m m mV R V RA A( ) + ( ) = 0

or = ( )A

AV RV R

( )m mm m


Under this condition, net deviation produced by the combination is = d + d = (m 1) A + (m 1) ARainbow : The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. It is due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain.

scattering of lightAs sunlight travels through the earths atmosphere, it gets scattered (changes its direction) by the atmospheric particles. Light of shorter wavelengths is scattered much more than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering.

refraction froM a spherical surfaceA spherical refracting surface is a portion of a refracting medium whose curved surface is a part of a sphere. Spherical refracting surface are of two types :

Convex refracting surface Concave refracting surface

For refraction from rarer to denser medium

+ =m m m m1 2 2 1

u v Rwhere u, v and R be the object distance, image distance and radius of curvature from the spherical surface respectively.When refraction occurs from denser to rarer medium we interchange m1 and m2

+ =m m m m2 1 1 2

u v Rrefraction by a lensA lens is a portion of a transparent refracting medium bound by two spherical surfaces, or one spherical surface and a plane surface.Lenses are of two types:

Convex lens is a lens that is thicker in the middle than at the edges as shown in figure (a).Concave lens is a lens that is thinner in the middle than at the edges as shown in figure (b).

R2

R1

R1 R2

(a)

(b)

R2R1

Biconvex Plano convex Concavo convex

Biconcave Plano concave Convexo concave

KEYPOINT

The sign convention for thin lenses are same as those of spherical mirrors except that instead of pole of the mirror, we take use of optical centre of the lens.

lens Makers formula1 1 1 1

1 2f R R= ( )m

when R1 and R2 are radii of curvature of the two surfaces of the lens and m is the refractive index of material of lens with respect to surrounding medium.When the refractive index of the material of the lens is greater than that of the surroundings, then biconvex lens acts as a converging lens and a biconcave lens acts as a diverging lens as shown in the figure.

When the refractive index of the material of the lens is smaller than that of the surrounding medium, then biconvex lens acts as a diverging lens and a biconcave lens acts as a converging lens as shown in the figure.


thin lens formula1 1 1v u f =

where u is the distance of the object from the optical centre of the lens, v is the distance of the image from the optical centre of the lens, f is the focal length of lens.

KEYPOINT

f is positive for converging or convex lens and f is negative for diverging or concave lens.

linear Magnification

mIO

vu= =

size of imagesize of object

( )( ) .

m is positive for erect image and m is negative for inverted image.

power of a lens

P = 1focal length in metres .

The SI unit of power of lens is dioptre (D).1 D = 1 m1.For a convex lens, P is positive, and for a concave lens, P is negative.

combination of thin lenses in contact When a number of thin lenses of focal length f1, f2, ...etc. are placed in contact coaxially, the equivalent focal length F of the combination is given by

1 1 1 11 2 3F f f f

= + + + ....

The total power of the combination is given by

P = P1 + P2 + P3 + ...

The total magnification of the combination is given by

m = m1 m2 m3 ....When two thin lenses of focal lengths f1 and f2 are placed coaxially and separated by a distance d, the focal length of a combination is given by

1 1 11 2 1 2F f f

df f= + .

In terms of power P = P1 + P2 dP1P2.

SELF CHECK

1. A thin convex lens made from crown glass

m =

32 has focal length f. When it is measured

in two different liquids having refractive

indices 43

and 53

, it has the focal lengths f1 and

f2 respectively. The correct relation between the focal lengths is(a) f1 and f2 both become negative (b) f1 = f2 < f(c) f1 > f and f2 becomes negative (d) f2 > f and f1 becomes negative

(JEE Main 2014)2. An object 2.4 m in front of a lens forms a sharp

image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?(a) 2.4 m (b) 3.2 m(c) 5.6 m (d) 7.2 m

(AIEEE 2012)

optical instruMents

simple Microscope It is used for observing magnified images of tiny objects. It consist of a converging lens of small focal length.Magnifying power of a simple microscope when the image is formed at infinity,

M Df=

where D is the least distance of distinct vision and f is the focal length of convex lens. When the image is formed at near point (at D),

M Df= +1 .

compound MicroscopeA compound microscope consist of two convex lenses coaxially separated by some distance. The


lens nearer to the object is called the objective and the lens through which the final image is viewed is called the eyepiece.

