Xam Idea Physics - Class 12 Term 1 and 2 Question Bank

Ans. (i) (a) XL=wL = 2πnL; graph XL of ν and ν is a straight line (b) XC = 1 = 1 , graph of XC and n is a rectangular hyperbola as shown in fig. ~C 2rνC νν (ii) Yes; because V = V 2 + (VC –VL) 2; R As VC and VL have opposite faces, VC or VL may be greater than V. The situation may be as shown in figure where VC>V. Long Answer Questions [5 marks] Q. 1. Explain the term inductive reactance. Show graphically the variation of inductive reactance with frequency of the applied alternating voltage. An ac voltage V = V0 sin ωt is applied across a pure inductor of inductance L. Find an expression for the current i, flowing in the circuit and show mathematically that the current r flowing through it lags behind the applied voltage by a phase angle of 2 . Also draw (i) phasor diagram (ii) graphs of V and i versus ωt for the circuit. [CBSE East 2016] Ans. Inductive Reactance: The opposition offered by an inductor to the flow of alternating current through it is called the inductive reactance. It is denoted by XL. Its value is XL. = ωL=2πfL where L is inductance and f is the frequency of the applied voltage. Obviously XL ∝ f Thus, the graph between XL and frequency f is linear (as shown in fig.). Phase Difference between Current and Applied Voltage in Purely Inductive circuit : AC circuit containing pure inductance: Consider a coil of self-inductance L and negligible ohmic resistance. An alternating potential difference is applied across its ends. The magnitude and direction of ac changes periodically, due to which there is a continual change in magnetic flux linked with the coil. Therefore according to Faraday’s law, an induced emf is produced in the coil, which opposes the applied voltage. As a result the current in the circuit is reduced. That is inductance acts like a resistance in ac circuit. The instantaneous value of alternating voltage applied V = V0 sin ωt ...(i) If i is the instantaneous current in the circuit and di the rate of dt change of current in the circuit at that instant, then instantaneous induced emf di dt f = –L According to Kirchhoff ’s loop rule V+f = 0 & V–L di = 0 dt 298 Xam idea Physics–XII

or V = L di or di = V dt dt L or di = V0sin~t or di = V0sin ~ t dt dt L L Integrating with respect to time ‘t’, i = V0 y sin ~t dt = V0 &– cos ~t 0 = – V0 cos ~t = – V0 sin a r –~tk L L ~ ~L ~L 2 or i = V0 sin a~t– r k …(ii) ~L 2 This is required expression for current or i = i0sin a~t– r k ...(iii) where 2 V0 i0 = ~L ...(iv) is the peak value of alternating current Also comparing (i) and (iii), we note that current lags behind the applied voltage by an angle r (Fig. b). 2 Phasor diagram: The phasor diagram of circuit containing pure inductance is shown in Fig. (b). Graphs of V and I versus ωt for this circuit is shown in fig. (c). Q. 2. Derive an expression for impedance of an ac circuit consisting of an inductor and a resistor.          [CBSE Delhi 2008] Ans. Let a circuit contain a resistor of resistance R and an inductor of inductance L connected in series. The applied voltage is V =V0 sin ωt. Suppose the voltage across resistor VR and that across inductor is VL . The voltage VthReacnudrrceunrtrebnytaInaraenginleth2re same phase, while the voltage VL leads . Thus, VR and VL are mutually perpendicular. The resultant of VR and VL is the applied voltage i.e., V= V 2 + V 2 But R L VR = RI, VL = XL I=ωLI ∴ where XL = ωL is inductive reactance ∴ V = (RI) 2 + (XL I)2 ∴ V Impedance, Z= I = R2 + X 2 & Z = R2 + (~L)2 L Alternating Current 299

Q. 3. (a) What is impedance? (b) A series LCR circuit is connected to an ac source having voltage V = V0 sin ωt . Derive expression for the impedance, instantaneous current and its phase relationship to the applied voltage. Find the expression for resonant frequency. [CBSE Delhi 2010] OR (a) An ac source of voltage V = V0 sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called? (b) In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. P1 Calculate P2 . [CBSE Delhi 2016] Ans. Impedance: The opposition offered by the combination of a resistor and reactive component to the flow of ac is called impedance. Mathematically it is the ratio of rms voltage applied and rms current produced in circuit i.e., Z = V . Its unit is ohm (Ω). I Expression for Impedance in LCR series circuit: Suppose resistance R, inductance L and capacitance C are connected in series and an alternating source of voltage V = V0 sin ωt is applied across it (fig. a). On account of being in series, the current (i) flowing through all of them is the same. Suppose the voltage across resistance R is VR voltage across inductance L is VL and voltage across capacitance C is VC. The voltage VR and current i are in the same phase, the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig. b). Clearly VC and VL are in opposite directions, therefore their resultant potential difference = VC –VL (if VC > VL). Thus VR and (VC – VL) are mutually perpendicular and the phase difference between them is 90°. As applied voltage across the circuit is V, the resultant of VR and (VC –VL) will also be V. From fig. V2 =V 2 + (VC – VL) 2 & V= V 2 + (VC – VL) 2 ...(i) R R But VR =R i , VC =XC i and VL = XL i ...(ii) 1 where XC = ~C = capacitance reactance and XL = ωL= inductive reactance V = (Ri)2 + (XC i – XL i)2 Impedance of circuit, Z= V = R2 + (XC –XL)2 i i.e., Z = R2 + (XC – XL)2 = R 2 + c 1 –~L 2 ~C m Instantaneous current I = V0 sin (~t + z) R 2 + c 1 – 2 ~C wL m 300 Xam idea Physics–XII

The phase difference (φ) between current and voltage is given by, tan z = XC –XL Resonant Frequency: For resonance φ= 0, so XC –XL =0 R 1 = ~L & ~2 = 1 ~C LC 1 ∴ Resonant frequency ~r = LC Phase difference (φ) in series LCR circuit is given by tan z = VC –VL = im (XC – XL) = (XC –XL) VR im R R When current and voltage are in phase φ=0 ⇒ XC –XL =0 ⇒ XC = XL This condition is called resonance and the circuit is called resonant circuit. Case I: XL = R ∴ Z= R2 + X 2 = R2 + R2 = 2R L R R =1 Power factor, P1 = cosz = Z = 2R 2 Case II: XL = XC ∴ Z = R2 + (XL –XC) 2 = R2 = R Power factor, P2 = R = R =1 Z R ∴ P1 = 1 P2 2 Q. 4. A device ‘X’ is connected to an ac source V = V0 sin ωt. The variation of voltage, current and power in one cycle is show in the following graph: Y A C B O ωt π 2π (a) Identify the device ‘X’. (b) Which of the curves, A, B and C represent the voltage, current and the power consumed in the circuit? Justify your answer. (c) How does its impedance vary with frequency of the ac source? Show graphically. (d) Obtain an expression for the current in the circuit and its phase relation with ac voltage. Ans. (a) The device ‘X’ is a capacitor. (b) Curve B : Voltage Curve C : Current Curve A : Power consumed in the circuit r 2 Reason : This is because current leads the voltage in phase by for a capacitor. Alternating Current 301

(c) Impedance: XC = 1 = 1 ⇒ ~C 2rνC XC \\ 1 ν XC (d) Voltage applied to the circuit is V = V0 sin ωt Due to this voltage, a charge will be produced which will charge the plates of the capacitor with positive and υ negative charges. V = Q & Q = CV C C Therefore, the instantaneous value of the current in the circuit is I = dQ = d (CV) = d (CV0 sin ~t) ∴ dt dt dt I = ~CV0 cos ~t = V0 sin c ~t + r m 1 2 ~C I = I0 sin c~t + r m 2 where, I0 = V0 = Peak value of current 1 ~C Hence, current leads the voltage in phase by r . 2 Q. 5. (a) State the condition for resonance to occur in series LCR ac circuit and derive an expression for resonant frequency. [CBSE Delhi 2010] (b) Draw a plot showing the variation of the peak current (im) with frequency of the ac source used. Define the quality factor Q of the circuit. Ans. (a) Condition for resonance to occur in series LCR ac circuit: For resonance the current produced in the circuit and emf applied must always be in the same phase. Phase difference (φ) in series LCR circuit is given by tan φ = XC –XL R For resonance φ =0 ⇒ XC – XL = 0 or XC = XL If ωr is resonant frequency, then XC = 1 ~rC and XC = ωr L ~1r C = ~r L & ~r = 1 LC Linear resonant frequency, or = ~r = 1 2r 2r LC 302 Xam idea Physics–XII

(b) The graph of variation of peak current im with frequency is shown in fig. Half power frequencies are the frequencies on either side of resonant frequency for which current reduces to half of its maximum value. In fig., are half power frequencies. ν1 and ν2 Quality Factor (Q): The quality factor is defined as the ratio of νν ν resonant frequency to the width of half power frequencies. ν i.e., Q = ~r = νr = ~rL ~2 – ~1 ν2 – ν1 R Q. 6. (a) An alternating voltage V = Vm sin ω applied to a series LCR circuit drives a current given by i = im sin (ωt +φ) . Deduce an expression for the average power dissipated over a cycle. (b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. [CBSE (F) 2011] OR A voltage V = V0 sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit? [CBSE (AI) 2014] Ans. (a) V = Vm sin ωt and i = im sin (ωt+φ) and instantaneous power, P =Vi = Vm sin ωt . im sin (ωt+ φ)         =Vm im sin ωt sin (ωt+ φ) = 1 Vm im 2 sin ~t. sin (~t + z) 2 From trigonometric formula 2 sin A sin B =cos (A – B) – cos (A+B) ∴ Instantaneous power, P = 1 Vm im [cos (~t – ~t – z) – cos (~t + z+ ~t)] 2 = 1 Vm im [cos z – cos (2~t + z)]  … (i) 2 Average power for complete cycle P = 1 Vm im [cos z – cos (2~t + z)] 2 where cos (~t + z) is the mean value of cos (2ωt+ φ) over complete cycle. But for a complete cycle, cos (2ωt+ φ) = 0 ∴ Average power, P= 1 Vm im cos z = V0 i0 cos z 2 2 2 P = Vrms irms cos z (i) If phase angle φ =90° (resistance R is not used in the circuit) then no power dissipated. (ii) If phase angle φ =0° or circuit is pure resistive (or XL=XC) at resonance then Max power P = Vrms × Irms = V0 I0 2 (b) The power is P=Vrms Irms cos φ. If cos φ is small, then current considerably increases when voltage is constant. Power loss, we know is I2R. Hence, power loss increases. Alternating Current 303

