Physics - 12 Sample Paper - 1

SAMPLE Paper 1 [Easy Concept] Time Allowed : 3 hrs.] [Max. Marks : 70 General Instructions (i) All questions are compulsory. There are 33 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains ten very short answer questions and four assertion reasoning MCQs of 1 mark each, Section B has two case based questions of 4 marks each, Section C contains nine short answer questions of 2 marks each, Section D contains five short answer questions of 3 marks each and Section E contains three long answer questions of 5 marks each. (iv) There is no overall choice. However internal choice is provided. You have to attempt only one of the choices in such questions. SECTION – A All questions are compulsory. In case of internal choices, attempt any one of them. 1. What is dielectric constant of metal? Or Where does the energy of a capacitor reside? 2. How does the electric field E vary with the distance r from a uniformly charged plane sheet of infinite extension? Or What does q1 + q2 = 0 signify? 3. What will be the terminal potential difference of a cell, when open circuited? Or How does the electrical resistivity of a conductor depend on temperature? 4. How can a galvanometer be converted into an ammeter? 5. How is the relative permeability of a material related to its susceptibility? Or What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal? 6. What happens to the wavelength of a photon after it collides with an electron? 7. What will happen if an electron instead of revolving becomes stationary in H-atom? 8. Which orientation of a magnetic dipole in a uniform magnetic field will correspond to its stable equilibrium? 9. Can a capacitor of suitable capacitance replace an inductor coil in an AC circuit? 10. Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends. 58 Together with®  EAD Physics—12

For question numbers 11, 12, 13 and 14, two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false and R is also false. 11. Assertion: The focal length of a concave mirror does not change when it is kept in water. Reason: The focal length of the concave mirror does not depend on nature of a medium. 12. Assertion: A convex mirror is used as a shaving mirror. Reason: A convex mirror forms magnified and erect image when held closer to the face. 13. Assertion: Magnifying power of a telescope won’t change on increasing the diameter of its objective. Reason: Magnifying power of a telescope is independent of the aperture of the objective. 14. Assertion: The angle of deviation produced by the prism is between angle of incidence and angle of emergence. Reason: The angle of minimum deviation can be used to determine the refractive index of the prism. SECTION – B All questions are compulsory. In case of internal choices, attempt any one. Questions 15 and 16 are Case Study based questions and are compulsory. Attempt any 4 sub parts from each question. Each question carries 1 mark. 15. When the nucleus of an atom disintegrates spontaneously and emits one or more penetrating radiations such as a-particles, b-particles or g-rays, the phenomenon is called radioactivity. It occurs naturally in a nucleus which is unstable such a nucleus is radioactive and the process of decaying of a nucleus is called radioactive decay a-decay results in the emission of a-particles which are doubly ionised Helium nucleus (42He2+), b-decay is when b-particles are emitted. b-particles are electrons or positrons (positively charged particles with mass same as that of electron) g-rays are the ones in which high energy photons are emitted. (i) Why do a-particles have large ionising power? (a) because of large mass. (b) because of small mass. (c) because of large mass and large nucleus. (d) Because of small mass and small nucleus. (ii) Why are g-rays also called electromagnetic waves? (a) because they consists of Hydrogen atoms. (b) they consist of photons. (c) they consists of b-particles. (d) None of the above. Sample Papers 59

