Physics for You March 2015

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Training for research starts in schools

W

hen 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 one’s 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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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 (a_m

g = 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 ^ ^ 2 3 2 .

The unit vector along the reflected ray will be (a) i^ ^j 2 3 2 + (b) i^ ^j 2 3 2 − (c) − 3 + 2 i j ^ ^ _{(d) −} 3 ₋ 2 1 2 i j ^ ^

3. A ray of light moving along the vector (^i− 2^j) undergoes 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 5

2 . The unit vector along the refracted ray is

(a) − −3 5 34 i j ^ ^ (b) −(4^−5^) 5 i j (c) − −3 4 5 i j ^ ^ (d) 4 3 5 i 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) 6psr}2

(c) 12psr2 _{(d) 24psr}2

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 cm–1_.

(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

P

hysics 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.

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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 R m (a) gR (b) gR 2 (c) 2gR (d) gR 3

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 s–1₎

(a) 800 cm3 _{(b) 1200 cm}3

(c) 1600 cm3 _{(d) 2000 cm}3

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) F 2 (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) 2mv2 r p (b) 2 2 2 2 mv r p (c) 2 2mv2 r p (d) mvr 2 p nn solution of february 2015 crossword

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 riya Kataria

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 Divya Acharya  Shubhneet Bhatia  Akash Kashyap

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1. A 15 g ball is shot from a spring gun whose spring has a force constant 600 N m–1. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression (Take g = 10 m s–2)

(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 w w w+ (b) 2 _{1 2} 1 2 w w w w+ (c) w w_{w w}1 2 1 2 − + (d) w w w w₁1 2₂ 2( + )

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 = kp–5/6r1/2E1/3 (b) T = kp–4/7r1/2E1/3

(c) T = kp–5/6r1/2E1/2 (d) T = kp–4/7r1/3E1/2

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 t2 2 2₂_m (b) 2E t_{qm (c)}2 2 Eqm₂_t (d) Eq m t 2 2 2 5. 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 400°C. 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 s–1 relative to the ground. Time taken by the man to complete one revolution with respect to disc is

(a) p s (b) 3₂ps (c) 2p s (d) p₂ s 8. A vessel contains oil (density = 0.8 g cm–3 over

mercury (density = 13.6 g cm–3). 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 cm–3 is

(a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 9. Maxwell’s 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 T₁ T2

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

PRACTICE PAPER 2015

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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 A 12. 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 × 10–7 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 earth’s surface. How much will the same body stretch the spring at a place 16400 km above the earth’s 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) _{ma F}mal₊ (b) 2(F ma l) ma + (c) l l F ma +     (d) ( ) F ma l ma + 2

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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 C–1. The average total energy density of the electromagnetic wave is

(a) 3.3 × 10–3 J m–3 (b) 4.58 × 10–6 J m–3 (c) 6.37 × 10–9 J m–3 (d) 81.35 × 10–12 J m–3 24. In Young’s 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) Im 3 1 2+ 22     cos f (b) Im 5 1 4+ 22     cos f (c) I₉m 1 8 2 2 +     cos f (d) I₉m 8 2 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) ₁₉G (c) 20G (d) ₂₀G 27. The power factor of the circuit as shown in

figure is R₁= 40  220 V 20 Hz X_C= 40  X_L= 100  R₂= 40  (a) 0.2 (b) 0.4 (c) 0.8 (d) 0.6 28. 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

(C_V = molar heat capacity at constant volume) (a) C_V = 3R

2 for a monoatomic gas (b) C_V > 3R

2 for a monoatomic gas (c) C_V > 5R

2 for a diatomic gas (d) C_V = 5R

2 for a diatomic gas

29. A body is projected vertically upwards with a velocity of 10 m s–1. It reaches the maximum height h in time t. In time t/2, the height covered is

(a) h

2 (b) 25h (c) 34h (d) 58h 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 q_q2

1 is

(a) 1 (b) 2 (c) 3 (d) 5

31. A uniform chain of mass m and length l is lying on a table with ₄l 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) mgl

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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 W 33. 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 × 10–6 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 m–1, 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−    − 10 100 10

5_sin _{m then the velocity of}_,

wave will be

(a) 100 m s–1 (b) 4 m s–1 (c) 1000 m s–1 (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 h–1 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 h–1 (b) 35 km h–1 (c) 64 km h–1 (d) 18.6 km h–1