Q. 7. Explain with the help of a labelled diagram, the principle and working of an ac generator. Write the expression for the emf generated in the coil in terms of speed of rotation. Can the current produced by an ac generator be measured with a moving coil galvanometer? OR Describe briefly, with the help of a labelled diagram, the basic elements of an ac generator. State its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf induced in the rotating loop. [CBSE Delhi 2010] OR State the working of ac generator with the help of a labelled diagram. The coil of an ac generator having N turns, each of area A, is rotated with a constant angular velocity ω. Deduce the expression for the alternating emf generated in the coil. What is the source of energy generation in this device? [CBSE (AI) 2011] Ans. AC generator: A dynamo or generator is a device which converts mechanical energy into electrical energy. Principle: It works on the principle of electromagnetic induction. When a coil rotates continuously in a magnetic field, the effective area of the coil linked normally with the magnetic field lines, changes continuously with time. This variation of magnetic flux with time results in the production of an alternating emf in the coil. Construction: It consists of the four main parts: (i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet. (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. It can revolve round an axle between the two poles of the field magnet. The drum or ring serves the two purposes: (a) It serves as a support to coils and (b) It increases the magnetic field due to air core being replaced by an iron core. (iii) Slip Rings: The slip rings R1 and R2 are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature. (iv) Brushes: These are two flexible metal plates or carbon rods (B1 and B2) which are fixed and constantly touch the revolving rings. The output current in external load RL is taken through these brushes. Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in clockwise direction, the wire ab moves downward and cd upward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B1 RLB2. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves upward and cd downward, so the direction of current is reversed and in external circuit it flows along B2 RLB1. Thus the direction of induced emf and current changes in the external circuit after each half revolution. Expression for Induced emf: When the coil is rotated with a constant angular speed ω , the angle θ between the magnetic field vector B and the area vector A of the coil at any instant 304 Xam idea Physics–XII

t is θ = ωt (assuming θ = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, the flux at any time t is φB= BA cos θ = BA cos ωt From Faraday’s law, the induced emf for the rotating coil of N turns is then, f = –N dz B = –NBA d (cos ~t) dt dt Thus, the instantaneous value of the emf is ε = NBA ω sin ωt where NBAω=2πυNBA is the maximum value of the emf, which occurs when sin ωt = ±1. If we denote NBAω as ε0, then ε= ε0 sin ωt ⇒ ε = ε0 sin 2πnt where ν is the frequency of revolution of the generator’s coil. Obviously, the emf produced is alternating and hence the current is also alternating. Current produced by an ac generator cannot be measured by moving coil ammeter; because the average value of ac over full cycle is zero. The source of energy generation is the mechanical energy of rotation of armature coil. Q. 8. (a) Describe briefly, with the help of a labelled diagram, the working of a step up transformer. (b) Write any two sources of energy loss in a transformer. [CBSE (F) 2012] (c) A step up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain. [CBSE Delhi 2011, 2009] OR Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances? [CBSE (AI) 2010, (East) 2016] Ans. (a) Transformer: A transformer converts low voltage into high voltage ac and vice-versa. Construction: It consists of laminated core of soft iron, on which two coils of insulated copper wire are separately wound. These coils are kept insulated from each other and from the iron-core, but are coupled through mutual induction. The number of turns in these coils are different. Out of these coils one coil is called primary coil and other is called the secondary coil. The terminals of primary coils are connected to ac mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. Transformers are of two types: ac mains ac mains Laminated Laminated iron core iron core Alternating Current 305

1. Step up Transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil (i.e., NS>NP). 2. Step down Transformer: It transforms the alternating high voltage to alternating low voltage and in this the number of turns in secondary coil is less than that in primary coil (i.e., NS<NP). Working: When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil; due to which the magnetic flux linked with the secondary coil changes continuously, therefore the alternating emf of same frequency is developed across the secondary. Let NP be the number of turns in primary coil, NS the number of turns in secondary coil and φ the magnetic flux linked with each turn. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. According to Faraday’s laws the emf induced in the primary coil fP = –NP Tz ...(i) Tt ...(ii) and emf induced in the secondary coil fS = –NS Tz Tt From (i) and (ii) fS = NS ...(iii) fP NP If the resistance of primary coil is negligible, the emf (εP) induced in the primary coil, will be equal to the applied potential difference (VP) across its ends. Similarly if the secondary circuit is open, then the potential differenceVS across its ends will be equal to the emf (εS) induced in it; therefore VS = fS = NS = r(say) ...(iv) VP fP NP where r= NS is called the transformation ratio. If iP and iS are the instantaneous currents NP in primary and secondary coils and there is no loss of energy; then For about 100% efficiency, Power in primary =Power in secondary VP iP =VS iS iiPS = VP = NP = 1 ...(v) VS NS r In step up transformer, NS > NP → r > 1; So VS > VP and iS < iP i.e., step up transformer increases the voltage, but decreases the current. In step down transformer, NS < NP → r < 1 so VS < VP and iS > iP i.e., step down transformer decreases the voltage, but increases the current. Laminated core: The core of a transformer is laminated to reduce the energy losses due to eddy currents, so that its efficiency may remain nearly 100%. In a transformer with 100% efficiency (say), Input power = output power VP IP =VS IS 306 Xam idea Physics–XII

(b) The sources of energy loss in a transformer are (i) eddy current losses due to iron core (ii) flux leakage losses. (iii) copper losses due to heating up of copper wires (iv) hysteresis losses due to magnetisation and demagnetisation of core. (c) When output voltage increases, the output current automatically decreases to keep the power same. Thus, there is no violation of conservation of energy in a step up transformer. Q. 9. With the help of a diagram, explain the principle of a device which changes a low voltage into a high voltage but does not violate the law of conservation of energy. Give any one reason why the device may not be 100% efficient. [CBSE Sample Paper 2018] Ans. Transformer changes a low voltage into a high voltage without voilating the law of conservation of energy. Principle: When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil; due to which the magnetic flux linked with the secondary coil changes continuously, therefore the alternating emf of same frequency is developed across the secondary. The device may not be 100% efficient due to following energy losses in a transformer: (i) Joule Heating: Energy is lost due to heating of primary and secondary windings as heat (I2Rt). (ii) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil. Q. 10. (a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device. (b) A small town with a demand of 1200 kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat. [CBSE 2019 (55/1/1)] Ans. (a) Refer to Q. 8, Page no. 305. (b) Demand of electric power = 1200 kW Distance of town from power station = 20 km Two wire = 20 × 2 = 40 km Total resistance of line = 40 × 0.5 = 20 Ω The town gets a power of 4000 volts Power = voltage × current I = 1200 ×103 = 1200 = 300 A 4000 4 The line power loss in the form of heat = I2 × R = (300)2 × 20 = 9000 × 20 = 1800 kW Alternating Current 307

Q. 11. A 2 µF capacitor, 100 W resistor and 8 H inductor are connected in series with an ac source. (i) What should be the frequency of the source such that current drawn in the circuit is maximum? What is this frequency called? (ii) If the peak value of emf of the source is 200 V, find the maximum current. (iii) Draw a graph showing variation of amplitude of circuit current with changing frequency of applied voltage in a series LRC circuit for two different values of resistance R1 and R2 (R1 > R2). (iv) Define the term ‘Sharpness of Resonance’. Under what condition, does a circuit become more selective? [CBSE (F) 2016] Ans. (i) For maximum frequency ~L = 1 ~C ⇒ 2rν ×8 = 2ro # 1 & (2rν) 2 = 16 1 2 ×10 –6 # 10–6 ⇒ 2rν = 1 & 2rν = 103 4 # 10–3 4 ⇒ ν = 250 = 39.80 s–1 2r This frequency is called resonance frequency. (ii) Maximum current, I0 = E0 = 200 =2A [E0 maximum emf] R 100 (iii) (iv) ~0 is measure of sharpness of resonance, where w0 is the resonant frequency and 2∆w is 2D~ the bandwidth. Circuit is more selective if it has greater value of sharpness. The circuit should have smaller bandwidth ∆w. Q. 12. (i) Draw a labelled diagram of ac generator. Derive the expression for the instantaneous value of the emf induced in the coil. (ii) A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s–1 in a uniform magnetic field of magnitude 3.0 × 10–2 T. Calculate the maximum value of the current in the coil. [CBSE Delhi 2017] Ans. (i) Refer to Q. 7, page 304. (ii) Given, N = 20 A = 200 cm2 = 200 × 10–4 m2 B = 3.0 × 10–2 T ω = 50 rad s–1 308 Xam idea Physics–XII

EMF induced in the coil ε = NBAω sin ωt Maximum emf induced εmax = NBAω = 20 × 3.0 × 10–2 × 200 × 10–4 × 50 = 600 mV Maximum value of current induced Imax = fmax = 600 mA R R Q. 13. (i) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils. (ii) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. [CBSE Delhi 2017] Ans. (i) Refer to Q. 8, Page 305. (ii) Given, VP = 2200 V NP = 3000 turns We have, VS = 220 V VS = NS VP NP NS = VS # NP VP = 220 # 3000 2200 NS = 300 turns Q. 14. (a) What do you understand by ‘sharpness of resonance’ for a series LCR resonant circuit? How is it related with the quality factor ‘Q’ of the circuit? Using the graphs given in the diagram, explain the factors which affect it. For which graph is the resistance (R) minimum? [CBSE 2019 (55/4/1)] R1 C B R2 I R3 A ω = ωr ω (b) A 2 µF capacitor , 100 Ω resistor and 8 H inductor are connected in series with an ac source. Find the frequency of the ac source for which the current drawn in the circuit is maximum. If the peak value of emf of the source is 200 V, calculate the (i) maximum current, and (ii) inductive and capacitive reactance of the circuit at resonance. Alternating Current 309

Ans. (a) The circuit would be set to have a high sharpness of resonance, if the current in the circuit drops rapidly as the frequency of the applied ac source shifts from its resonant value. Sharpness of resonance is measured by the quality factor Q = 1 L R C Sharpness of resonance for given value of L and C or value of ~r depends on R. R is minimum for C. (b) ν = 1 2r LC =1 2×3.14 8×2×10–6 = 1000 = 39.81 or 40 Hz (approximately) 8 × 3.14      V0 = 200 V (i) i0 = V0 = V0 (a Z = R at resonance) Z R = 200 =2 A 100 (ii) At resonance XL = XC XL = wL = 2pnL = 2p × 39.81 × 8 = 2000 X Self-Assessment Test Time allowed: 1 hour Max. marks: 30 1. Choose and write the correct option in the following questions. (3 × 1 = 3) (i) The average power dissipation in a pure capacitance is: (a) 1 CV2 (b) CV2 2 (d) zero 1 (c) 4 CV2 (ii) In an ac series circuit, the instantaneous current is maximum when the instantaneous voltage is maximum. The circuit element connected to the source will be (a) pure inductor (b) pure capacitor (c) pure resistor (d) combination of a capacitor and an inductor (iii) R, L and C represent the physical quantities resistance, inductance and capacitance respectively. Which one of the following combinations has dimension of frequency? (a) 1 (b) R RC L (c) 1 (d) C LC L 310 Xam idea Physics–XII