(iii) Natural radioactive nuclei are (a) Nuclei with higher mass. (b) Nuclei with lower mass. (c) It does not depend upon nuclear mass. (d) Nuclei with lower number of neutrons. (iv) In nuclear decay reaction, where X is a positron 11H  →  1 n + QP X 0 (a) P = 0  Q = 0 (b) P = 0  Q = 1 (c) P = 1  Q = 0 (d) P = 1  Q = 1 (v) g-rays are not deflected by electric and magnetic fields because (a) g-rays are made up of neutrons. (b) g-rays are made up of electrons. (c) g-rays are made up of neutral particles. (d) g-rays made up of helium atoms. 16. Coherent sources emit light waves with same frequency or same wavelength, with a phase difference which is either zero or constant on the other hand non-coherent sources donot emit light waves which have a constant or zero phase difference. Interference pattern can be produced only when the light emitting sources are coherent. The intensity of pattern is maximum when phase difference is integral even multiple of p and minimum for odd multiple of p. (i) The maximum intensity of interference pattern is (a) (I1 + I2)2 (b) (I1 – I2)2 (c) I1 + I2 (d) ` I1 + I2 j2 (ii) A phase difference of 2p corresponds to the path difference of (a) l/2 (b) l (c) 3l (d) 3l/2 (iii) The intensity of light at a point on screen where the path difference is l is k units. Find the intensity where path difference is λ . 6 k 3k 3k (a) k (b) 2 (c) 4 (d) 2 (iv) For a sustained interference pattern, which of the following is true? (a) Sources should be coherent. (b) Sources should be monochromatic. (c) Sources should be narrow. (d) All of the above. (v) What happens to the interference pattern in Young’s double slit experiment when a monochromatic source is replaced by a white source of light? (a) The pattern disappears. (b) The fringe width decreases. (c) Coloured fringe pattern is obtained. (d) The pattern does not change. 60 Together with®  EAD Physics—12

SECTION – C All questions are compulsory. In case of internal choices, attempt any one. 17. When a resistance of 2 Ω is placed across the terminals of a battery, the current is 0.5 A; when the resistance across the terminals is 5 Ω, the current is 0.25 A. Calculate the emf of the battery. 18. A 10 D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification for a normal eye? Or Write the principle on which optical fibres work. Give their two uses. 19. Which one of the following will experience maximum force when projected with the same velocity –v, perpendicular to the magnetic field B (i) a-particle or (ii) b-particle? Or Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. Y E FE v X B Z FB 20. A capacitor C, a variable resistor R and a bulb B are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance? CR B Mains Or Explain why current flows through an ideal capacitor when it is connected to an a.c. source but not when it is connected to a d.c. source in a steady state. 21. Define intensity of magnetisation of a magnetic material. How does it vary with temperature for a paramagnetic material? 22. How is an n-type semiconductor formed? Name the majority change carriers in it. Draw the energy band diagram of an n-type semiconductor. 23. What is meant by depletion region in a junction diode? How is this region formed? 24. Carbon and silicon both have four valence electrons each. How then are they distinguished? 25. Out of the two magnetic materials, A has relative permeability slightly greater than unity while B has less than unity. Identify the nature of the materials A and B. Will their susceptibilities be positive or negative? Sample Papers 61

SECTION – D All questions are compulsory. In case of internal choices, attempt any one. 26. An equiconvex lens of focal length 15 cm is cut into two halves along its aperture. What is the focal length of each part? 27. Explain, giving necessary reactions, how energy is released during (i) fission and (ii) fusion. Or (a) In a nuclear reaction, 23He+32He → 42He+11H+11H+12.86MeV, though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How? Explain. (b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the regions where potential energy is (i) positive and (ii) negative. 28. The figure shows two identical parallel plate capacitors A and  B connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the electrostatic energies stored in both the capacitors before and after the introduction of the dielectric. 29. Establish a relation between current density and electric field. 30. Explain the origin of different spectral lines of hydrogen spectrum on the basis of Bohr’s theory. Or In an experiment on a-particle scattering by a thin foil of gold, draw a plot showing the number of particles scattered versus the scattering angle q. Why is it that a very small fraction of the particles are scattered at q > 90°? Write two important conclusions that can be drawn regarding the structure of the atom from the study of this experiment. SECTION – E All questions are compulsory. In case of internal choices, attempt any one. 31. (a) Find out capacitance of a capacitor with a dielectric slab of thickness t between the capacitor plates having separation d. (b) What happens when the dielectric slab is replaced by a conducting slab of same thickness? (c) How would you connect two capacitors across a battery in series or parallel, so that they store greater (i) total charge and (ii) total energy? 62 Together with®  EAD Physics—12