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 s–1. 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 × 10–3. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m2, its Young’s modulus is

(a) 1.6 × 108 N m–2 (b) 2.5 × 1010 N m–2 (c) 1.25 × 1011 N m–2 (d) 16 × 109 N m–2 42. In an experiment on photoelectric emission

from a metallic surface, wavelength of incident light is 2 × 10–7 m and stopping potential is 2.5 V. The threshold frequency of the metal is approximately

(Charge of electron e = 1.6 × 10–19 C, Planck’s constant h = 6.6 × 10–34 J s)

(a) 12 × 1015 Hz (b) 9 × 1015 Hz (c) 9 × 1014 Hz (d) 12 × 1013 Hz

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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 2

A

B C

(a) 3 : 1 (b) 1 : 3 (c) 2 : 3 (d) 3 : 2 44. 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 s–2) (a) 2.5 m s–1 (b) 5 m s–1

(c) 10 m s–1 (d) 20 m s–1

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 g mu mg max= = × 2 2 1 2 2 But 1 2 1 2 2 2 mu = kx \ R = kx × = mg kx mg max 1₂ 2 2 2 = × × = 600 0 05 0 015 10 10 2 ( . ) . m

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

m₁g = w₁ m2g = w2 T w₂ m₂a₀ a T w₁ m₁a₀ a m₂ m₁

From force diagram,

T – m₂a₀ – w₂ = m₂a

or T – m2g – w2 = m2a ( a0 = g)

or T – 2w2 = m2a ... (i)

From force diagram,

m1a0 + w1 – T = m1a

or m1g + w1 – T = m1a ( a0 = g)

or 2w₁ – T = m₁a ... (ii)

From eqs. (i) and (ii), we get

T w 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] = [ML–1T–2]a[ML–3]b[ML2T–2]c [T] = [Ma+b+c][L–a–3b+2c][T–2a–2c] Comparing the powers, we obtain

a + b + c = 0 –a – 3b + 2c = 0 –2a – 2c = 1 On solving, we get a= −5 b= c= 6 1 2 1 3 , , 4. (a) : Here, u a F m qE m =0, = = v u at qE mt = + = +0 KE =1 = = 2 2 ₂ 2 2 2 2 2 2 2 2 mv mq E t m E q t m

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 d B M d 1 0 13 2 0 23 4 2 4 =m = p m p , As B1 = B2 \ m = p m p 0 13 0 23 4 2 4 M d M d d₁3=2d₂3_;d d1₂ =21 3/

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18 physics for you | march ‘15 6. (b) : Here, n = 5, g = 7/5, T1 = 0°C, T2 = 400°C dU=nRdT − g 1 = × × − − = 5 8 31 400 0 7 5 1 41550 . ( ) ( / ) J = 41.55 kJ

7. (c) : Using angular momentum conservation,

Li = 0, Lf = mvR – Iw, so, mvR = I w w =mvR= × × = I 50 1 2 200 1 2 For one complete revolution,

(v+wR t) =2pR 1 1 2 2 2 2 + ×    t= p× ⇒ t = 2p s. 8. (c) : Weight = Buoyant force

V g Vr_m = r g V+ r g 2 Hg 2 oil r_m=r +r = + = = Hg oil 2 13 6 0 8 2 14 4 2 7 2 . . . _. Oil Mercury 9. (b) : Higher is the temperature, greater is the

most probable velocity. 10. (d) : 1 1 1 1 25 1 100 5 100 1 20 1 2 C_s =C +C = + = = Cs = 20 mF = 20 × 10–6 F U₁ 1C V_s 2 6 2 2 1 2 20 10 120 = = ( × − )( ) = 144 × 10–3_J

Charge on each capacitor,

q1 = q2 = Cs ⋅ V = 20 × 120 =2400 mC In parallel, Cp = C1 + C2 = 25 + 100 = 125 mF = 125 × 10–6 F \ = = + × × × − − U Q C_p 2 2 6 2 6 2 2400 2400 10 2 125 10 [( ) ] = 92.16 × 10–3 J Loss of energy = U₁ – U₂ = (144 – 92.16) × 10–3 J = 51.84 × 10–3 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 3 2 3 2 8 1 2 4 . . . W \ Main current, I = =6 4 3 2A Potential difference across A and B

