Time: 2 Hours<\/span> General Instructions to candidates:<\/span><\/p>\n The given value of constants can be used wherever necessary.<\/span><\/p>\n Charge of proton = 1.6 x 10-19<\/sup>C Answer any three questions from 1 to 4. Each carries 1 score. (3 \u00d7 1 = 3)<\/span><\/p>\n Question 1. Question 2. Question 3. Question 4. Answer any six questions from 5 to 11. Each carries 2 scores. (6 \u00d7 2 = 12)<\/span><\/p>\n Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Answer any six questions from 12 to 18. Each carries 3 scores. (6 \u00d7 3 = 18)<\/span><\/p>\n Question 12. Then, according to Gauss’s theorem we can write, Question 13. Question 14. Question 15. Question 16. Hence the path difference, S1<\/sub>O – S2<\/sub>O = 0 Question 17. Working: The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal. The nickel crystal scatters the electron beam to different angles. The crystal is fixed at an angle of \u03a6 = 50\u00b0 to the incident beam. The detector current for different values of the accelerating potential \u2018V\u2019 is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.<\/p>\n Analysis of graph: Experimental wavelength of electron: The wave length of the electron can be found from the formula Theoretical wave length of electron: Discussion: The experimentally measured wave-length is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.<\/p>\n Question 18. Answer any three questions from 19 to 22. Each carries 4 scores. (3 \u00d7 4 = 12)<\/span><\/p>\n Question 19. b) Radius of the hydrogen atom: Consider an electron of charge \u2018e\u2019 and mass m revolving round the positively charged nucleus in circular orbit of radius ‘r’. Question 20. b) Consider a solenoid (air core) of cross sectional area A and number of turns per unit length n. Another coil of total number of turns N is closely wound over the first coil. Let I be the current flow through the primary. Question 21. Question 22. Answer any three questions from 23 to 26. Each carries 5 scores. (3 \u00d7 5 = 15)<\/span><\/p>\n Question 23. Constructional Details: Cyclotron consists of two semicircular dees D1<\/sub> and D2<\/sub>, enclosed in a chamber C. This chamber is placed in between two magnets. An alternating voltage is applied in between D1<\/sub> and D2<\/sub>. An ion is kept in a vacuum chamber.<\/p>\n Working: At certain instant, let D1<\/sub> be positive and D2<\/sub> be negative. Ion (+ve) will be accelerated towards D2<\/sub> and describes a semicircular path (inside it). When the particle reaches the gap, D1<\/sub> becomes negative and D2<\/sub> becomes positive. So ion is accelerated towards D1<\/sub> and undergoes a circular motion with larger radius. This process repeats again and again.<\/p>\n Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate. Mathematical expression: Let \u2018v\u2019 be the velocity of ion, q the charge of the ion and B the magnetic flux density. Resonance frequency (cyclotron frequency): The condition for resonance is half the period of the accelerating potential of the oscillator should be \u2018t\u2019. (i.e., T\/2 = t or T = 2t). Hence period of AC Question 24. Question 25. Working: During the +ve half cycle of the a.c signal at secondary, the diode D1<\/sub> is forward biased and D2<\/sub> is reverse biased. So that current flows through D1<\/sub> and RL<\/sub>.<\/p>\n During the negative half cycle of the a.c signal at secondary, the diode D1<\/sub> is reverse biased and D2<\/sub> is forward biased. So that current flows through D1<\/sub> and RL<\/sub>. Thus during both the half cycles, the current flows through RL<\/sub> in the same direction. Thus we get a +ve voltage across RL<\/sub> for +ve and -ve input. This process is called full wave rectifcation.<\/p>\n b) 100 Hz<\/p>\n Question 26. b) c) Remain the same<\/p>\n
\nCool off time : 15 Minutes
\nMaximum : 60 Score
\n<\/span><\/p>\n\n
\nMass of proton = 1.67 x 10-27<\/sup>kg<\/p>\n
\nA charged particle enters a uniform magnetic field at an angle of 40\u00b0. It\u2019s path becomes …………
\nAnswer:
\nHelical<\/p>\n
\nFigure shows the symbolic representation of ……….
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-1.png\")
\ni) OR gate
\nii) NAND gate
\niii) NOR gate
\niv) NOT gate
\nAnswer:
\niii) NOR gate<\/p>\n
\nWrite the unit of mobility.
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-2.png\")
<\/p>\n
\nIf \u2018h\u2019 is Planck\u2019s constant, the momentum of a photon of wavelength 1 A\u00b0 is
\ni) h
\nii) 10-10<\/sup>
\niii) 1010<\/sup> h
\niv) 10 h
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-3.png\")
<\/p>\n
\na) The ratio of electric field on the equatorial point and at the axial point at equal distances from the centre of a short electric dipole is ………..