Or (a) Two cells of emfs e1 and e2 having internal resistances r1 and r2 are joined in parallel. Find out an expression for an equivalent emf and an internal resistance. (b) Find out the current drawn by the circuit as shown below. 32. (a) Explain the production of the electrical oscillations in L-C circuit. Under what conditions are free oscillations produced in the circuit? (b) Radio waves of wavelength 360 m are being transmitted from a transmitter. Calculate the inductance of the coil to be connected with a condenser of capacity 1.20 mF in a resonant circuit for receiving these waves. Or Two identical charges separated by a distance r are moving in the same direction with velocity v. Prove v 2 that the ratio of magnetic and electric forces acting between them is b c where c is speed of light in vacuum. l 33. Describe the displacement method of finding the focal length of a convex lens. Derive the necessary formula. If I1 and I2 are the linear sizes of the image. Prove that I0 = I1 × I2 Or (a) What are the basic conditions for obtaining sustained and well defined interference fringes on the screen? (b) The interference fringes for sodium light (l = 5890 Å) in a double slit experiment have an angular width 0.20°. For what wavelength will this width be 10% greater? Sample Papers 63

Answers 1. Dielectric constant of metal is infinitely large. Or Electrical energy resides in the space, within the plates. 2. Electric field due to uniformly charged plane sheet remains constant as it depends only on the charge density. Or q1 + q2 = 0 ⇒ q1 = – q2 \\  q1 and q2 are the two charges of an electric dipole. 3. Terminal potential difference will be equal to e.m.f. of cell, when circuit is open. Or Electrical resistivity (r) increases linearly with increase in temperature. 4. By connecting a small resistance known as shunt in parallel with the galvanometer. 5. mr = 1 + cm Or Angle of dip is 45°. 6. The wavelength of a photon increases. 7. Then the electrostatic field of the nucleus will attract the electron into the nucleus itself. 8. In stable equilibrium, the dipole moment vector and the magnetic field vector are in same direction. 9. Yes, because average power consumed in both is least while controlling an AC. 10. (i) Band gap (ii) Biasing 11. (a) Both A and R are true and R is the correct explanation of A. 12. (d) A is false and R is also false. 13. (a) Both A and R are true and R is the correct explanation of A. 14. (b) Both A and R are true but R is NOT the correct explanation of A. 15. (i) (c) A heavy nucleus will have more number of neutrons and hence higher ionising power. (ii) (b) g-rays are called electromagnetic waves because they consist of photons of short wavelength and show properties similar to electromagnetic waves. (iii) (a) Nuclei with higher mass has higher number of neutrons and hence, is unstable. (iv) (b) By conservation mass, m(LHS) = m(RHS) \\ P = 0 By conservation of charge No. of protons (LHS) = No. of protons (RHS) \\ Q = 1 (v) (c) g-rays are made up of photons. Photons are neutral particle and therefore are not deflected by electric and magnetic fields. 64 Together with®  EAD Physics—12