= ×3 =

2 1 2 1 8. . V

\ Current through 2 W =1 8=

2 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 1

20 1 30 1 60 v 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

d

n=(2 1− )l ⇒₂ Y3= l5₂ _dD

For bright fringes,

Y n D d n= l ⇒Y5= l5_dD Y = Y₅ – Y₃ Y D d = = × × × × − − 5 2 5 6 5 10 1 2 10 7 3 l . = 1.625 × 10–3 m = 1.63 mm

15. (d) : Here, power of the bulb, P = 100 W Supply voltage, e = 230 V

Let R be the resistance of the bulb. As P= e_R2 ⇒ =R e_P2 =(230₁₀₀)2 =529W Changed supply voltage, e′ = 115 V

Heat and light energy produced by the bulb in 20 min.

=e′2 =115 20 602× × 529

t

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16. (b) : In equilibrium, weight of the suspended body = stretching force.

\ At the earth’s surface, mg = k × x At a height h, mg′ = k × x′ g g x x R R h_e e ′ = ′ = + = + 2 2 2 2 6400 6400 1600 ( ) ( ) ( ) =    = 6400 8000 16 25 2 x′ =16× =x × 25 16 25 1cm = 0.64 cm 17. (b) : As VC = e(1 – e–t/RC) or 1 – e–t/RC =VC = = e 120 200 3 5 or et/RC = 2.5 or log et/RC = loge 2.5 or t RC=2 3026. log102 5 0 92. = . or R t C = = × × − = × 0 92 5 0 92 2 10 6 2 7 10 6 . _. . W 18. (a) : e= −df= − + = − dt d dt(50t2 4) 100t When t = 2 s, |e| = 200 V Induced current at t = 2 s, I R =| |e =200 = . 400 0 5 A

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

m (refractive index) = c_v=1₁/_/e m0 0_em = e mr r

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

Thus, lmedium = _ml=l₂ 20. (b) : Y F A L L = × ∆

The breaking strength F ∝ A.

\ F = = = F A A D D D D 2 1 2 1 22 12 2 2 12 4 4 p p ( / ) ( / ) or D D F cm F 2 1 2 1 1 2 1 2 3 600 150 6 =       = _ _ = / /

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 C

F₁ 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 l 2 1 2 ( ) 2 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=1K x

(

−x

)

2 22 12 =1 − = 2K(20 102 2) 150kJ

which is less than work done in stretching it from 20 cm to 30 cm. viz.1 kJ 2K(30 202− 2)=250    

23. (b) : Total average energy density of electro-magnetic wave is < > =u 1 E + B 2 1 2 0 2 0 2 e m rms rms = +         _ = _ 1 2 1 2 0 2 0 2 2 e m E E c B E c rms rms  rms rms =1 + 2 1 2 0 2 0 2 0 0 e m e m E_rms E_rms =1 + = 2 1 2 0 2 0 2 0 2 e E_rms e E_rms e E_rms = (8.85 × 10–12) × (720)2 = 4.58 × 10–6 J m–3 24. (c) : Here, A2 = 2A1  Intensity ∝ (Amplitude)2 \ =      =      = I I A A A A 2 1 2 1 2 1 1 2 2 ₄

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20 physics for you | march ‘15 I₂ = 4I1 Maximum intensity, I_m=

(

I₁+ I₂

)

2 =

(

I₁+ 4I₁

)

2=

( )

3 I₁ 2=9I₁ I₁=Im 9 or ...(i)

Resultant intensity, I I I= + +₁ ₂ 2 I I_{1 2}cosf = +I₁ 4I₁+2 I I₁( ) cosf4₁ = 5I1 + 4I1cosf = I1 + 4I1 + 4I1cosf = I₁ + 4I₁(1 + cosf) = +I₁ 8I₁ 2 2 cos f 1 2 2 2 + =     cosf cos f =  +    I₁ 1 8 2 2 cos f

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

I=Im +    9 1 8cos22 f 25. (a) : Here, uo = –5 cm, fo = 4 cm f_e = 10 cm, D = 20 cm

According to lens formula, 1 1 4 1 5 1 4 1 5 1 20 v_o = +− = − = v_o = 20 cm Magnification, M v u D f o o e =  +      1 =  +   = 20 5 1 2010 12

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 S I G I IG _G = − ( ) S (I – I ) G G I G I I Here, I_G= 5 I 100 \ = − = S IG I I G 5 100 5 100 19

27. (c) : Resistance of the circuit,

R = R1 + R2 = 40 W + 40 W = 80 W

Impedance of the circuit,

Z= R2+(X_L−X_C)2 = ( ) (802+100 40− )2 = ( ) ( )802+ 602 =100W

Power factor, cosf = =R = .