2. Fill in the blanks. (2 × 1 = 2) (i) One complete set of positive and negative values of alternating current or emf is called ______________. (ii) The core of transformer, if laminated, ______________ eddy currents. 3. The power factor of an ac circuit is 0.5. What is the phase difference between voltage and current in this circuit? 1 4. Draw a graph to show variation of capacitive-reactance with frequency in an ac circuit. 1 5. A device ‘X’ is connected to an ac source V = V0. The variation of voltage, current and power in one complete cycle is shown in the following figure. (i) Which curve shows power consumption over a full cycle? (ii) Identify the device ‘X’. 1 6. Prove that an ideal capacitor, in an ac circuit does not dissipate power. 2 7. Derive an expression for the impedance of an ac circuit consisting of an inductor and a resistor. 2 8. A 15.0 µF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current. 2 9. How much current is drawn by the primary coil of a transformer which steps down 220 V to 22 V to operate a device with an impedance of 220 Ω? 2 10. You are given three circuit elements X, Y and Z. When the element X is connected across an ac source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across the source, voltage is ahead of the current in phase by p/4. But the current is ahead of the voltage in phase by p/4 when Z is connected in series with X across the source. Identify the circuit elements X, Y and Z. When all the three elements are connected in series across the same source, determine the impedance of the circuit. Draw a plot of the current versus the frequency of applied source and mention the significance of this plot. 3 11. A voltage v = vm sin ~t applied to a series LCR circuit, drives a current in the circuit given i = im sin (~t + z) . Deduce the expression for the instantaneous power supplied by the source. Hence, obtain the expression for the average power. Define the terms ‘power factor’ and ‘wattless current’, giving the examples where power factor is maximum and the circuit where there is wattless current. 3 Alternating Current 311

12. A series LCR circuit with L = 4.0 H, C = 100 µF and R = 60 Ω is connected to a variable frequency 240 V source as shown in figure. 3 Calculate: (i) The angular frequency of the source which drives the circuit at resonance; (ii) The current at the resonating frequency; (iii) The rms potential drop across the inductor at resonance. 13. (a) Using phasor diagram for a series LCR circuit connected to an ac source of voltage v = v0 sin wt, derive the relation for the current flowing in the circuit and the phase angle between the voltage across the resistor and the net voltage in the circuit. (b) Draw a plot showing the variation of the current I as a function of angular frequency ‘w’ of the applied ac source for the two cases of a series combination of (i) inductance L1, capacitance C1 and resistance R1 and (ii) inductance L2, capacitance C2 and resistance R2 where R2 > R1. Write the relation between L1, C1 and L2, C2 at resonance. Which one, of the two, would be better suited for fine tuning in a receiver set? Give reason. 5 Answers 1. (i) (d) (ii) (c) (iii) (b) 2. (i) cycle (ii) decreases 9. 0.1 A, 0.01 A 12. (i) ω = 50 rad/s; (ii) I = 4 A; (iii) VL = 800 V zzz 312 Xam idea Physics–XII

Electromagnetic Chapter –8 Waves 1. Need for Displacement Current Ampere’s circuital law for conduction current during charging of a capacitor was found inconsistent. Therefore, Maxwell modified Ampere’s circuital law by introducing displacement current. It is dz E given by Id = f0 dt Modified Ampere’s circuital law is: y \" . d \" = n0 dI + f0 dz E n dt B l where φE = electric flux. 2. Electromagnetic Waves The waves propagating in space through electric and magnetic fields varying in space and time simultaneously are called electromagnetic waves. The electromagnetic waves are produced by an accelerated or decelerated charge or LC circuit. The frequency of EM waves is ν = 1 2r LC 3. Characteristics of Electromagnetic Waves (i) The electromagnetic waves travel in free-space with the speed of light (c = 3 × 108 m/s) irrespective of their wavelength. (ii) Electromagnetic waves are neutral, so they are not deflected by electric and magnetic fields. (iii) The electromagnetic waves show properties of reflection, refraction, interference, diffraction and polarisation. (iv) In electromagnetic wave the electric and magnetic fields are always in the same phase. (v) The ratio of magnitudes of electric and magnetic field vectors in free space is constant equal to c. E = 1 = c = 3 # 108 m/s B n0 f0 (vi) The speed of electromagnetic waves in a material medium is given by v= 1= c = c , where n is the refractive index. nf nrfr n (vii) In an electromagnetic wave the energy is propagated by means of electric and magnetic field vectors in the direction of propagation of wave. [Note : We also use µ for refractive index] Electromagnetic Waves 313

(viii) In electromagnetic wave the average values of electric energy density and magnetic energy density are equal c 1 f0 E2 m = e B2 o 2 2n0 av av (ix) The electric vector of electromagnetic wave is responsible for optical effects and is also called the light vector. (x) Electromagnetic waves carry energy and momentun E= hc , p = U = mc 4. Transverse Nature of Electromagnetic Waves m c The electromagnetic waves are transverse in nature. In electromagnetic waves the electric and magnetic fields are mutually perpendicular and also perpendicular to the direction of wave propagation, such that E , B and K form a right handed set ( K is propagation vector along the direction of propagation). 5. Electromagnetic Spectrum The electromagnetic waves have a continuous wavelength starting from short gamma rays to long radiowaves. The orderly distribution of wavelength of EM waves is called the electromagnetic spectrum. The complete spectrum is given in the following table: S. No. Name Wavelength Range (m) Frequency Range (Hz) i. Gamma rays 10–13 – 10–10 3 × 1021 – 3 × 1018 ii. X-rays 10–10 – 10–8 3 × 1018 – 3 × 1016 iii. Ultraviolet rays 10–8 – 4×10–7 3 × 1016 – 7.5 × 1014 iv. Visible light 4 × 10–7 – 7.5 × 10–7 7.5 × 1014 – 4 × 1014 v. Infra red light 7.5 × 10–7 – 10–3 4 × 1014 – 3 × 1011 vi. Microwaves 10–3 – 10–1 3 × 1011 – 1010 vii. Radio waves 10–1 – 104 1010 – 3 × 104 6. Wavelength Range of Visible Spectrum Visible light has a continuous wavelength starting from 400 nm to 750 nm; for convenience it is divided into 7 colours. V Violet 400 nm — 420 nm I Indigo 420 nm — 450 nm B Blue 450 nm — 500 nm G Green 500 nm — 570 nm Y Yellow 570 nm — 600 nm O Orange 600 nm — 650 nm R Red 650 nm — 750 nm 7. Uses of Electromagnetic Spectrum (i) γ-rays are highly penetrating, they can penetrate thick iron blocks. Due to high energy, they are used to initiate some nuclear reactions. γ-rays are produced in nuclear reactions. In medicine, they are used to destroy cancer cells. (ii) X-rays are used in medical diagnostics to detect fractures in bones, tuberculosis of lungs, presence of stone in gallbladder and kidney. They are used in engineering to check flaws in bridges. In physics X-rays are used to study crystal structure. (iii) Ultraviolet rays provide vitamin D. These are harmful for skin and eyes. They are used to sterilise drinking water and surgical instruments. They are used to detect invisible writing, forged documents, finger prints in forensic lab and to preserve food items. 314 Xam idea Physics–XII

(iv) Infrared rays are produced by hot bodies and molecules. These waves are used for long distance photography and for therapeutic purposes. (v) Radiowaves are used for broadcasting programmes to distant places. According to frequency range, they are divided into following groups (1) Medium frequency band or medium waves 0·3 to 3 MHz (2) Short waves or short frequency band 3 MHz — 30 MHz (3) Very high frequency (VHF) band 30 MHz to 300 MHz (4) Ultrahigh frequency (UHF) band 300 MHz to 3000 MHz (vi) Microwaves are produced by special vacuum tubes, namely; klystrons, magnetrons and gunn diodes. Their frequency range is 3 GHz to 300 GHz. They are used in RADAR systems for aircraft navigation and microwave used in homes. Selected NCERT Textbook Questions Q. 1. Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff ’s first rule function rule valid at each plate of the capacitor? Explain. Ans. Here, I = 0.15 A r = 12 cm = 12 × 10–2 m d = 5.0 mm = 5 × 10–3 m A = rr2 (a) Capacitance C = f0 A = f0 rr2 d d   = 8.85×10–12 × 22×(12×10–2)2 7× 5×10–3   = 28036.8×10–16 = 801.05 × 10–13 F 35×10–3   = 80.1×10–12 F   = 80.1 pF Let C be the capacitance of capacitor and q the instantaneous charge on plates, then q = CV ∴ dq = C dV    ⇒      dV = I dt dt dt C ∴ = 0.15 80.1×10–12 i.e., = 0.00187 × 1012 Vs–1 ∴ = 1.87 × 109 Vs–1 Electromagnetic Waves 315

(b) Displacement current Id = f0 A dE = f0 A I =I = conduction current = 0.15 A. dt f0 A (c) Yes, Kirchhoff ’s law holds at each plate of capacitor since displacement current is equal to conduction current. Q. 2. A parallel plate capacitor (fig.) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an angular frequency of 300 rad/s. (a) What is the rms value of the conduction current? (b) Is conduction current equal to the displacement current? (c) Determine the amplitude of magnetic field induction B at a point 3.0 cm from the axis between the plates. Ans. Given R = 6.0 cm, C=100 pF = 1 × 10–10 F, w = 300 rad/s, Vrms = 230 V 1 (a) Impedance of circuit Z = capacitance reactance XC = ~C Root mean square current, Irms = Vrms = Vrms # ~C Z = 230 × 300×10–10 = 6.9 ×10–6 A = 6.9 µA (b) Yes, the conduction current is equal to the displacement current. (c) The whole space between the plates occupies displacement current which is equal in magnitude to the conduction current. Magnetic field B= n0 Ir 2rR2 Here r = 3 cm = 3×10–2 m, R = 6 cm = 6 × 10–2 m Amplitude of displacement current = Peak value of conduction current = I0 = Irms 2 Amplitude of magnetic field B = n0 I0 r = n0 Irms 2 r 2rR2 2rR2 = 4r×10–7 # 6.9 # 10–6 # 1.41 # (3 # 10–2) 2r× (6×10–2)2 = 1.63 × 10–11 T Q. 3. A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the directions of electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength ? \"\" Ans. In an electromagnetic wave’s propagation, vector K , electric field vector E and magnetic field \" vector B form a right handed system. As the propagation vector is along Z-direction, electric field vector will be along X-direction and magnetic field vector will be along Y-direction. Frequency ν = 30 MHz= 30 × 106 Hz Speed of light, c = 3 ×108 ms–1 Wavelength, m = c = 3 # 108 = 10 m ν 30 # 106 316 Xam idea Physics–XII