16. (i) (d) When cos f = +1, the intensity will be maximum I = I1 + I2 + 2 I1I2 cos φ = I1 + I2 + 2 I1I2 = ` I1 + I2 j2 (ii) (b) l.  A wave completes l wavelength white traversing phase difference of 2p. (iii) (a) 3k 4 I = I1 + I2 + 2 I1I2 cos φ I1 = I2 = I0 Let I = 2I0 + 2I0 cos f = 4I0 cos2 φ 2 For l = path difference f = 2p I = 4I0 = k = units 2π For 6 λ = path difference, f = = π 6 3 π 3k I = 4I0 cos2 6 = 4 (iv) (d) All of the above. All the stated conditions are important for a sustained interference pattern. (v) (c) Coloured fringes are obtained in place of the dark and light fringes. 17. Given: R1 = 2 W, I1 = 0.5 A, R2 = 5 W, I2 = 0.25 A When a resistance 2 W is placed across the terminals of a battery and current is 0.5 A, then e = (R1 + r) I1 …(i) e = (2 + r) 0.5 …(ii) When the resistance across the terminals is 5 W and current is 0.25 A, then e = (R2 + r) I2 e = (5 + r) 0.25 From equations (i) and (ii), we get (2 + r) 0.5 = (5 + r) 0.25 r = 1 W From equation (i), we get e = (2 + 1) 0.5 = 1.5 V 18. Given: P = 10 D i.e. f = + 10 cm For a normal eye, the magnification is maximum when an object is at least distance of distinct vision (≈ 25 cm). In case of simple microscope, the magnification is maximum when an image of the object is at 25 cm. So, f = + 10 cm, u = ?, v = – 25 cm \\ 1 = 1 – 1 f v u Sample Papers 65

⇒ 1 = 1 – 1 = 1 – 1 =– 1 + 1 o = – 35 u v f – 25 10 e 25 10 250 \\ u = – 7.14 cm Or Optical fibres work on the principle of multiple total internal reflection. It is used in endoscopy and high speed communication. 19. Radius of circular path when velocity is perpendicular to the magnetic field is given by, r = mv ,  r∝ m for same v and B qB q for a-particle, m = 4mP = 2mP q 2e e for b-particle, m = 1 d mP n q 1840 e Since, b-particle has smallest value of m , therefore, if will describe the smallest circle. q Or Let an electric charge ‘q’ is moving with velocity v in a region where uniform electric field E and magnetic field B are applied as shown in the figure F = q (E + v × B) when F = 0; then electric charge will go undeflected. \\ q (E + v × B) = 0 Etj + vit × Bkt = 0 ⇒ Etj − vBtj = 0 8a it × kt = − tjB tj  (E – vB) = 0 \\ E = v.B or v = E B 20. It is a CR circuit connected with ac mains where the inductance of bulb remains constant. The current is i = E = E Z R2 + XC2 (i) If slab is inserted between the plates, capacitance will increase and as XC = 1 , XC will decrease so Z will decrease making the current in bulb 2πfC more and hence, the bulb will shine brighter. (ii) If R is increased, Z will increase making the current less and again the bulb will glow dimly. Or As capacitive reactance, Xc = 1 = 1 wC 2πvC V for d.c. v = 0, so Xc is infinity and the current, I = Xc =0 66 Together with®  EAD Physics—12

No, current will flow through a capacitor, if it is connected to a d.c. source. For a.c., frequency v is finite so Xc has finite value and current, I = V is finite. Thus, current flows Xc through an ideal capacitor when it is connected to an a.c. source in a steady state. 21. It is defined as the magnetic dipole moment developed per unit volume of the specimen when it is placed in the magnetic field. The intensity of magnetisation of a paramagnetic substance varies inversely with its temperature T before saturation. 22. An n-type semiconductor is formed by doping a pure semiconductor with pentavalent impurity. Electrons are the majority charge carriers in n-type semiconductor. Energy level diagram at T > 0°K Conduction band – hole electron Donor energy level Valence band 23. Depletion region: It is the region around the junction which is devoid of free electron and holes and has immobile ions is called depletion region. When a junction of p-type and n-type of semiconductor is formed, due to difference in density of electrons and holes, the electrons from n side diffuse toward p-type and holes diffuse from p to n side. Near the junction, there is recombination of electrons and holes, thus, depleting the region of its free electrons and holes. There are positive and negative immobile ions in this region. 24. Although both carbon and silicon have same lattice structure, the four bonding electrons of carbon and silicon are respectively in the second and third orbits. Thus, ionisation energy is less for silicon than carbon. Hence, the number of free electrons in silicon is more than of carbon. Thus, they can be distinguished on the basis of their conductivity. 25. The relative permeability µr = (1 + χm), where χm = magnetic susceptibility. For paramagnetic substances, µr >1 and χm is positive. For diamagnetic substances µr < 1 and χm is negative. ∴ Magnetic material A is paramagetic, its χm will be positive and B is diamagnetic, its χm will be negative. 26. For an equiconvex lens, R1 = + R; R2 = – R Q 1 = (n – 1) e 1 + 1 f R Ro \\ 1 = (n – 1) = 2 G …(i) f R When it is cut into two equal halves along its aperture, then R1 = +R R2 = ∞ Q 1 = (n – 1) e 1 – 1 o = (n – 1) e 1 o …(ii) fl R 3 R Sample Papers 67