Z

80 100 0 8 28. (c) : Monoatomic gas C_V = 3R

2

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

C_V = 5R

2 (at low temperature) Also, C_V> 5R

2 (at high temperature due to vibrational kinetic energy)

29. (c) : As v2 – v₀2 = 2gh, 0 – (10)2 = 2(–10) h or h = 5 m

Also, v = v0 + at, 0 = 10 + (–10) t

or t = 1 s

Height covered in time t/2, i.e., (1/2 s),

′ = + − = × − × ×    h v t₀ 1 g t2 2 2 10 12 1 2 10 12 ( ) = 3.75 m = (3/4) h 30. (c) : As q w₁ ₀ 1a 2 a 2 a 2 0 12 2 2 = t+ t = + ( ) = (q q₁ ₂) w₀ 1a 2 a( )2 a 2 0 12 4 8 + = t+ t = + = Thus, q2 = 6a or q_q2 1 =3 31. (c) : Mass of ₄l  

 length of the chain = m4 The weight of this part of the chain acts as its

CG which is at a distance l

8  



 from the edge of the table. Work done = m g l mgl₄ ₈ ₃₂       = 32. (c) : As RD V_PD D = 2 = 0 5 2 = 0 1 2 5 ( . ) . . V W W I V R D D D = = 0 2. A

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R V I_D

eq= =_{0 2}1 5_.. =7 5. W

Resistance of the series resistor, R = R_eq – R_D = 7.5 – 2.5 = 5 W 33. (c) : As N_NX N N N Y = X+ Y = X 1 15, 16 Thus, N N_X+XN_Y = 116 or N_X = 1 N_X+N_Y = N_X+N_Y 16 1 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 × 10–6 m, f0 = 0.1 eV

Energy of the incident photon, E h= u=hc l or E = × × J × − − ( . )( ) . 6 6 10 3 10 6 2 10 34 8 6 = × × × × = − − − 6 6 3 10 6 2 10 1 6 10 0 2 26 6 19 . ( . )( . )eV . eV AsE K= +f₀,K E= −f₀ = 0.2 eV – 0.1 eV = 0.1 eV 35. (b) : u p p = 1 = = 2 1 2 200 0 2 5 k m . Hz. In equilibrium, kx = mg or x mg k = =0 2 10× = 200 0 01 . _{. m}

When mass is raised till the spring is unstretched, the work =1 =

2kx2 mgx

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

mgx′ =(mgx) 1+ kx = mgx

2 2 2

or x′ = 2x = 2 × 0.1 = 0.02 m

As 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−    − 10 100 10 5_sin

Comparing it with, standard equation of wave motion Y r T t x =  −    sin 2p 2p l 2 ₁₀₀ 2 100 50 p p p T = ,T= = s ; 2 1 10 20 p l = ,l= p Velocity, v T = =l p = − p 20 50 1000 1 / m s

37. (a) : On the surface of earth, the force on a mass of 1 kg is F GMm R GM R = ₂ = ₂× =1 10 ... (i)

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

F GMm r GM R = ₂ ′= ×_{2 2}100 3 2 ( / ) = × ×10 4 100= × 9 4 44 10. 2 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 1 f v u= − ⇒ = + = − − 1 1 1 1 6 0 1 18 0 v f u . . ⇒ v = – 4.5 m \ The image size,

h h v u ′ = _ _= × − −   = 2 4 5 18 0 0 50 . . . m

39. (a) : Here, displacement = 8.4 + 2 = 10.4 km Total time taken = 1

2 8 4

70 0 62

+ . = . h

Average velocity = Displacement Total time taken

=10 4 = − 0 62 16 8 1 . . . km h km h

40. (c) : Let u be frequency of standard fork. The frequency of A, u_A= + 2u u

100

and the frequency of B, u_B = − 3u u 100 According to question,

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22 physics for you | march ‘15 \  +   − −    = u 2 u u u 100 3 100 6 or 5 or Hz 100 6 600 5 120 u= u= = The frequency of A u_A =u+ u   = + × = 2 100 120 2 100 120 122 4. Hz 41. (a) : Poisson’s ratio = _{Longitudinal strain}Lateral strain