Magnifying power of a compound microscopeWhen the final image is formed at infinity (normal adjustment),

Mvu

Df

oo e

=

Length of tube, L = vo + feWhen the final image is formed at least distance of distinct vision,

Mvu

Df

oo e

= +

1

where uo and vo represent the distance of object and intermediate image from the objective lens, fe is the focal length of an eye lens.

Length of the tube, L vf D

f Doe

e= + +

astronomical refracting telescopeIt is used for observing astronomical bodies. Refracting telescope use lenses as their main components. The one facing the object is called objective or field lens and has large focal length, while the other facing the eye is called eye-piece or ocular has small focal length.Magnifying power of a astronomical telescopeWhen the final image is formed at infinity (normal adjustment),

Mffoe

=

Length of tube, L = fo + feWhen the final image is formed at least distance of distinct vision,

Mff

fD

oe

e= +1

Length of tube, L ff D

f Doe

e= + +

terrestrial telescopeIt is used for observing far off objects on the ground. The essential requirement of such a telescope is that

final image must be erect with respect to the object. To achieve it, an inverting convex lens (of focal length f) is used in between the objective and eye piece of astronomical telescope. This lens is known as erecting lens.In normal adjustment,

Magnifying power, Mffoe

=

Length of the tube, L = fo + 4f + fereflecting type telescopeReflecting type telescope was designed by Newton in order to overcome the drawbacks of refracting type telescope. In a reflecting type telescope, a concave mirror of large aperture is used as objective in place of a convex lens. It possesses a large light gathering power and a high resolving power. Due to this, it enables us to see even faint stars and observe their minute details.In normal adjustment

Magnifying power, Mff

R

foe e

= =( )2

where R is the radius of curvature of concave mirror.Reflecting type telescope is free from chromatic aberration because light does not undergo refraction.

wave optics

Wave optics deals with the theories of the nature of light and provides an explanation for different phenomena like reflection, refraction, interference, diffraction and polarisation.

wavefrontParticles of light wave which are equidistant from the light source and vibrate in the same phase constitute a wavefront.Depending on the type of light source, wavefronts are of three types:

Spherical wavefront : It is formed by a point source of light.Cylindrical wavefront : It is formed by a linear source of light.


Plane wavefront : It is obtained from a point source of light when the observation point is far away from the light source.

huygens principleAccording to this principle, 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. 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 secondary wavefront.

coherent sourcesSources of light which emit continuous light waves having the same wavelength, same frequency and in same phase or having a constant phase difference are known as coherent sources of light.

interference of lightIt is the phenomenon of redistribution of energy on account of superposition of light waves from two coherent sources. Interference pattern produce points of maximum and minimum intensity. Points where resultant intensity is maximum, interference is said to be constructive and at the points of destructive interference, resultant intensity is minimum.

conditions for sustained interference of lightThe two sources should continously emit waves of the same wavelength or frequency. The amplitudes of waves from two sources should preferably be equal. The waves emitted by the two sources should either be in phase or should have a constant phase difference. The two sources must lie very close to each other. The two sources should be very narrow.

intensity DistributionIf a, b are the amplitudes of interfering waves due to two coherent sources and f is constant phase difference between the two waves at any point P, then the resultant amplitude at P will be

R a b ab= + +2 2 2 cos f

If a2 = I1, b2 = I2, then resultant intensity, I = R2 = a2 + b2 + 2 ab cos f

I I I I I= + +1 2 1 22 cosf

If I1 = I2 = I0, then I = I0 + I0 + 2I0 cos f = 4I0 cos2f2

I I I I I= + +1 2 1 22 cosf

When cos , maxf= = + + = +( )1 21 2 1 2 1 2 2I I I I I I IWhen cos , minf = = ( )1 1 2 2I I I

or II

I I

I I

a ba b

maxmin

( )( )

=+( )( )

= +

1 22

1 22

2

2

If I1 = I2 = I0, then Imax = 4I0, Imin = 0.

Resultant intensity, I = 4I0 cos2 f2

If the sources are incoherent, I = I1 + I2youngs double slit experimentYoungs double slit experiment was the first to demonstrate the phenomenon of interference of light. Using two slits illuminated by monochromatic light source, he obtained bright and dark bands of equal width placed alternately. These were called interference fringes.

For constructive interference (formation of bright fringes)For nth bright fringe,

Path difference = =xdD nn l

where n = 0 for central bright fringen = 1 for first bright fringe, n = 2 for second bright fringe and so ond = distance between the two slitsD = distance of slits from the screen xn = distance of nth bright fringe from the centre.