\nb) A closed surface encloses an electric dipole. What is the electric flux through the surface?
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-4.png\")
<\/p>\n
\nA series LCR circuit connected to an ac source is shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-5.png\")
\na) Write an expression for impedance offered by its circuit.
\nb) Under what condition this circuit is used for tuning radio?
\nAnswer:
\na) Impedence Z = \\(\\sqrt{R^{2}+\\left(x_{L}-x_{C}\\right)^{2}}\\)
\nb) at XL<\/sub> = XC<\/sub> this circuit is used as tuner circuit.<\/p>\n
\nWhich electromagnetic waves are used for the following purposes?
\ni) Diagnostic tool in medicine.
\nii) Kill germs in water purifiers.
\niii) Cellular phones.
\niv) In remote switches of household electronic systems.
\nAnswer:
\ni) X-rays
\nii) Uv rays
\niii) Radio Waves
\niv) IR rays<\/p>\n
\nCalculate the effective capacitance between \u2019a\u2019 and \u2018b\u2019 from the figure given below:
\nC1<\/sub> = C3<\/sub> = 100 \u00b5F, C2<\/sub> = C4<\/sub> = 200 \u00b5F
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-6.png\")
<\/p>\n
\nWrite any two uses of polaroids.
\nAnswer:<\/p>\n\n
\nThe temperature dependence of resistivity of a material is shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-7.png\")
\na) Identify the type of material.
\nb) Write the relation between resistivity and average collision time for electron.
\nAnswer:
\na) Semi conductor
\nb) \u03c1 = \\(\\frac{\\mathrm{m}}{\\mathrm{ne}^{2} \\tau}\\)<\/p>\n
\nWhat is meant by half life of a radioactive substance? Write its relation with decay constant.
\nAnswer:
\nHalf life is the time taken for a radio active substance to reduce half of its initial value.
\n\\(T_{1 \/ 2}=\\frac{0.693}{\\lambda}\\)<\/p>\n
\nA spherical shell of radius R\u2019 is uniformly charged with charge +q. By Gauss\u2019stheorem, find the electric field intensity at a point \u2018p\u2019.
\na) Outside the spherical shell and
\nb) Inside the spherical shell.
\nAnswer:
\nField Due To A Uniformly Charged Thin Spherical Shell: Consider a uniformly charged hollow spherical conductor of radius R. Let \u2018q\u2019 be the total charge on the surface.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-8.png\")
\nTo find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius ‘r’.<\/p>\n
\n\\(\\int \\overrightarrow{\\mathrm{E}} \\cdot \\mathrm{d} \\overrightarrow{\\mathrm{s}}=\\frac{1}{\\varepsilon_{0}} \\mathrm{q}\\)
\nThe electric field is constant,at a distance T. So we can write,
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-9.png\")
\nb) E = 0.<\/p>\n
\nThe equipotential surface through a point is normal to the electric field at that point.
\na) What is meant by equipotential surface?
\nb) What is the work done to move a charge on an equipotential surface?
\nc) Draw the equipotential surfaces for a uniform electric field.
\nAnswer:
\na) The surface over which potential is constant is equipotential surface.
\nb) Workdone = pd \u00d7 charge
\n= 0 \u00d7 q = 0
\nc)
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-10.png\")
<\/p>\n
\nThe elements of earth\u2019s magnetic field at a place are declination, dip and horizontal intensity.
\na) A magnetic needle free to move in horizontal plane is shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-11.png\")
\nWhich element of earth\u2019s magnetic field is represented by \u03b8 in the figure?
\nb) The vertical component of earth\u2019s magnetic field at a given place is \\(\\sqrt{3}\\) times its horizontal component. If total intensity of earth\u2019s magnetic field at the place is 0.4 G find the value of horizontal component of earth\u2019s magnetic field.
\nAnswer:
\na) Declination
\nb) tan \u03b8 = \\(\\frac{B_{V}}{B_{H}}=\\sqrt{3}\\)
\n\u2234 \u03b8 = 60\u00b0
\nBH<\/sub> = B cos \u03b8, B = 0.4 G
\nBH<\/sub> = 0.4 \u00d7 cos 60 = 0.2 G
\n= 0.2 \u00d7 10-4\u00a0<\/sup>T<\/p>\n
\nA transformer is used to change the alternating voltage to a high or low value.
\na) What is the principle of a transformer?