Q. 4. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band ? Ans. Speed of wave c = 3 × 108 ms–1 When frequency ν1 = 7.5 MHz = 7.5 × 106 Hz, Wavelength m1 = c = 3 # 108 = 40 m ν1 7.5 # 106 When frequency ν2 = 12 MHz, wavelength m2 = c = 3 # 108 = 25 m ν2 12 # 106 Wavelength band is from 25 m to 40 m. Q. 5. The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave ? Ans. The relation between magnitudes of magnetic and electric field vectors in vacuum is E0 = c & E0 = B0 c B0 Here, B0 = 510 nT = 510×10–9 T, c = 3×108 ms–1 E0 = 510 # 10–9 ×3×108 = 153 N/C. Q. 6. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/\"C and t\"hat its frequency ν = 50.0 MHz. (a) Determine B0, w, k and λ (b) Find expressions for E and B. Ans. (a) We have E0 = c & B0 = E0 = 120 = 4 # 10–7 T B0 c 3 # 108 ~ = 2rν = 2 # 3.14 # 50 # 106 = 3.14 # 108 rads–1 k = ~ = 3.14 # 108 = 1.05 radm–1 c 3 # 108 Wavelength, m = c = 3 # 108 = 6.00 m. ν 50.0 # 106 (b) If wave is propagating along X-axis, electric field will be along Y-axis and magnetic field along Z-axis. \" E = E0 sin (kx – ~t)tj where x is in m and t in s ⇒ \" = 120 sin (1.05 x – 3.14×108 t) tj N/C E \" = B0 sin (kx – ~t) kt B = (4×10–7) sin (1.05 x – 3.14×108 t) kt tesla. Q. 7. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 Vm–1. (a) What is the wavelength of a wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the electric field equals the average energy density of the B field. [c = 3 × 108 ms–1] Ans. (a) Wavelength m = c = 3 # 108 = 1.5 # 10–2 m ν 2 # 1010 (b) B0 = E0 = 48 = 1.6 # 10–7 tesla c 3 # 108 Electromagnetic Waves 317

(c) Energy density of electric field is UE = 1 f0 E2 …(i) 2 Energy density of Magnetic field UB = 1 B2 …(ii) 2n0 where e0 is permittivity of free space and m0 is permeability of free space We have, E = cB ...(iii) ∴ UE = 1 f0 (cB)2 2 = c2 d 1 f0 B2 n 2 But c= 1 n0 f0 ∴ UE = 1 d 1 f0 B2 n n0 f0 2 = 1 B2 2n0 ∴ UE = UB Q. 8. Suppose that the electric field of an electromagnetic wave in vacuum is E ={(3.1 N/C) cos (1.8 rad/m) y + (5.4×106 rad/s) t} it (a) What is the direction of propagation? (b) What is the wavelength λ? (c) What is the frequency ν? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. Ans. (a) Wave is propagating along negative y-axis. (b) Standard equation of wave is \"E = E0 cos (ky + ~t) it Comparing the given equation with standard equation, we have E0= 3.1 N/C, k = 1.8 rad/m, w = 5.4 × 106 rad/s. Propagation constant k= 2r m m = 2r = 2 # 3.14 m = 3.49 m (c) We have k 1.8 w = 5.4 × 106 rad/s Frequency, ν= ~ = 5.4 # 106 Hz = 8.6×105 Hz 2r 2 # 3.14 (d) Amplitude of magnetic field, B0 = E0 = 3.1 = 1.03 # 10–8 T c 3 # 108 318 Xam idea Physics–XII

\"\"\" (e) The magnetic field is vibrating along Z-axis because K, E, B form a right handed system – tj × it × kt ∴ Expression for magnetic field is \" B = B0 cos (ky + ~t) kt = [1.03×10–8T cos {1.8 rad/m) y + (5.4×106 rad/s) t}] kt Q. 9. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotopically and neglect reflection. Ans. Power in visible radiation, P = 5 ×100 = 5 W 100 For a point source, Intensity I = P , where r is distance from the source. 4rr2 (a) When distance r = 1 m, I = 5 = 5 = 0.4 W/m2 4r (1)2 4×3.14 (b) When distance r = 10 m, I = 5 = 5 = 0.004 W/m2 4r (10)2 4×3.14×100 Multiple Choice Questions [1 mark] Choose and write the correct option(s) in the following questions. 1. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in [NCERT Exemplar] (a) visible region (b) infrared region (c) ultraviolet region (d) microwave region 2. A plane electromagnetic wave travelling along X-axis has a wavelength 10.0 mm. The electric field points along Y-direction and has peak value of 30 V/m. Then the magnetic field in terms of x in metre and t in second may be expressed as [NCERT Exemplar] (a) 30 sin 200r ( ct – x) (b) 10–7 sin 200r (ct– x) (c) 30 sin 2r (ct – x) (d) 10–7 sin 2r (ct – x) 10 10 3. Out of the following options which one can be used to produce a propagating electromagnetic wave? (a) A chargeless particles (b) An accelerating charge (c) A charge moving at constant velocity (d) A stationary charge 4. A linearly polarised electromagnetic wave given as E = E0it cos (kz – ~t) is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as [NCERT Exemplar] (a) Er = –E0 it cos (kz – ~t) (b) Er = E0 it cos (kz + ~t) (c) Er = – E0 it cos (kz + ~t) (d) Er = E0 it sin (kz – ~t) Electromagnetic Waves 319

5. Light with an energy flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered (for complete absorption) during 30 minutes is  [NCERT Exemplar] (a) 36 × 10–5 kg m/s (b) 36 × 10–4 kg m/s (c) 108 × 104 kg m/s (d) 1.08 × 107 kg m/s 6. A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 22 V source. When the capacitor is 50% charged, the peak value of the displacement current is (a) 2.2 A (b) 11 A (c) 4.4 A (d) 11 2 A 7. An LC circuit contains inductance L = 1mH and capacitance C = 0.01 µF. The wavelength of the electromagnetic wave generated is nearly (a) 0.5 m (b) 5 m (c) 30 m (d) 188 m 8. The radiowaves of wavelength 360 m are transmitted from a transmitter. The inductance of the coil which must be connected with capacitor of capacitance 3.6 mF in a resonant circuit to receive these waves will be nearly (a) 103 H (b) 102 H (c) 10–4 H (d) 10–8 H 9. What is the amplitude of electric field produced by radiation coming from a 100 W bulb at a distance of 4 m? The efficiency of bulb is 3.14% and it may be assumed as a point source. (a) 2.42 V/m (b) 3.43 V/m (c) 4.2 × 104 V/m (d) 14 × 104 V/m 10. The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is [NCERT Exemplar] (a) E (b) 2E 2 (d) 2 E (c) E 2 11. If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along [NCERT Exemplar] (a) E (b) B (c) B × E (d) E × B 12. An electromagnetic wave travelling along z-axis is given as: E = E0 cos (kz – ~t) . Choose the correct options from the following; [NCERT Exemplar] (a) The associated magnetic field is given as B = 1 k # E = 1 (kt # E) c ~ (b) The electromagnetic field can be written in terms of the associated magnetic field as E = c (B # kt) . (c) kt . E = 0, kt . B= 0 (d) kt × E = 0, kt × B= 0 320 Xam idea Physics–XII

13. If we want to produce electromagnetic waves of wavelength 500 km by an oscillating charge; then frequency of oscillating charge must be (a) 600 Hz (b) 500 Hz (c) 167 Hz (d) 15 Hz 14. Electromagnetic waves travelling in a medium having relative permeability nr = 1.3 and relative permittivity fr = 2.14 . The speed of electromagnetic waves in medium must be (a) 1.8 × 108 m/s (b) 1.8 × 104 m/s (c) 1.8 × 106 m/s (d) 1.8 × 102 m/s 15. Electromagnetic waves travelling in a medium has speed 2 × 108 m/s. If the relative permeability is 1, then the relative permittivity of medium must be (a) 2 (b) 2.25 (c) 2.5 (d) 1.5 16. An electromagnetic wave of frequency 3.0 MHz passes from vacuum into a dielectric medium with relative permittivity fr= 4.0 . Then (a) wavelength is doubled and frequency remains unchanged (b) wavelength is doubled and frequency becomes half (c) wavelength is halved and frequency remains unchanged (d) wavelength and frequency both remains unchanged 17. An electromagnetic wave radiates outwards from a dipole antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as [NCERT Exemplar] (a) 1 (b) 1 r3 r2 1 (c) r (d) remains constant 18. A plane electromagnetic wave of energy U is reflected from the surface. Then the momentum transferred by electromagnetic wave to the surface is (a) 0 (b) U c (c) 2U (d) U c 2c 19. The rms value of the electric field of light coming from the sun is 720 N/C. The average total energy density of the electromagnetic wave is : (a) 4.58 × 10–6 J/m3 (b) 6.37 × 10–9 J/m3 (c) 1.35 × 10–12 J/m3 (d) 3.3 × 10–3 J/m3 20. A plane electromagnetic wave propagating along x direction can have the following pairs of E and B [NCERT Exemplar] (a) Ex, By (b) Ey, Bz (c) Bx, Ey (d) Ez, By Electromagnetic Waves 321

Answers 2. (b) 3. (b) 4. (b) 5. (b) 6. (a) 8. (d) 9. (b) 10. (c) 11. (d) 12. (a), (b), (c) 1. (c) 14. (a) 15. (b) 16. (c) 17. (c) 18. (c) 7. (d) 20. (b), (d) 13. (a) 19. (a) Fill in the Blanks [1 mark] 1. In case of electromagnetic wave, the vibrating electric field vector ( E ) and magnetic field vector ( B ) are mutually perpendicular to each other and both are perpendicular to the direction of _________________. 2. The current which comes into play in the region, whenever the electric field and hence the electric flux is changing with time is called _________________. 3. The orderly distribution of electromagnetic radiations according to their frequency or wavelength is called _________________. 4. The displacement current is precisely equal to the conduction current, when the two are present in different parts of the circuit. These currents are individually discontinuous, but the two currents together posses the property of _________________ through any closed circuit. 5. Electromagnetic wave is _________________ in nature as the electric and magnetic fields are perpendicular to each other and to the direction of propagation of the wave. 6. Electromagnetic waves are not _________________ by electric and magnetic waves. 7. The _________________ of electromagnetic waves does not change when it goes from one medium to another but its wavelength changes. 8. _________________ particles radiate electromagnetic waves. 9. The shortest wavelength radio waves are called _________________. 10. Ozone layer in the atmosphere plays a protective role, and hence its depletion by _____________ gas is a matter of international concern. Answers 1. propagation 2. displacement current 3. electromagnetic spectrum 8. Accelerated charged 4. continuity 5. transverse 6. deflected 7. frequency 9. micro-waves 10. chlorofluorocarbons (CFCs) Very Short Answer Questions [1 mark] Q. 1. How is the speed of EM-waves in vacuum determined by the electric and magnetic fields? [CBSE Delhi 2017] Ans. Speed of EM waves is determined by the ratio of the peak values of electric field vector and magnetic field vector. c= E0 B0 Q. 2. Do electromagnetic waves carry energy and momentum? [CBSE (AI) 2017; 2019, (55/4/1)] Ans. Yes, EM waves carry energy E and momentum p. As electromagnetic waves contain both electric and magnetic fields, there is a non-zero energy density associated with it. 322 Xam idea Physics–XII