From equations (i) and (ii), we get, f′ = 2f \\ f′ = 30 cm [Q f = 15 cm] 27. (i) Nuclear Fission: It is a process in which a heavy nucleus splits up into two lighter nuclei of nearly equal masses. It is found that the sum of the masses of the product nuclei and particles is less than the sum of the masses of the reactants, i.e. there is some mass defect. This mass defect appears as energy. One such fission reaction is given below. 29325U + 10n  →  15461Ba + 3962Kr + 310n + Q 15461Ba, 9326Kr The Q value of the above reaction is about 200 MeV. The sum of the masses of and 3 neutrons is less than the sum of the masses of 29325U and one neutron. (ii) Nuclear Fusion : It is the process in which two lighter nuclei combine together to form a heavy nucleus. For fusion, a very high temperature of the order of 107K is required. One such fusion reaction is given below. 23He + 23He  →  42He + 2 11H + Q The Q value of this nuclear reaction is 12.9 MeV. It the energy equivalent of the mass defect in the above reaction. The energy released per fusion is much less than in fission but the energy released per unit mass is much greater than that released in fission. Or (a) On product side, 42He is having higher value of binding energy per nucleons, which is a result of conversion of mass into energy. (b) Potential energy (MeV) Positive potential energy 100 Negative potential energy where r0 ≈ 0.8 fm 0 For distances larger than 0.8 fm, force is attractive and is –100 repulsive for distances less than 0.8 fm. r0 0.8 1 2 3 r(fm) 28. In 1st case when switch is closed then Charge on capacitor A, qA = CV Charge on capacitor B, qB = CV \\ Ui = 1 CV2 + 1 CV2 = CV2 2 2 When switch is opened (i) Potential difference across capacitor A remains same. (ii) Charge on the capacitor B remains same. New capacitance of capacitor A is given by C′ = erC 68 Together with®  EAD Physics—12

SRFor capacitor B VW SSa W SS` ClV l = CV W 1 1 C V2 S CV CV W Q Uf = UA + UB = 2 (εrC) V2 + 2 εr SS& Vl = Cl = W T er ε C W r V Vl = εr W W X Uf = 3 CV2 + 1 CV2 (Q = 3) 2 6 Uf = 5 CV2 3 \\ Ui = CV2 = 3 Uf 5 5 CV2 3 29. Consider a conductor of length l, area of cross-section A and carrying current I, on application of potential difference V. I = V (Ohm’s law)  …(i) R As J = I …(ii) and A …(iii) E = V l where E is the electric field established within the conductor. From equations (i), (ii) and (iii), we get JA = El R J = El ea R = ρ l AR Ao J = E ea σ = 1 o ρ ρ J = sE where s is conductivity of conductor. The above equation shows the relation between current density and electric field. 30. According to Bohr’s theory, the energy of an electron in hydrogen atom in an orbit is given by, En = −2π2 mK2 e4 , n is quantum number n2 h2 when an electron makes a transition from higher energy level (n2) to lower energy level (n1), the difference of energy appears in form of a photon. So, the energy of emitted photon is given by hn = En2 – En1 = 2π2 mK2 e4 > 1 − 1 h2 n12 n22H n = 2π2 mK2 e4 1 − 1 h3 > n12 n22H Sample Papers 69