Longitudinal strain = _{Poisson’s ratio}Lateral strain = 0 01 10 0 4 3 . . × − _{... (i)}

Young’s modulus, Y = _{Longitudinal strain}Normal stress

Y F A = × ×  0 01 10. _{0 4}_. −3_ (Using (i)) = × × × − − 100 0 4 0 025 0 01 10 3 2 . . . N m = 1.6 × 10 8_{N m}–2

42. (c) : Energy of incident photon, E hc= l E = × × × × − − 6 6 10 3 10 2 10 34 8 7 . = 9.9 × 10–19 J = 9 9 10 1 6 10 6 2 19 19 . . . × × = − − eV eV K_max = eV_s = e × 2.5 V = 2.5 eV

According to Einstein’s photoelectric equation

K_max=hc−

l f0

where the symbols have their usual meanings. or f l 0=hc K− max = 6.2 eV – 2.5 eV = 3.7 eV Threshold frequency, u0=f_h0 u₀ 3 7 1 6 10₃₄19 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 e e e e 1 1 2 1 2 1 2 1 2 1 l l l l , or 1 100 or 300 1 13 2 1 2 1 −e = = − e e e or e or e e e 2 1 1 2 2 3 3 2 = = 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 = × = g 1 2 2 2 5 10 1 2 _or _s R = ut or 10 = u × 1 or u = 10 m s–1 45. (c) nn Form IV

1. Place of Publication : New Delhi 2. Periodicity of its publication : Monthly 3. Printer’s and Publisher’s Name : Mahabir Singh Nationality : Indian Address : Physics for You,

406, Taj Apartment,

New Delhi - 110029.

4. Editor’s 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 capital

I, Mahabir Singh, hereby declare that particulars given above are true to the best of my knowledge and belief.

Mahabir Singh Publisher

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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 cm–1

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}90_{Sr is}

10 mCi. How many decays per second will be taking place after 84 years. The half life of 90_Sr

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

55_{Co 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*

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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 Li}2+_.

Energy of nth_{state of a hydrogen like atom with}

atomic number Z is given by,

E Z n n= −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} 4_C₂_{= 6 lines.} l_max . n n hc E = → = = =  ₋   = 4 3 1242 13 6 4 9 4 16 470 ∆ nm l_min . . n n hc E = → = = =  ₋   = 4 1 1242 13 6 4 4 16 24 4 ∆ nm

Similarly, in spectrum of Li2+_{there will be} 6_C₂_{= 15 lines.} l_max . . n= → =n =  ₋    = 6 5 1242 13 6 9 25 9 36 830 2 nm l_min . . n= → =n =  ₋   = 6 1 1242 13 6 9 9 36 10 4 nm

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

I x I e( ) = ₀ − x \ I0 =I e₀ − x

2

m m

e−mx =1 =emx

2 or 2

Taking log both sides, mx = ln 2

x= x= = = × − = ln ln . . . 2 2 55 0 693 55 1 26 10 2 0 126 m or mm cm

Hence, 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 × l The kinetic energy is

K p m h m p h e _e = 2 = 2 ₂ _ = _ 2 _{2 l}  l \ = = × × × × × × = − − − l h m K_e 2 6 63 10 2 9 1 10 100 1 6 10 0 123 34 31 19 . ( . ) . . nm

\ On the surface, interatomic spacing of atoms is d = = °= l q sin . sin . 0 123 24 0 30 nm

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

84 years.

So the activity will be only 1 8 1 2 1 2 1 2 8 as × × =1   .