E = hc m ⇒ p = U = mc c Here, c = speed of EM wave in vacuum m = wavelength of EM wave U = total energy transferred to the surface. Q. 3. In which situation is there a displacement current but no conduction current? [CBSE South 2016] Ans. During charging or discharging there is a displacement current but no conduction current between plates of capacitor. Q. 4. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates? [CBSE (F) 2016] Ans. The displacement current is equal to the charging current. So, displacement current is also 0.25 A. Q. 5. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves? [CBSE (AI) 2012] Ans. Both electric field and magnetic fields are electromagnetic waves. These waves are perpendicular to each other and perpendicular to the direction of propagation. Q. 6. Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum. [CBSE Delhi 2012] Ans. Velocity (c = 3×108 m/s) This is because both are electromagnetic waves. Q. 7. Write the expression for speed of electromagnetic waves in a medium of electrical permittivity f and magnetic permeability n . [CBSE (F) 2017] Ans. The speed of electromagnetic waves in a material medium in given by v= 1 nf Q. 8. The speed of an electromagnetic wave in a material medium is given by v = 1 , n being the nf permeability of the medium and ε its permittivity. How does its frequency change? [CBSE (AI) 2012] Ans. The frequency of electromagnetic waves does not change while travelling through a medium. Q. 9. A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the direction of electric and magnetic field vectors ? [CBSE Delhi 2011] Ans. Electric field vector along X-axis Magnetic field vector along Y-axis. Q. 10. To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019 Hz belong? [CBSE (AI) 2014] Ans. X-rays or γ-rays Q. 11. To which part of the electromagnetic spectrum does a wave of frequency 3 × 1013 Hz belong? [CBSE (AI) 2014] Ans. Infrared radiation Q . 12. Arrange the following electromagnetic waves in order of increasing frequency: c -rays, microwaves, infrared rays and ultraviolet rays. [CBSE (F) 2014] Ans. Microwave < Infrared < Ultraviolet < c -rays Electromagnetic Waves 323

Q. 13. Arrange the following electromagnetic waves in decreasing order of wavelength: c -rays, infrared rays, X-rays and microwaves. [CBSE (F) 2014] Ans. Microwave > Infrared > X-rays > c -rays Q . 14. Which part of the electromagnetic spectrum is used in operating a RADAR? [CBSE Delhi 2010; 2019 (55/2/1)] Ans. Microwaves with frequency range between 1010 to 1012 Hz are used in operating a RADAR. Q. 15. Why are microwaves considered suitable for radar systems used in aircraft navigation? [CBSE Delhi 2016] Ans. Microwaves are considered suitable for radar systems used in aircraft navigation due to their short wavelength or high frequency. Q. 16. Which part of the electromagnetic spectrum is absorbed from sunlight by ozone layer? [CBSE Delhi 2010] Ans. Ultraviolet light is absorbed by the ozone layer. Q . 17. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency. [CBSE (AI) 2013] Ans. Ultraviolet radiations. Frequency range 1015 – 1017Hz. Hint: Frequency of visible light is of the order of 1014 Hz. Q . 18. Name the electromagnetic waves, which (i) maintain the Earth’s warmth and (ii) are used in aircraft navigation. [CBSE (F) 2012] Ans. (i) Infrared rays (ii) Microwaves Q. 19. Why are infra-red radiations referred to as heat waves? Name the radiations which are next to these radiations in the electromagnetic spectrum having (i) shorter wavelength (ii) longer wavelength. [CBSE (F) 2013] Ans. Infrared waves are produced by hot bodies and molecules, so are referred to as heat waves. (i) Electromagnetic wave having short wavelength than infrared waves are visible, UV, X-rays and γ-rays. (ii) Electromagnetic wave having longer wavelength than infrared waves are microwaves, radio waves. Q. 20. How are X-rays produced? [CBSE (AI) 2011] Ans. X-rays are produced when high energetic electron beam is made incident on a metallic target of high melting point and high atomic weight. Q . 21. Write the following radiations in ascending order in respect of their frequencies: X-rays, microwaves, ultraviolet rays and radiowaves and gamma rays. [CBSE Delhi 2010] Ans. In ascending order of frequencies: radiowaves, microwaves, ultraviolet rays, X-rays and gamma rays. Q . 22. It is necessary to use satellites for long distance T.V. transmission. Why? [CBSE Delhi 2014] Ans. T.V. signals are not properly reflected by ionosphere. Therefore, signals are made to be reflected to earth by using artificial satellites. Q. 23. Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from a satellite orbiting the earth, why? [CBSE (AI) 2009] Ans. The visible radiations and radiowaves can penetrate the earth’s atmosphere but X-rays are absorbed by the atmosphere. 324 Xam idea Physics–XII

Q. 24. Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery.  [CBSE 2018] Ans. (a) Ultraviolet rays (b) Ultraviolet rays/laser Q. 25. How are electromagnetic waves produced by accelerating charges? [CBSE 2019 (55/2/1)] Ans. Accelerated charge produces an oscillating electric field which produces an oscillating magnetic field, which is a source of oscillating electric field, and so on. Thus electromagnetic waves are produced. Q. 26. Why did Maxwell introduce displacement current in Ampere’s circuital law? Ans. Ampere’s circuital law was found inconsistent when applied to the circuit for charging a capacitor. Therefore, Maxwell added displacement current to usual conduction current. The displacement current is Id = f0 dz E where φE is the electric flux. dt Q. 27. From the following, identify the electromagnetic waves having the (i) Maximum (ii) Minimum frequency. (a) Radio waves (b) Gamma-rays (c) Visible light (d) Microwaves (e) Ultraviolet rays, and (f) Infrared rays. Ans. (i) The waves of maximum frequency are gamma rays. (ii) The waves of minimum frequency are radio waves. Q. 28. Why is the orientation of the portable radio with respect to broadcasting station important? [NCERT Exemplar] [HOTS] Ans. As electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric/magnetic part of the wave. Q . 29. The charge on a parallel plate capacitor varies as q = q0 cos 2pvt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor? [NCERT Exemplar] [HOTS] Ans. Conduction current IC = Displacement current ID IC = ID = dq = d (q0 cos 2rot) = –2rq0 o sin 2rot dt dt Q . 30. A variable frequency ac source is connected to a capacitor. How will the displacement current change with decrease in frequency? [NCERT Exemplar] [HOTS] Ans. On decreasing the frequency, reactance XC = 1 will increase which will lead to decrease in ~C conduction current. In this case ID = IC, hence displacement current will decrease. Q. 31. Professor C.V. Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of em waves was he exhibiting? Give one more example of this property. [NCERT Exemplar] [HOTS] Ans. Electromagnetic waves exert radiation pressure. Tails of comets are due to solar radiation. Q. 32. How are infrared waves produced? Ans. Infrared waves are produced by hot bodies and molecules. Electromagnetic Waves 325

Short Answer Questions–I [2 marks] Q. 1. State two properties of electromagnetic waves. How can we show that EM waves carry momentum? [CBSE South 2016] Ans. Properties of electromagnetic waves: (i) Transverse nature (ii) Does not get deflected by electric or magnetic fields (iii) Same speed in vacuum for all waves (iv) No material medium required for propagation (v) They get refracted, diffracted and polarised Electric charges present on a plane, kept normal to the direction of propagation of an EM wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. Q. 2. How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. [CBSE Delhi 2017] Ans. During charging, electric flux between the plates of capacitor keeps on changing; this results in the production of a displacement current between the plates. Id = f0 e dz E o dt Q. 3. Write the generalised expression for the Ampere’s circuital law in terms of the conduction current and the displacement current. Mention the situation when there is: (i) only conduction current and no displacement current. (ii) only displacement current and no conduction current. [CBSE (F) 2013] Ans. Generalised Ampere’s circuital Law— y \"B. d\"l = n0 IC + n0 f0 dz E dt Line integral of magnetic field over closed loop is equal to µ0 times sum of conduction current and displacement current. (i) In case of steady electric field in a conducting wire, electric field does not change with time, conduction current exists in the wire but displacement current may be zero. So, y \"B.d\"l = n0 IC . (ii) In large region of space, where there is no conduction current, but there is only a displacement current due to time varying electric field (or flux). So, \"\" = n0 f0 dz E . zB.dl dt Q. 4. (a) How does oscillating charge produce electromagnetic waves? (b) Sketch a schematic diagram depicting oscillating electric and magnetic fields of an em wave propagating along + z-direction. [CBSE (F) 2014, Delhi 2016] Ans. (a) An oscillating charge produces an oscillating electric field in space, which produces an oscillating magnetic field. The oscillating electric and magnetic fields regenerate each other, and this results in the production of em waves in space. (b) Electric field is along x-axis and magnetic field is along y-axis. 326 Xam idea Physics–XII

Q. 5. (a) An EM wave is travelling in a medium with a velocity \" = vit . Draw a sketch showing the v propagation of the EM wave, indicating the direction of the oscillating electric and magnetic fields. (b) How are the magnitudes of the electric and magnetic fields related to the velocity of the EM wave? [CBSE Delhi 2013] \"\" Ans. The direction of propagation of electromagnetic wave is given by E×B (a) it = tj × kt. (b) The speed of electromagnetic wave c= E0 B0 Q. 6. Name the part of the electromagnetic spectrum whose wavelength lies in the range 10–10 m. Give its one use. [CBSE (AI) 2010] Ans. The electromagnetic waves having wavelength 10–10 m are X-rays. X-rays are used to study crystal structure. Q. 7. (i) How are infrared waves produced? Write their one important use. (ii) The thin ozone layer on top of the stratosphere is crucial for human survival. Why? [CBSE East 2016; 2019 (55/4/1)] Ans. (i) Infrared waves are produced by hot bodies and molecules. Important use: (a) To treat muscular strains (b) To reveal the secret writings on the ancient walls (c) For producing dehydrated fruits (d) Solar heater (e) Solar cooker (Any one) (ii) Ozone layer protects us from harmful UV rays. Q. 8. (i) Which segment of electromagnetic waves has highest frequency? How are these waves produced? Give one use of these waves. (ii) Which EM waves lie near the high frequency end of visible part of EM spectrum? Give its one use. In what way this component of light has harmful effects on humans? [CBSE (F) 2016] Ans. (i) Gamma rays have the highest frequency. These are produced during nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. (ii) Ultraviolet rays lie near the high frequency end of visible part of EM spectrum. They are used to sterlise drinking water and surgical instruments. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. Electromagnetic Waves 327