\nb) A power transmission line feeds input power of 2300 V to a stepdown transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
\nAnswer:
\na) Mutual Induction
\nb) Vp<\/sub> = 2300 v, Np<\/sub> = 4000 turns
\nVs<\/sub> = 230 v
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-12.png\")
<\/p>\n
\nDescribe Young\u2019s double slit experiment and derive an expression for the band width of the interference band.
\nAnswer:
\nExpression for band width
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-13.png\")
\nS1<\/sub> and S2<\/sub> are two coherent sources having wave length \u03bb. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. \u2018O\u2019 is a point on the screen equidistant from S1<\/sub> and S2<\/sub>.<\/p>\n
\nSo at \u2018O\u2019 maximum brightness is obtained.
\nLet \u2018P\u2019 be the position of nth<\/sup> bright band at a distance xn<\/sub> from O. Draw S1<\/sub>A and S2<\/sub>B as shown in figure.
\nFrom the right angle \u0394S1<\/sub>AP
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-14.png\")
\nBut we know constructive interference takes place at P, So we can take
\n(S2<\/sub>P – S1<\/sub>P) = n\u03bb
\nHence eq (1) can be written as
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-15.png\")
\nLet xn + 1<\/sub> be the distance of (n + 1)th<\/sup> bright band from centre O, then we can write
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-16.png\")
\nThis is the width of the bright band. It is the same for the dark band also.<\/p>\n
\nThe schematic diagram of an experimental setup to study the wave nature of electron is shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-17.png\")
\na) Identify the experiment.
\nb) Explain how this experiment verified the wave nature of electrons.
\nAnswer:
\na) Davisson and Germer Experiment
\nb)
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-18.png\")
\nExperimental setup: The Davisson and Germer Experiment consists of filament \u2018F\u2019, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. \u2019N\u2019 is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.<\/p>\n
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-19.png\")
\nThe graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of crystal). Thus wave nature of electron is established.<\/p>\n
\n2d sin\u03b8 = n\u03bb …….. (1)
\nFrom the figure, we get
\n\u03b8 + \u03a6 + \u03b8 = 180\u00b0
\n2\u03b8 = 180 – \u03a6, 2\u03b8 = 180 – 50\u00b0
\n\u03b8 = 65\u00b0
\nfor n = 1
\nequation (1) becomes
\n\u03bb = 2d sin\u03b8 ……….. (2)
\nfor Ni crystal, d = 0.91 A\u00b0
\nSubstituting this in eq. (2), we get
\nwavelength \u03bb = 1.65 A\u00b0<\/p>\n
\nThe accelerating voltage is 54 V
\nEnergy of electron E = 54 \u00d7 1.6 \u00d7 10-19<\/sup>\u00a0J
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-20.png\")
<\/p>\n
\nThe energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus.
\na) Write an expression for binding energy in terms of mass defect.
\nb) Draw the graph showing the variation of binding energy per nucleon as a function of mass number.
\nc) Which nucleus possess maximum binding energy per nucleon?
\nAnswer:
\na) BE = \u0394 mc2<\/sup> or
\nBE = (Zmp<\/sub> + (A – Z) mn<\/sub> – M)c2<\/sup>
\nb) Graph
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-21.png\")
\nc) Fe (nucleus of iron)<\/p>\n
\nNiels Bohr made certain modification in Rutherford\u2019s model by adding the ideas of quantum hypothesis.
\na) State Bohr\u2019s second postulate of quantisation of angular momentum.
\nb) Derive an expression for the radius and energy of the electron in the nth orbit of hydrogen atom.
\nAnswer:
\na) The orbital angular momentum of electron is an integral multiple of \\(\\frac{\\mathrm{h}}{2 \\pi}\\)<\/p>\n
\nThe force of attraction between the nucleus and the electron is
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-22.png\")
\nThis force provides the centripetal force for the orbiting electron
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-23.png\")
\nAccording to Bohr\u2019s second postulate, we can write
\nAngular momentum, mvr = \\(\\frac{\\mathrm{nh}}{2 \\pi}\\)
\ni.e., v = \\(\\frac{\\mathrm{nh}}{2 \\pi \\mathrm{mr}}\\)
\nSubstituting this value of V in equation (2), we get
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-24.png\")
\nEnergy of the hydrogen atom:
\nThe K.E. of revolving electron is
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-25.png\")
\nSubstituting the value of equation (5) in equation (9)
\nwe get
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-26.png\")
<\/p>\n
\nTwo long co-axial solenoids of same length are shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-27.png\")
\na) Define mutual inductane of the pair of coils.
\nb) Derive an expression for mutual inductance of two co-axial solenoids.