1 = 2π2 mK2 e4 1 − 1 λ ch3 > n12 n22H or 1 = R> 1 − 1 R – Rydberg constant. λ n12 n22H,  Or According to the Rutherford’s nuclear model of an atom, most of the space in an atom is empty with a positively charged nucleus occupying only a small space at its centre. Thus, for most of the a-particles, the impact parameter being very large, they pass through the atom without being affected by the positively charged nucleus and suffer scattering at an angle less than 90°. Only a few gets scattered at q > 90°. Number of scattered particles detected 107 106 105 104 103 102 10 20 40 60 80 100 120 140 160 180 0 Scattering angle q (in degree) Conclusions: (i) The mass of the atom is concentrated in a small volume. (ii) The entire positive charge of the atom is with the nucleus and the negatively charged electrons are distributed around it. 31. (a) If A is area of capacitor plates and Q is the magnitude of charge on each plate, d we have potential difference between the capacitor plates that is given by t V = Emt + E0(d – t) Em Where Em is the electric field inside the dielectric slab and E0 is the electric field in the vacuum left. V = E0 t + E0(d – t) fa k= E0 p k Em \\ V = σ e t + (d – t)o ea E0 = σ o ...(i) ε0 k ε0 Charge on the capacitor plates is given by Q = sA ...(ii) We know, C = Q \\ V C = σ [t σAε0k t)] = kε0A + k (d – t + k (d – t) 70 Together with®  EAD Physics—12

(b) Net electric field inside a slab of conducting material is zero when kept in an external electric field. So, V = σ (d – t) Q ε0 and Q = sA \\ C = ε0A (d – t) (c) Total charge q = CV Total energy u = 1 CV2 2 As V is constant and CP > CS So the capacitors must be connected in parallel for storing greater charge and energy. Or I1 e1 r1 (a) I I2 e2 r2 V So, V = e1 – I1r1 ⇒ I1 = ε1 – V and r1 As \\ I2 = ε2 – V r2 I = I1 + I2 I = e ε1 – V o + e ε2 − V o r1 r1 r2 r2 I = e ε1 + ε2 o − Ve 1 + 1 o fa 1 = 1 + 1 r1 r2 r1 r2 rp r1 r2 p I = e ε1 + ε2 o – V × 1 r1 r2 rp Irp = e ε1 + ε2 orp – V r1 r2 ⇒ V = e ε1 + ε2 orp – Irp ...(i) r1 r2 V = eeq – Ireq ...(ii) Comparing above equations (i) and (ii), we get eeq = e ε1 + ε2 orp and r1 r2 \\ req = rp 1 = 1 + 1 rp r1 r2 Sample Papers 71

(b) As all three resistances are in parallel combination. So, 1 = 1 + 1 + 1   ⇒  rp = 1 and rp 1 2 2 2 Req = 1 +4 = 9 Ω 2 2 2 3 6 1 1 eeq = e 1 + 2 − 2o × 2 = 4 \\ I = εeq = 1 × 2 = 1 A Req 4 9 18 32. (a) When a capacitor is discharged through an inductor, the current in the inductor establishes a magnetic field. This magnetic field decreases and thereby, inducing an emf in the inductor in the same direction as that of the current. This charges the capacitor in the opposite sense to its initial polarity. In the absence of energy loses, current oscillates back and forth indefinitely. The process is called electrical oscillations. The frequency of oscillations is determined by the values of L and C, i.e. n = 1 1 2π LC (b) Frequency of radio waves is given by n = c = 3 × 108 ms−1 = 8.33 × 105 s–1 [Q  l = 360 m] λ 360 m For resonance L = 1 2C = (2 3.14 8.33 1 1.20 # 10−6 [\\  C = 1.20 mF] (2πν) # 105)2 # # # = 3.043 × 10–8 H Or When two identical charges are in motion, forces experienced by them are:  The electric force which exists whether the particles are in motion or at rest.  The magnetic force which exists only when both charged particles are in motion. \\  Electric force, Fe = 1 q1q2 ...(i) 4πε0 r2 If charges q1 and q2 are in motion in the same direction perpendicular to the line joining two charges with velocity v1 and v2 respectively, then a magnetic force acts between them. Magnetic field at the location of charge q2 due to motion of charge q1 is q1 µ0 i ∆l sin θ µ0 ∆t (v1 ∆t sin 90°) 4π r2 4π B1 = = r2 B1 = µ0 q1v1 4π r2 Due to this field, Lorentz (magnetic) force acting on charge q2 in motion with velocity v2 is given by Fm = q2B1v2 = µ0 q1q2v1v2 4π r2 72 Together with®  EAD Physics—12