\ =x n Ddn l

For destructive interference (formation of dark fringes).

For nth dark fringe,

path difference = = x dD nn ( )2 1 2l

where, n = 1 for first dark fringe, n = 2 for 2nd dark fringe and so on.


xn = distance of nth dark fringe from the centre

\ = x n Ddn ( )2 1 2l

Fringe width : The distance between any two consecutive bright or dark fringes is known as fringe width.

Fringe width, l= Dd

Angular fringe width, q l= =D dIf W1, W2 are widths of two slits, I1, I2 are intensities of light coming from two slits; a, b are the amplitudes of light from these slits, then

WW

II

ab

12

12

2

2= =

II

a b

a bmaxmin

( )

( )=

+

2

2

Fringe visibility, VI II I=

+

max min

max min

When entire apparatus of Youngs double slit experiment is immersed in a medium of refractive index m, then fringe width becomes

= = = l lmm

Dd

Dd

When a thin transparent plate of thickness t and refractive index mis placed in the path of one of the interfering waves, fringe width remains unaffected but the entire pattern shifts by

Dx t Dd t= = ( ) ( )m ml1 1

This shifting is towards the side in which transparent plate is introduced.

KEYPOINT

For constructive interference at a point, the phase difference between the two waves reaching that point should be zero or an even integral multiple of p.For destructive interference at a point the phase difference between the two waves reaching that particular point should be an odd integral multiple of p.

DiffractionIt is the phenomenon of bending of light around the corners of an obstacle placed in its path, on account of which it penetrates into the region of geometrical shadow of the obstacle.Diffraction of light at a single slitIn this case, the diffraction pattern obtained on the screen consists of a central bright band, having alternate dark and weak bright bands of decreasing intensity on both sides.Condition for nth secondary maximum in terms of

path difference = = a nnsin ( )ql2 1 2

where n = 1, 2, 3,.......Condition for nth secondary minimum in terms ofpath difference = asinqn = nl where n = 1, 2, 3,.......Width of secondary maxima or minima

l l= =Daf

awhere, a is the width of slit, D is the distance of screen from the slit, f is the focal length of lens for diffracted light.

Width of central maximum = =2 2l lD

afa .

Angular fringe width of central maximum =2la .

Angular fringe width of secondary maxima or

minima =la .

resolving powerIt is the ability of an optical instrument to produce distinctly separate images of two close objects i.e. it is the ability of the instrument to resolve or to see as separate, the images of two close objects.

limit of resolutionThe minimum distance between two objects which can just be seen as separate by the optical instrument is known as the limit of resolution of the instrument. Smaller the limit of resolution of the optical instrument, greater is its resolving power and vice-versa.


the refractive index of the medium is equal to the tangent of the polarising angle.

m = tan ip

Malus lawAccording to this law, when a beam of plane polarised light is incident on the analyzer, the intensity of light transmitted from the analyser is directly proportional to the square of the cosine of the angle between the planes of transmission of the polariser and analyser.

I a cos2qIf the intensity of plane polarised light incident on analyser is I0, then intensity of light emerging from analyser is

I = I0 cos2qpolaroidsA polaroid is a type of plastic sheet which polarises light. It can be used to control the intensity of light in sunglasses, windowpanes, photographic cameras and 3D movie cameras.

KEYPOINT

When light is incident at polarising angle, the reflected and refracted rays are perpendicular to each other.

SELF CHECK

3. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30 makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB

respectively, then IIAB

equals

(a) 13 (b) 3 (c)

32 (d) 1

(JEE Main 2014)

4. A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its

resolving power of a Microscope It is defined as the reciprocal of the minimum distance d between two point objects, which can just be seen through the microscope as separate.

Resolving power = =1 2dm qlsin

where mis refractive index of the medium between object and objective lens, qis half the angle of cone of light from the point object, d represents limit of resolution of microscope and msinq is known as the numerical aperture.

resolving power of a telescope It is defined as reciprocal of the smallest angular separation (dq) between two distant objects, whose images are just seen in the telescope as separate.

Resolving power = =1 1 22dD

q l.

where D is diameter or aperture of the objective lens of the telescope, dq represents limit of resolution of telescope.

polarisationThe phenomenon of restricting vibrations of light to a single plane is known as polarisation of light.Angle of polarisation is the angle of incidence for which an ordinary light is completely polarised in the plane of incidence when it gets reflected from a transparent medium.