Q. 9. Explain briefly how electromagnetic waves are produced by an oscillating charge. How is the frequency of EM waves produced related to that of the oscillating charge?  [CBSE (F) 2012, 2019 (55/2/3)] Ans. An oscillating or accelerated charge is supposed to be source of an electromagnetic wave. An oscillating charge produces an oscillating electric field in space which further produces an oscillating magnetic field which in turn is a source of electric field. These oscillating electric and magnetic field, hence, keep on regenerating each other and an electromagnetic wave is produced The frequency of EM wave = Frequency of oscillating charge. Q . 10. Identify the electromagnetic waves whose wavelengths vary as (a) 10–12m < l < 10–8 m (b) 10–3 m < l < 10–1 m Write one use for each. [CBSE (AI) 2017] Ans. (a) X-rays: Used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. (b) Microwaves: Used in radar systems for aircraft navigation. Q. 11. Identify the electromagnetic waves whose wavelengths lie in the range (a) 10–11m < l < 10–8 m (b) 10–4 m < l < 10–1 m Write one use of each. [CBSE (AI) 2017] Ans. (a) X-rays / Gamma rays (b) Infrared / Visible rays / Microwaves (i) X-rays are used as a diagnostic tool in medicine. (ii) Gamma rays are used in medicine to destroy cancer cells. (iii) Infrared are used in green houses to warm plants. (iv) Visible rays provide us information about the world. (v) Microwaves are used in RADAR system for aircraft navigation. Q. 12. In a plane electromagnetic wave, the electric field oscillates with a frequency of 2 × 1010 s–1 and an amplitude of 40 Vm–1. (i) What is the wavelength of the wave? (ii) What is the energy density due to electric field? [HOTS] Ans. (i) Wavelength c 3 # 108 o 2 # 1010 m = = = 1.5 # 10–2 m = 1.5 cm (ii) Given E0 = 40 Vm–1 1 2 Energy density due to electric field = f0 E 2 rms = 1 f0 f E0 2 = 1 f0 E02 2 2 4 p = 1 # 8.86 # 10–12 # (40) 2 = 3.5 # 10–9 J/m3 4 Q . 13. (a) Why are infra-red waves often called heat waves? Explain. (b) What do you understand by the statement, ‘‘Electromagnetic waves transport momentum’’? Ans. (a) Infra-red waves are often called heat waves because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is they heat up and heat their surroundings. (b) Electromagnetic waves can set and sustain electric charges in motion by their electric and magnetic fields. The charges thus acquire energy and momentum from the waves. Since it carries momentum, an electro magnetic wave also exerts pressure, called radiation pressure. Hence they are said to transport momentum. 328 Xam idea Physics–XII

Short Answer Questions–II [3 marks] Q. 1. How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves.  [CBSE North 2016] Ans. EM waves are produced by oscillating charged particle. Mathematical expression for electromagnetic waves travelling along z-axis: Ex =E0 sin (kz – wt) and [For electric field] [For magnetic field] By =B0 sin (kz – wt) Properties (i) Electromagnetic waves have oscillating electric and magnetic fields along mutually perpendicular directions. (ii) They have transverse nature. Q. 2. Arrange the following electromagnetic waves in the order of their increasing wavelength: (a) g-rays (b) Microwaves (c) X-rays (d) Radiowaves How are infra-red waves produced? What role does infra-red radiation play in (i) maintaining the earth’s warmth and (ii) physical therapy? [CBSE Panchkula 2015] Ans. g-rays < X-rays < Microwaves < Radiowaves Infra red rays are produced by the vibration of atoms and molecules. (i) Maintaining Earth’s Warmth: Infrared rays are absorbed by the earth’s surface and reradiated as longer wave length infrared rays. These radiations are trapped by green house gases such as CO2 and maintain earth’s warmth. (ii) Physical Therapy: Infrared rays are easily absorbed by water molecules present in body. After absorption, their thermal motion increases causing heating which is used as physical therapy. Q. 3. When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current? [CBSE Delhi 2012] Ans. When an ideal capacitor is charged by dc battery, charge flows (momentarily) till the capacitor gets fully charged. Ic = dq keep on flowing in the When an ac source is connected then conduction current dt connecting wire. Due to changing current, charge deposited on the plates of the capacitor changes with time. This causes change in electric field between the plates of the capacitor which causes the electric flux to change and gives rise to a displacement current in the region between the plates of the capacitor. As we know, displacement current Id = f0 dz E dt and Id = Ic at all instants. Q. 4. Why does a galvanometer when connected in series with a capacitor show a momentary deflection, when it is being charged or discharged? How does this observation lead to modifying the Ampere’s circuital law? Hence write the generalised expression of Ampere’s law. [CBSE (F) 2015] Electromagnetic Waves 329

Ans. During charging or discharging of the capacitor, displacement current between the plates is produced. Hence, circuit becomes complete and galvanometer shows momentary deflection. I According to Ampere’s circuital Law y \" . \" = n0 I B dl At surface P, At surface S, y \" . \" = n0 Ic ∴ y B . dl = 0 \" \" B dl \" \" \" \" yp B. dl ! ys B . dl This contradicts Ampere’s circuital law. This law must be missing something. Hence the law needs modification. Modified form of Ampere’s circuital law y \"\" = n0 <Ic + f0 d z EF B . dl dt Q. 5. A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor. [CBSE (AI) 2013] Ans. In Fig. conduction current is flowing in the wires, causes charge on the plates So, Ic = dq         ...(i) dt According to Maxwell, displacement current between plates, Id = f0 dz E , where φE= Electric flux ...(ii) dt Using Gauss’s theorem, if one of the plate is inside the tiffin type Gaussian surface, then zE = q So f0 Id = f0 d d q n & Id = dq ...(iii) dt f0 dt From equation (i) and (iii), Both conduction current and displacement current are equal. 330 Xam idea Physics–XII

Q. 6. Write the expression for the generalised form of Ampere’s circuital law. Discuss its significance and describe briefly how the concept of displacement current is explained through charging/ discharging of a capacitor in an electric circuit. [CBSE Allahabad 2015] Ans. The generalisation in Ampere’s circuital law was modified by Maxwell, as \"\" y B.dl = n0 (Ic + Id) = n0 Ic + n0 Id = n0 Ic + n0 f0 dz E dt dz E where Id = f0 dt is displacement current. Significance: This expression signifies that the source of magnetic field is not just due to the conduction current in the metallic conductors, but also due to the time rate of change of electric flux called displacement current. During charging and discharging of a capacitor, electric field between the plates will change. Hence there will be a change in electric flux, called displacement current, between the plates. Q. 7. Considering the case of a parallel plate capacitor being charged, show how one is required to generalise Ampere’s circuital law to include the term due to displacement current. [CBSE (AI) 2014] Ans. l(t) During charging capacitor C, a time varying current I(t) flows through the conducting wire, so \"\" on applying Ampere’s circuital law (for loop A) y B.dl = n0 I (t) … (i) l(t) Now we consider a pot like surface enclosing the positively charged plate and nowhere touches the conducting wire, y \"\" = 0 … (ii) B.dl From equation (i) and (ii), we have a contradiction \" If surfaces A and B forms a tiffin box, and electric field E is passing through the surface (B); constitute an electric flux z= E A = v A = Q A = Q … (iii) f0 Af0 f0 Electromagnetic Waves 331

If the charge on the plate in the tiffin box is changing with time, there must be a current between the plates. From equation (iii) I = dQ = d (f0 z) = f0 dz dt dt dt This is the missing term in Ampere’s circuital law. l(t) The inconsistency may disappear if displacement current is included between the plates. So generalised Ampere’s circuital law can be given as y \"\" = n0 Ic + n0 Id = n0 Ic + n0 f0 dz B.dl dt Q. 8. (a) Which one of the following electromagnetic radiations has least frequency: UV radiations, X-rays, Microwaves? (b) How do you show that electromagnetic waves carry energy and momentum? (c) Write the expression for the energy density of an electromagnetic wave propagating in free space. [CBSE Bhubaneswar 2015] Ans. (a) Microwave (b) When a charge oscillates with some frequency. It produces an oscillating electric field and magnetic field in space. So, an electromagnetic wave is produced. The frequency of the EM wave is equal to the frequency of oscillation of the charge. Hence energy associated with the EM wave comes at the expense of the energy of the source. If the em wave of energy U strikes on a surface and gets completely absorbed, total momentum delivery to the surface is p = U . E Hence em wave also carry momentum. (c) The EM wave consists of oscillating electric and magnetic fields, So net energy density of EM wave is U = UE + UB U = 1 f0E2 + 1 B2 2 2 n0 Q. 9. (a) How are electromagnetic waves produced by oscillating charges? (b) State clearly how a microwave oven works to heat up a food item containing water molecules. (c) Why are microwaves found useful for the radar systems in aircraft navigation? [CBSE (F) 2013] 332 Xam idea Physics–XII

Ans. (a) If a charge particle oscillates with some frequency, produces an oscillating electric field in space, which produces an oscillating magnetic field, which inturn, is a source of electric field, and so on. Thus oscillating electric fields and magnetic fields regenerate each other, and an electromagnetic wave propagates in the space. (b) In microwave oven, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves get transferred efficiently to the kinetic energy of the molecules. This kinetic energy raises the temperature of any food containing water. (c) Microwaves are short wavelength radio waves, with frequency of order of few GHz. Due to short wavelength, they have high penetrating power with respect to atmosphere and less diffraction in the atmospheric layers. So these waves are suitable for the radar systems used in aircraft navigation. Q. 10. Name the parts of the electromagnetic spectrum which is (i) suitable for radar systems used in aircraft navigation. (ii) used to treat muscular strain. (iii) used as a diagnostic tool in medicine. Write in brief, how these waves can be produced. [CBSE Delhi 2015] Ans. (i) Microwave, (ii) Infrared, (iii) X-rays Microwave are produced by special vacuum tubes, like klystorms, magnetrons and gunn diodes. Infrared are produced by the vibrating molecules and atoms in hot bodies. X-rays are produced by the bombardment of high energy electrons on a metal target of high atomic weight (like tungsten). Q. 11. (i) Identify the part of the electromagnetic spectrum which is: (a) Suitable for radar system used in aircraft navigation. (b) Produced by bombarding a metal target by high speed electrons. (ii) Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor? Write the necessary expression to explain this observation. [CBSE Central 2016] Ans. (i) (a) Microwaves (b) X-rays (ii) Due to conduction current in the connecting wires and the production of displacement current between the plates of capacitor on account of changing electric field. Current inside the capacitor is given by Id = f0 dz E dt Q. 12. Answer the following questions: (a) Name the EM waves which are produced during radioactive decay of a nucleus. Write their frequency range. (b) Welders wear special glass goggles while working. Why? Explain. (c) Why are infrared waves often called as heat waves? Give their one application. [CBSE Delhi 2014] Ans. (a) EM waves : γ-rays Range : 1019 Hz to 1023 Hz Electromagnetic Waves 333

(b) This is because the special glass goggles protect the eyes from large amount of UV radiations produced by welding arcs. (c) Infrared waves are called heat waves because water molecules present in the materials readily absorb the infrared rays and get heated up. Application: They are used in green houses to warm the plants. Q. 13. Answer the following: (a) Name the EM waves which are used for the treatment of certain forms of cancer. Write their frequency range. (b) Thin ozone layer on top of stratosphere is crucial for human survival. Why? (c) Why is the amount of the momentum transferred by the em waves incident on the surface so small? [CBSE Delhi 2014] Ans. (a) X-rays or γ-rays Range: 1018 Hz to 1022 Hz. (b) Ozone layer absorbs the ultraviolet radiations from the sun and prevents it from reaching the earth’s surface. p = U (c) Momentum transferred, c where U = energy transferred, and c = speed of light Due to the large value of speed of light (c), the amount of momentum transferred by the em waves incident on the surface is small. Q. 14. Electromagnetic waves with wavelength (i) l1 is used in satellite communication. (ii) l2 is used to kill germs in water purifier. (iii) l3 is used to detect leakage of oil in underground pipelines. (iv) l4 is used to improve visibility in runways during fog and mist conditions. (a) Identify and name the part of electromagnetic spectrum to which these radiations belong. (b) Arrange these wavelengths in ascending order of their magnitude. (c) Write one more application of each. [NCERT Exemplar] Ans. (a) l1 → Microwave, l2 → UV l3 → X-rays, l4 → Infrared (b) l3 < l2 < l4 < l1 (c) Microwave – RADAR UV – LASIK eye surgery X-ray – Bone fracture identification (bone scanning) Infrared – Optical communication Q. 15. Show that during the charging of a parallel plate capacitor, the rate of change of charge on each plate equals ε0 times the rate of change of electric flux ‘φE’ linked with it. What is the name given to the term f0 dz E ? [HOTS] dt Ans. Charge on each plate of a parallel plate capacitor q(t) = σ (t) A But σ(t) = e0 E (t) ∴ q(t) = e0 AE (t) where σ(t) instantaneous charge per unit area 334 Xam idea Physics–XII