If both the charges are moving with same velocity (v1 = v2 = v), then magnetic force is given by q1q2v2 Fm = µ0 r2 ...(ii) 4π Dividing equation (ii) by equation (i), we get Fm = µ0 q1q2v2 # 4πε0 # r2 Fe 4π r2 q1q2 Now, Fm = m0e0v2 Fe µ0 4π 10−7 NA−2 m0e0 = µ0 ε0 # = µ0 # 4πε0 = 4π = 9 # 109 Nm2 C−2 4π 4π 1 f 4πε0 p m0e0 = 9 1016 1 = 1 mC−1) 2 = 1 A2m2C−2 (3 # 108 A (3 # 108 ms−1)2 # i.e. m0e0 = 1 c2 Fm So, Fe = v2 = b v 2 c2 c l As v << c, the magnetic force Fm is very small in comparison to electric force Fe. 33. If the distance between object and image needle is kept more than 4f (where f is focal length of lens), then by placing lens at two positions, L1 and L2, the real and inverted images of needle are obtained. Since object needle and image needle are conjugate to each other so, AL2 = A2L1 = u u′ v′ If L1L2 = x and AA1 = D, then L2 D = 2u + x ⇒ u = D− x A2 L1 A 2 A1 v For position L2 of the lens D− x x AL2 = u = 2 D u and L2A2 = v = D – u v = D – e D– x o = D+ x 2 2 From lens formula, 1 = 1 – 1 f v u 1 = 1 – 1 = 2 + 2 f D+ D– x e D+ x o –e D– x o x 2 2 \\ 1 = 4D x2 f D2 – \\ If x = 0, D = 4f ; f = D2 – x2 4D Sample Papers 73

When lens is at position L2  , AL2 = u, L2A2 = v If linear size of an object and an image is I0 and I1 respectively, then I1 m1 = I0 =– v ....(i) u When lens is in position L1, then I2 m2 = I0 =– vl ul where A1L1 = v′ = u AL1 = u′ =v m2 = – u v ....(ii) From equations (i) and (ii), I1 = I0 =fa m1 II=10 and m2 I2 p I0 I2 I0 I0 = I1× I2 Or (a) The basic conditions for obtaining sustained and well defined interference fringes on the screen are: (i) two interference sources should be coherent, i.e. they should have constant phase difference. (ii) the sources should emit light of same frequency or wavelength, i.e. it should not change with time. (iii) the amplitude of the interfering waves should be equal or nearly equal. (iv) the separation between the sources should be small. (v) width of the sources should be comparable to the wavelength of light. (vi) distance of the source from the screen should be large. (b) We know For maximum interference, d sin q = nl; n = 0, 1, 2, ... For minimum interference, d sin q = (2n + 1) λ 2 For 1st bright fringe n = 1 sin q = λ d When angle is small then q= λ radians d Let q1 and q2 be angular widths at wavelengths l1 and l2, then q1 = λ1 , q2 = λ2 d d \\ q1 = l1   ⇒  l2 = θ2 × λ1 q2 l2 θ1 Given: l1 = 5890 Å; q1 = 0.20° ; q2 = 110 × 0.20 = 0.22° 100 \\ l2 = 5890 × 0.22 = 6479 Å 0.20 74 Together with®  EAD Physics—12