\nc) Write the dimension of mutual inductance.
\nAnswer:
\na) \u03a6 = MI, when I = 1 A, \u03a6 = M
\nThe mutual inductance of a pair of coils is numerically equal to the magnetic flux linked with one coil when unit current flows through the other.<\/p>\n
\nFlux density of the first coil B = \u03bc0<\/sub>nI
\nFlux linked with second coil, \u03a6 = BAN
\n\u03a6 = \u03bc0<\/sub>nIAN ………. (1)
\nBut we know \u03a6 = MI ………. (1)
\nFrom eq (1) and eq (2) weget
\n\u2234 MI = \u03bc0<\/sub>nIAN
\nM = \u03bc0<\/sub>nAN
\nIf the solenoid is covered over core of relative per-meability \u03bcr<\/sub>
\nthen M = \u03bcr<\/sub>\u03bc0<\/sub>nAN
\nc) ML2<\/sup>T-2<\/sup>A-2<\/sup><\/p>\n
\nA small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when
\na) the telescope is in normal adjustment.
\nb) the final image is formed at the least distance of distinct vision.
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-28.png\")
<\/p>\n
\nIn Amplitude Modulation, the amplitude of the carrier wave is varied in accordance with the information signal.
\na) What is meant by modulation index?
\nb) A message signal of frequency 10 kHz and peak value of 10 V used to modulate a carrier of frequency 1 MHz and peak voltage of 20 V. Determine the modulation index.
\nc) The block diagram of a transmitter is shown below. Identify the elements labelled X and Y.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-29.png\")
\nAnswer:
\na) Modulation index, \u03bc = \\(\\frac{A_{m}}{A_{c}}\\)
\nb) Am = 10 V, Ac = 20 V
\n\u2234 \u03bc = \\(\\frac{A_{m}}{A_{c}}=\\frac{10}{20}\\) = 0.5<\/p>\n
\nCyclotron is a device used to accelerate charged particles.
\na) With a suitable diagram briefly explain the working of a cyclotron and obtain an expression for cyclotron frequency.
\nb) Acyclotron oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons?
\nAnswer:
\na) Principles: Cyclotron is based on two facts<\/p>\n\n
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-30.png\")
<\/p>\n
\nIf the ion moves along a semicircular path of radius Y, then we can write
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-31.png\")
\nEq. (2) shows that time is independent of radius and velocity.<\/p>\n
\nT = 2t
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-32.png\")
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-33.png\")
<\/p>\n
\nThe experimental set up to find an unknown resistance using a metre bridge is shown below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-34.png\")
\na) What is the principle of a metre bridge?
\nb) If the balance point is found to be at 39.5 cm from the end \u2018A\u2019, the resistor ‘S\u2019 is of 12.5 \u03a9. Determine the resistance \u2018R\u2019. Why are the connections between resistors in a metre bridge made of thick copper strips?
\nc) If the galvanometer and cell are interchanged at the balance point of the bridge would the galvanometer show any current?
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-35.png\")
\nThe resistors in metre bridge are made of thick copper strips to minimise the resistance of connection.
\nc) No. The galvanometer will not show any current.<\/p>\n
\nThe circuit used to change alternating voltage to direct voltage is called rectifier.
\na) With a neat diagram, explain the working of a full wave rectifier having two diodes.
\nb) What is the output frequency of a full wave rectifier if the input frequency is 50 Hz?
\nc) Draw the output wave form across the load resistance connected in the full wave rectifier circuit.
\nAnswer:
\na) Full wave rectifier:
\nCircuit details
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-36.png\")
\nFull wave rectifier consists of transformer, two diodes and a load resistance RL<\/sub>. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1<\/sub> and D2<\/sub>. The output is taken across RL<\/sub>.<\/p>\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-37.png\")
<\/p>\n
\nA ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure below:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-38.png\")
\na) Identify the type of mirror used in the diagram.
\nb) By drawing a suitable ray diagram, obtain the mirror equation.
\nc) If the mirror is immersed in water, its focal length ………….
\nAnswer:
\na) Concave mirror<\/p>\n
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-39.png\")
\nLet points P, F, C be pole, focus and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F. Another ray of light from B incident at pole P and then reflected. These two rays meet at M. The ray of light from point B is passed through C. Draw EN perpendicular to the principal axis.
\n\u0394 IMF and \u0394 ENF are similar.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-40.png\")
\nbut IF = PI – PF and NF = PF (since aperture is small)
\nhence eq. (1) can be written as
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2019-41.png\")
\nThis is called mirror formula or mirror equation,<\/p>\nPlus Two Physics Previous Year Question Papers and Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"