So the activity of source will be,

A = × =10 1 = × − 8 1 25. mCi 1 25 10. 3Ci (Q 1 Ci = 3.70 × 1010_decays/sec) \ = × × × = × − A ( . ) ( . ) . 1 25 10 3 70 10 4 63 10 3 10

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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

dN dt N dN N dt = − − = (a l ) a l or Integrating, we get dN N dt N t N N t e _NN (a l− )= l log (a l ) − ₋ ₌

∫

0 0 0 1 or or (a l− N) (= a l− N e₀) −lt

which gives the number of nuclei at time t,

N=1 − −N₀ e− t

l[a a( l) l ] …(i)

(b) For a = 2lN0, eqn. (i) becomes

N=2N₀−N e₀ −lt …(ii)

To obtain the number of nuclei after one half life of A, put t T= _{1 2}_/ =0 693.

l in eqn. (ii) N= N −N e−  = N −N e   − 2 ₀ ₀ 2 0 693 0 0 0 693 l ._l _. \ N=2N −N

{

e− =

}

2 1 2 0 0  0 693. = 3 2N0

The limiting value of N as t → ∞ is

N N N e N N e N t t =  −  = − = →∞ − −∞ lim 2 ₀ ₀ l 2 ₀ ₀ 2 ₀ 6. Initial activity, A0 = lN0

Activity of nuclei at time t,

A = lN = lN0e–lt = A0e–lt …(i)

Since activity decreases at h = 4% per hour, so activity of 55_{Co radio nuclide at t = 1 h,}

A = A0 – hA0 = A0 (1 – h) …(ii)

Taking natural log of equation (i), we get ln A A₀ = − lt lnA ( ) A 0 0 1− = − h l (using (ii)) or l = –ln(1 – h) ≈ –(–h) \ l = h

Put the value, we get l ≈ 1.1 × 10–5_s–1

Mean life time, t = 1

l = 9 × 104 s

nn

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26 physics for you | march‘15

The apparent changes in frequency detected due to relative motion between source and observer is termed as Doppler’s 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 Doppler’s effect.

Case-I : Source(S) moving, observer(O) stationary

v_w = wave velocity

f0 = frequency of source towards observer vS = velocity of source towards observer

In 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

0

Let, 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.

\ l_app=v T v T v− = −v f w S ( w S) 1 0 ⇒ v = − f_appw (vw vS) 1f₀ ⇒ = −     f v v_wwv_S f app 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

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28 physics for you | march‘15

f v

v_wwv_s f 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 = f v f w O appl0 app w 0 \ = −    f v v v f w O w app 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 v

v f

w O

w 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 v

vw_w vO_S f 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). \ = − −     f v v vw_w vO_S f 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 direction

v 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 v

vw_w v f f

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.

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\ = + −     f v v vw_w vO_S O_S f

app cos_cosθ_θ 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 v

v_wwv f

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 f v v_wwv f ref 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).

\ = +    f v v v f w w app ref = +     −     v v v v v v f w w w w 0 f v v vw_w v f app=_ +₋ _ 0

where fapp is the apparent frequency detected

by the observer A after reflection from wall. \ Beat frequency detected by him (A) is

f_b = fref – f0 = + −     − v v vw_w v f0 f0 = −     2 0 v v_w v f

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.

\ = + −     f v v vw_w v f 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. Don’t 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 Earth’s 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

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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 reflection

The 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 surface

In 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 360_ _q°_ is an even integer, then the number of images (n) is given by n =_360_q°_{ −}1

If 360_ _q°_ 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_q°_{ −}1.

If the object lies unsymmetrically, then

•

n =_360_{q .}°_

If 360_{q 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 surface

A 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.

C F P f Concave Mirror C F P Convex Mirror f

Optics and Modern Physics

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Here, P = pole of mirror, F = principal focus f = focal length, C = centre of curvature

New 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 1

u 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=_{size of object( )}size of image( )_OI _{= − = − =}v_u _{f u}f f v_f−

m is positive for erect image and m is negative for

inverted image.

Axial Magnification max= − uv_

2 Areal Magnification

mar =_{area of object}area of image

Newton’s 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 light

Refraction 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 refraction

The 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. sin sin i r= 1m2 where 1_m

2 is the refractive index, of the medium

2 with respect to medium 1. This is also known as Snell’s law.

If

• 1_m₂_{> 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 1_m

2 < 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 = =_vc _{speed of light in medium}speed of light in vacuum

General expression for Snell’s law 1 2 2 1 2 1 1 2 m =m_m =       = = c v c v v v i r sin sin

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 : 1 2 ₂ 1 1 m m = lateral shift

When 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 i

Medium 1 Medium 2

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