Unpolarisedlight

Plane of vibration

Plane of vibration Plane polarisedlight

The plane in which the vibrations of polarised light are confined is known as plane of vibration and plane perpendicular to the plane of vibration is known as plane of polarization.

Brewsters law According to this law, when unpolarised light is incident at polarising angle on the interface separating air from a medium of refractive index m, the reflected light is fully polarised, provided


principal plane makes an angle of 45 relative to that of A. The intensity of the emergent light is

(a) I0/8 (b) I0 (c) I0/2 (d) I0/4

(JEE Main 2013)

Dual nature of Matter anD raDiation

Phenomena like interference, diffraction and polarisation can be explained only on the basis of wave nature of radiation whereas phenomena like black body radiation, photoelectric effect, compton effect can be explained only on the basis of particle (quantum) nature of radiation. Thus, radiation has dual nature i.e. particle and wave.

photoelectric effectIt is the phenomenon of emission of electrons from the surface of metals, when light radiations of suitable frequency fall on them. The emitted electrons are known as photoelectrons and the current so produced is known as photoelectric current.

work function The minimum energy needed by an electron to come out from a metal surface is known as work function of the metal. It is denoted by f0 or W0 and measured in electron volt (eV). The work function depends on the properties of the metal and the nature of its surface.

laws of photoelectric emissionThe laws of photoelectric effect are as follows : For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light.For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is known as threshold frequency. Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of incident light but depends only upon the frequency (or wavelength) of the incident light. The photoelectric emission is an instantaneous process.

hertzs and lenards observationHertz found that high voltage sparks across a detector loop were enhanced when an emitter plate was illuminated by ultraviolet light from an arc lamp.Hallwachs and Lenard found that when ultraviolet radiation was allowed to fall on the emitter plate of an evacuated glass tube enclosing two metal plates, current flowed in the circuit. After the discovery of electrons, it became evident that the incident light causes electrons to be emitted from the emitter plate. It was also observed that no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value. experimental features and observations of photoelectric effectExperimental features and observations of photoelectric effect are as follows :

For a given photosensitive material and frequency of incident radiation (above the threshold frequency) the photoelectric current is directly proportional to the intensity of incident light.For a given photosensitive material and frequency of incident radiation, saturation current (the maximum value of photoelectric current) is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity.

Retardingpotential

Collector platepotential

Phot

oele

ctric

curr

ent I I I> > 13 2

I3I2I1

OV0

Stoppingpotential

For a given photosensitive material, there exists a certain minimum cut-off frequency

Phot

oele

ctric

curr

ent

Intensity of light


of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of incident radiation, but is independent of its intensity.

Stoppingpotential

( )V0

Metal AMetal B > 0

> 0

0Frequency of incident radiation

( )0O

The photoelectric emission is an instantaneous process without any apparent time lag (~ 109 s or less), even when the incident radiation is made exceedingly dim.

particle nature of light : the photonEinstein proposed that electromagnetic radiation is quantized and exists in elementary amounts we now call photons.The photon picture of electromagnetic radiations and the characteristic properties of photons are as follows :

In the interaction of radiation with matter, radiation behaves as if it is made of particles like photons.Each photon has energy E (= hu = hc/l) and

momentum p hch= =

ul , where h is Plancks

constant, u and l are the frequency and wavelength of radiation and c is the velocity of light.Irrespective of the intensity of radiation, all the photons of a particular frequency have the same energy and same momentum. The photon energy is independent of the intensity of radiations.All the photons emitted from a source of radiations travel through space with the same speed c.

The frequency of photon gives the radiation, a definite energy (or colour) which does not change when photon travels through different media.The velocity of photon in different media is different which is due to change in its wavelength.The rest mass of a photon is zero. According to theory of relativity, the mass m of a particle moving with velocity v, comparable with the velocity of light c is given by

mm

v cm m v c=

= 0

2 2 02 2

11

/or / ...(i)

where m0 is the rest mass of particle. As a photon moves with the speed of light, v = c, so from (i), m0 = 0. Photons are not deflected by electric and magnetic fields. This shows that photons are electrically neutral.In a photon-particle collision (such as photoelectron collision), the energy and momentum are conserved. However the number of photons may not be conserved in a collision.

Matter wavesThe waves associated with the moving material particles are called matter waves or de Broglie waves.

de Broglie wavelengthThe wavelength associated with moving particle is called de Broglie wavelength and it is given by

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Physics for You March 2015