E (t) = electric field strength But E(t) A = electric flux φE (t) ∴ q(t) = e0 φE (t) ∴ Rate of change of charge dq (t) = f0 dzE (t) dt dt ∴ Rate of change of charge = e0 × rate of change of electron flux |φE | The quantity f0 dzE (t) is named as displacement current. dt Self-Assessment Test Time allowed: 1 hour Max. marks: 30 1. Choose and write the correct option in the following questions. (3 × 1 = 3) (i) The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is (a) c : 1 (b) c2 : 1 (c) 1 : 1 (d) c : 1 (ii) The quantity n0 f0 represents (b) speed of light in vacuum (a) speed of sound (c) speed of electromagnetic waves (d) inverse of speed of light in vacuum (iii) The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to (a) the speed of light in vacuum (b) reciprocal of speed of light in vacuum (c) the ratio of magnetic permeability to the electric susceptibility of vacuum (d) unity 2. Fill in the blanks. (2 × 1 = 2) (i) Displacement current is the electric current which flows in the gap between the plates of capacitor during its ______________, which originates due to time varying electric field in the space between the plates of capacitor. (ii) The basic different between various types of electromagnetic waves lies in their ______________ since all of them travel through vacuum with the same speed. 3. In which directions do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the x-axis? 1 4. Name the electromagnetic radiation to which waves of wavelength in the range of 10–2 m belong. Give one use of this part of electromagnetic spectrum. 1 5. Name the electromagnetic radiation which can be produced by klystron or a magnetron valve. 1 6. The oscillating electric field of an electromagnetic wave is given by Ey = 30 sin (2×1011 t + 300rx) Vm–1 Electromagnetic Waves 335

(a) Obtain the value of wavelength of the electromagnetic wave. (b) Write down the expression for oscillating magnetic field. 2 7. The oscillating magnetic field in a plane electromagnetic wave is given by Bz = (8 × 10–6) sin [2×1011t + 300πx] T (i) Calculate the wavelength of the electromagnetic wave. (ii) Write down the expression for the oscillating electric field. 2 8. How are microwaves produced? Write their two important uses. 2 9. Answer the following questions : (a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why? (b) The small ozone layer on top of the stratosphere is crucial for human survival. Why? 2 10. A capacitor of capacitance ‘C’ is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor. 3 11. How are electromagnetic waves produced? What is the source of the energy carried by a propagating electromagnetic wave? Identify the electromagnetic radiations used (i) in remote switches of household electronic devices; and (ii) as diagnostic tool in medicine. 3 12. (a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range. (b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. 3 13. (a) A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. (b) Find the wavelength of electromagnetic waves of frequency 6 × 1012 Hz in free space. Give its two applications. 5 Answers 1. (i) (c) (ii) (d) (iii) (b) 2. (i) charging (ii) wavelengths or frequencies 6. (a) 6.67 × 10–3 m (b) 10–7 sin (2 × 1011t + 300px) T 7. (i) 6.67 × 10–3 m (ii) 2.4 × 103 sin (2 × 1011t + 300px) Vm–1 zzz 336 Xam idea Physics–XII

Ray Optics Chapter –9 and Optical Instruments 1. Optics The study of nature and propagation of light is called optics. Ray optics deals with particle nature of light whereas wave optics considers light as a wave. 2. Reflection of Light When a light ray incident on a smooth surface bounces back to the same medium, it is called reflection of light. Laws of regular Reflection (i) Angle of incidence is equal to the angle of reflection. i.e., i = r (ii) The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane. These laws hold for any reflecting surface whether plane or curved. There is no change in wavelength and frequency during reflection. Spherical Mirror: A spherical mirror is simply a part cut off from the surface of a hollow sphere which has been made smooth and silver polished on one side. Spherical mirrors are of two types: (i) Concave mirror: If outer side or bulging side of the spherical surface is silver polished, it is called a concave mirror. (ii) Convex mirror: If inner side of a spherical surface is silver polished, it is called a convex mirror. Relation between focal length and radius of curvature: The distance between centre (C) of spherical surface and its pole (P) is called the radius of curvature. It is denoted by R. The rays parallel to the principal axis (CP) after striking the mirror meet at a point (F) (in concave mirror) or appear to be meeting at a point F (in convex mirror). This point is called the principal focus (F) of mirror. The distance of focus (F) from pole (P) of a mirror is called the focal length of Ray Optics and Optical Instruments 337

the mirror. It is denoted by f. The focal length f is half of the radius of curvature. i.e., f = R 2 Mirror formula: The mirror formula is 1 = 1 + 1 f v u where u = distance of object from mirror; v= distance of image from mirror; and f = focal length of mirror. Magnification produced by mirror: The ratio of the size of image to the size of object is called linear magnification produced by the mirror. Magnification M= hl =– v =– f = f –v h u u– f f Where hl is the height of image and h is the height of object. 3. Refraction of Light When a ray of light enters from one transparent medium into another, there is a change in speed and direction of the ray in the second medium. This phenomenon is called refraction of light. Laws of refraction: (i) The incident ray, the refracted ray and the normal to the surface separating the two media, all lie in the same plane. (ii) Snell’s Law: For two media, the ratio of sine of angle of incidence to the sine of the angle of refraction is constant for a beam of particular wavelength, i.e., n2 sin i = constant = n1 = 1 n2 ...(i) sin r where n1 and n2 are absolute refractive indices of I and II media respectively and 1n2 is a refractive index of second (II) medium with respect to first (I) medium. Due to principle of reversibility of light, sin r sin i = 2 n1 …(ii) Multiplying (i) by (ii), we get 1 1n2 1 = 2 n1 # 1 n2 or 2 n1 = …(iii) The frequency of light remains unchanged while passing from one medium to the other. Refractive Index: The refractive index of a medium is defined as the ratio of speed of light in vacuum to the speed of light in a medium. 338 Xam idea Physics–XII

i.e., n = Speed of light in vacuum = c Speed of light in medium v = omair = mair …(iv) o m medium m medium λair and λmedium being wavelengths of light in air and medium respectively. ∴ sin i = n2 f= c/v2 p = v1 = m1 ...(v) sin r n1 c/v1 v2 m2 Formation of image due to refraction: According to Snell’s law, if n2 > n1, i > r. That is, if a ray of light enters from rarer medium to a denser medium, it is deviated towards the normal and if n2 < n1, i < r that is, if the ray of light enters from denser to a rarer medium it is deviated away from the normal. Accordingly, if the ray of light starting from object O, in the given diagram in a denser medium travels along OP, it is deviated away from the normal along PQ. The ray PQ appears to come from I. Thus I is the virtual image of O. It can be shown that n = Real depth (OM) = t x …(vi) Apparent depth (MI) t– where x is the apparent shift. 1 n ∴ The apparent shift, x = a1– kt …(vii) Refraction through a number of media: Let us consider the refraction of light ray through a series of media as shown in fig. The ray AB is incident on air-water interface at an angle i. The ray is deviated in water along BC towards the normal. Then it falls on water-glass interface and is again deviated towards normal along CD. If the last medium is again air, the ray emerges parallel to the incident ray. Let r1 and r2 be angles of refraction in water and glass respectively, then from Snell’s law, ssssiiiinnnnri1rr12==nnnnwa wg==anwwng sin r2 na …(viii) sin i = ng = g na …(ix) …(x) TRSSSSSSSSSnnnwag = refractive index of agwilraats=esr1WWWWWVWWWW = refractive index of X = refractive index of Multiplying (viii), (ix) and (x), we get anw × wng × gna=1 ang wng = 1 = anw …(xi) anw # gna 4. Critical Angle When a ray of light is incident on the interface from denser medium to rarer medium, it is deviated away from the normal. When angle of incidence is increased, angle of refraction also increases and at a stage it becomes 90°. The angle of incidence in denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle (C) for the pair of media. Ray Optics and Optical Instruments 339

If nr and nd are refractive indices for rarer and denser media, then n2 ∴ sin i = n1 gives sin r sin C = nr = d nr sin 90° nd sin C = d nr = 1 = 1 rnd n where rnd = n and n is the refractive index of a denser medium with respect to a rarer medium. 5. Total Internal Reflection When angle of incidence in the denser medium is greater than the critical angle, the incident ray does not refract into a rarer medium but is reflected back into the denser medium. This phenomenon is called total internal reflection. The conditions for total internal reflection are (i) The ray must travel from a denser into a rarer medium. (ii) The angle of incidence i> critical angle C. The critical angle for water-air, glass-air and diamond-air interfaces are 49°, 42° and 24° respectively. 6. Spherical Lenses There are two types of spherical lenses. (i) Convex lens (Converging lens) (ii) Concave lens (Diverging lens) Rules of Image Formation in Lenses (i) The ray incident on lens parallel to the principal axis, after refraction through the lens, passes through the second focus (in convex lens) or appear to come from second focus (in concave lens). (ii) The ray incident on lens through optical centre C, after refraction, pass straight without any deviation. (iii) A ray directed towards the first focus incident on the lens, after refraction becomes parallel to the principal axis. 7. Thin Lens Formula If u and v are object and image distances from a lens of focal length f, then thin lens formula is 1 = 1 – 1 f v u This equation holds for convex and concave lenses both, but proper signs of u, v and f are to be used according to sign convention of coordinate geometry. Focal length of a convex lens is taken as positive and of a concave lens is taken as negative. 340 Xam idea Physics–XII

Magnification produced by a lens f m = hl = v = u+f h u where hl is the size of image and h is the size of object. 8. Lens Maker’s Formula If R1 and R2 are the radii of curvature of first and second refracting surfaces of a thin lens of focal length f, then lens makers formula is 1 = (1 n2 – 1) × d 1 – 1 f R1 R2 n = (n – 1)×d 1 – 1 R1 R2 n where 1n2=n is refractive index of material of lens with respect to surrounding medium. 9. Power of a Lens The power of a lens is its ability to deviate the rays towards its principal axis. It is defined as the reciprocal of focal length in metres. 1 Power of a lens, P= f (in metre) Its unit is diopter and is represented as ‘D’. 10. Lens Immersed in a Liquid If a lens of refractive index ng is immersed in a liquid of refractive index nl then its focal length (fl) in liquid, is given by 1 1 1 fl = (l ng – 1) × d R1 – R2 n where lng = ng nl ng –1 If fa is the focal length of lens in air, then fl = ng –1 # fa nl Three cases arise: (i) If ng > nl , then fl and fa are of same sign but fl > fa. That is, the nature of lens remains unchanged, but its focal length increases and hence the power of lens decreases. In other words the convergent lens becomes less convergent and divergent lens becomes less divergent. (ii) If ng = nl, then fl = ∞. That is, the lens behaves as a glass plate. (iii) If ng < nl, then fl and fa have opposite signs. That is, the nature of lens changes. A convergent lens becomes divergent and vice versa. 11. Thin Lenses in Contact If two or more lenses of focal lengths f1, f2 are placed in contact, then their equivalent focal length F is given by 1 = 1 + 1 + ... F f1 f2 The power of combination P = P1 + P2 + ... Ray Optics and Optical Instruments 341

12. Combination of a Lens and a Mirror Consider a coaxial arrangement of a lens and a mirror. Let an object be placed in front of the lens. The incident rays, from the object, first undergo refraction at lens and are then incident on the mirror. To obtain the position of the image due to the combination, we can proceed as follows: (i) Using refraction formula, we can calculate where the image would have been formed, had there been only the lens. We then consider this image as an object for the mirror. (ii) Using the mirror formula, we can then locate the position of its final image formed by the mirror. This final position, would be the position of the image due to the combined effect of refraction at the lens and reflection at the mirror. 13. Refraction Through a Prism A prism is a transparent medium enclosed by two plane refracting surfaces. Let EF be the monochromatic ray incident on the face PQ of prism PQR of refracting angle A at angle of incidence i1. This ray is refracted along FG, r1 being angle of refraction. The ray FG is incident on the face PR at angle of incidence r2 and is refracted in air along GH. Thus GH is the emergent ray and i2 is the angle of emergence. The angle between incident ray EF and emergent ray GH is called angle of deviation δ. For a prism if A is the refracting angle of prism, then r1 + r2 = A …(i) and i1 + i2 = A + δ …(ii) Clearly, deviation δ = i1+ i2 – A, i1 and i2 may be inter-changed, therefore, there are two values of angles of incidence for same deviation δ. If n is the refractive index of material of prism, then from Snell’s law n = sin i1 = sin i2 . …(iii) sin r1 sin r2 If angle of incidence is changed, the angle of deviation δ changes as shown in fig. For a particular angle of incidence the deviation is minimum. This is called angle of minimum deviation δm. Minimum deviation: At minimum deviation the refracted ray within a prism is parallel to the base. Therefore, i1 = i2 = i (say) r1 = r2 = r (say) Then from equations (i) and (ii), r + r = A or r = Ai/2= A +2 d m ……((viv)) i + i = A + dm or ∴ The refractive index of material of prism sin e A + dm .o 2 n = sin i = …(vi) sin r sin (A/2) 342 Xam idea Physics–XII

For a thin prism, viz. A ≤ 10° δm= (n – 1) A. 14. Scattering of Light The light is scattered by air molecules. According to Lord Rayleigh the intensity of scattered light 1 1 I ? (wavelength) 4 & I ? m4 As λblue < λred, accordingly blue colour is scattered the most and red the least, so sky appears blue. At the time of sunrise and sunset, blue colour is scattered the most and red colour enters our eyes, so sunrise and sunset appear red. 15. Optical Instruments (Microscopes and Telescopes) A microscope is an optical instrument to see very small objects. (i) Simple Microscope: It consists of a convex lens of small focal length f. If β = angle subtended by an image on eye a = angle subtended by an object on eye, when object is at a distance of distinct vision (D) Magnifying power, M = b = D c1 + v m a v f If the final image is at ∞, v = ∞ then M = Df. D f If the final image is at a distance of distinct vision, v = D, M = 1 + . (ii) Compound Microscope: A compound microscope essentially consists of two co-axial convex lenses of small focal lengths. The lens facing the object is called an objective lens while that towards eye is called the eye lens (eyepiece). ∴ Magnifying power of microscope, M = b (= mo # me) = vo D e1 + ve o a uo ve fe Separation between lenses, L = v0 + ue Ray Optics and Optical Instruments 343

Special cases: (a) When final image is formed at a distance of distinct vision, ve = D M = – vo d1 + D n and L = v0 + ue uo fe The distance between second focal point of objective and first focal point of eye lens is called the tube length denoted by L,then vo = L uo f0 So, M = – L d1 + D n f0 fe (b) When final image is formed at infinity, ve= ∞, then M = – vo # D uo fe = – L . D and L = vo + fe fo fe Telescope: It is an optical instrument to see distant objects. (iii) Astronomical Telescope (Refracting Telescope): It is used to see magnified images of distant objects. An astronomical telescope essentially consists of two co-axial convex lenses. The lens facing the object has a large focal length and a large aperture and is called objective, while the lens towards eye has a small focal length and small aperture and is called eye lens. The magnifying power of telescope is M= Angle subtended by final image at eye = b Angle subtended by object on eye a = (m0 × me) = – f0 f1 + fe p fe ve and Length of telescope L=f0+ue where ue = distance of real image from eye lens ve = distance of final image A′ B′ from eye lens f0 = focal length of objective, fe= focal length of eye lens 344 Xam idea Physics–XII

α = angle subtended by an object at eye = h f0 β = angle subtended by an image at eye = h fe Special cases: (a) When final image is formed at a distance of distinct vision, then ve=D M = – fo d1 + fe nand L = fo + ue fe D (b) When final image is formed at infinity, then ve= ∞ M = – fo and L = fo + fe fe Reflecting Telescope: In this telescope, a concave mirror is used as an objective in place of a convex lens. It is free from chromatic aberration and it has larger resolving power than refracting telescope. 16. Magnifying Power of Optical Instruments The size of an object depends on the angle subtended by the object on eye. This angle is called visual angle. Greater the visual angle, greater the size of object. Stars are bigger than sun; but appear smaller because stars are much farther away than sun and they subtend smaller angles on eye. The angle subtended on eye may be increased by using telescopes and microscopes. The telescopes and microscopes form the image of an object. The image subtends larger angle on eye; hence the object appears big. The magnification produced by optical instrument (telescope/microscope) is defined as the ratio of angle (β) subtended by image on eye and the angle (a) subtended by object on eye. b a i.e., Angular magnification M= Selected NCERT Textbook Questions Reflection, Refraction and Total Internal Reflection Q. 1. A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how should the screen be moved? Ans. Given u =–27 cm, h = 2.5 cm | R | = | 2f | = 36 cm & f = – 36 = –18 cm (with sign convention) 2 1 1 1 f = u + v 1 = 1 – 1 = – 1 + 1 = –3 + 2 & v = –54 cm v f u 18 27 54 That is, image is formed in front of mirror at a distance 54 cm from the mirror. Therefore the screen must be placed at a distance 54 cm from the mirror. Size of the image hl = – v × h = – ^–54h ×2.5 cm. = –5 cm u –27 The image is real, inverted and 5 cm long. If the candle is moved closer, the screen should have to be moved farther and farther. If the candle is brought less than 18 cm, the image will be virtual and cannot be collected on the screen. Ray Optics and Optical Instruments 345

Q. 2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens if the needle is moved farther from the mirror. Ans. Given u = –12 cm , f = + 15 cm (convex mirror) 1 = 1 + 1 & 1 = 1 – 1 f v u v f u 1 = 1 + 1 = 4+5 & v = 60 = 20 = 6.67 cm v 15 12 60 9 3 That is image is formed at a distance 6.67 cm behind the mirror. Magnification m= – v = – 20 = 5 u –3×12 9 Size of image hl = mh = 5 ×4.5 = 2.5 cm 9 The image is erect, virtual and has a size 2.5 cm. Its position is 6.67 cm behind the mirror when needle is moved farther, the image moves towards the focus and its size goes on decreasing. Q. 3. A tank is filled with water to a height of 12.5 m. The apparent depth of the needle lying at the bottom of the tank as measured by a microscope is 9.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope be moved to focus on the needle again? Ans. Refractive index, n= Real depth (H) Apparent depth (h) Given H = 12.5 cm, h = 9.4 cm 12.5 9.4 ∴ Refractive index of water, nw = = 1.33 Refractive index of liquid, nl = 1.63 h= H = 12.5 = 7.7 cm ∴ Apparent height with liquid in tank, nl 1.63 ∴ Displacement of microscope, x = 9.4 – 7.7 = 1.7 cm Q. 4. Fig. (a) and (b) show refraction of an incident ray in air at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle (r) of refraction of an incident ray in water at 45° with the normal to a water-glass interface [fig. (c)]. Ans. Snell’s law of refraction is sin i nn=12 1 n2 sin=r Fig. (a) sin 60° = ng = a ng sin 35° na ⇒ Refractive index of glass with respect to air, a ng = sin 60° = 0.8660 =1.51 sin 35° 0.5736 nw Fig. (b) sin 60° = na = a nw sin 41° 346 Xam idea Physics–XII

Refractive index of water with respect to air, a nw = sin 60° = 0.8660 =1.32 sin 41° 0.6561 sin 45° a ng Fig.(c) sin r = a nw & sinr = a nw # sin 45° = 1.32 # 0.7071= 0.6181 a ng 1.51 ⇒ r = sin – 1 (0.6181) = 38° Q. 5. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive 4 index of water is 3 . Ans. The light rays starting from bulb can pass through the surface if angle of incidence at surface is less than or equal to critical angle (C) for water-air interface. If h is depth of bulb from the surface, the light will emerge only through a circle of radius r given by r = h tan C, where h = 80 cm = 0.80 m But sin=C 1 3 ∴ a=nw 4 tan C = 3 7 ∴ r = 0.80 # e 3 7o ∴ Area of circular surface of water, A = rr2 = 3.14 # d0.8 # 3 2 = 3.14 # 0.64 # 9 = 2.6 m2 7 7 n Q. 6. Use the mirror equation to show that (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. [CBSE Delhi 2015, (F) 2017, 2019 (55/3/3)] (b) a convex mirror always produces a virtual image independent of the location of the object. (c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [CBSE (AI) 2011] Ans. (a) Mirror equation is 1 = 1 + 1 or 1 = 1 – 1 f v u v f u For a concave mirror, f is negative, i.e., f < 0. For a real object (on the left of mirror), u < 0 1 1 1 ∴ 2f < u < f or 2f > u > f or – 1 < – 1 < – 1 or 1 – 1 < 1 – 1 < 1 – 1 2f u f f 2f f u f f or 1 < – 1 < 0 i.e, 1 is negative. 2f v v This implies that v is negative. Also from above inequality 2f > v or |2f| < |v| ( a 2f and v are negative) Hence, the real image is formed beyond 2f. (b) For a convex mirror, f is positive, i.e., f > 0. For a real object on the left, u is negative Ray Optics and Optical Instruments 347