{"id":22666,"date":"2024-01-20T09:48:36","date_gmt":"2024-01-20T04:18:36","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22666"},"modified":"2024-01-22T11:35:03","modified_gmt":"2024-01-22T06:05:03","slug":"ap-inter-2nd-year-physics-study-material-chapter-11","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-physics-study-material-chapter-11\/","title":{"rendered":"AP Inter 2nd Year Physics Study Material Chapter 11 Electromagnetic Waves"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material<\/a> 11th Lesson Electromagnetic Waves Textbook Questions and Answers.<\/p>\n

AP Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nWhat is the average wavelength of X-rays? (A.P. Mar. ’16 )
\nAnswer:
\nThe wavelength range of X-rays is from 10-8<\/sup>m(10nm) to 10-13<\/sup> m (10-4<\/sup> nm).
\nAverage wavelength of X – rays = \\(\\frac{10+0.0001}{2}\\) = 5.00005nm.<\/p>\n

Question 2.
\nGive anyone use of infrared rays. (T.S. Mar. ’19)
\nAnswer:<\/p>\n

    \n
  1. Infrared radiation plays an important role in maintaining the Earth warm.<\/li>\n
  2. Infrared lamps are used in physical therapy.<\/li>\n
  3. Infrared detectors are used in Earth Satellites.<\/li>\n
  4. These are used in taking photographs during conditions of fog, smoke, etc.<\/li>\n<\/ol>\n

    \"AP<\/p>\n

    Question 3.
    \nIf the wavelength of electromagnetic radiation is doubled, what happens to the energy of photon? (T.S. Mar. ’16)
    \nAnswer:
    \nPhoton energy (E) = hv = \\(\\frac{\\mathrm{hc}}{\\lambda}\\)
    \nE \u221d \\(\\frac{1}{\\lambda}\\)
    \nGiven \u03bb1<\/sub> = \u03bb, \u03bb2<\/sub> = 2\u03bb, E1<\/sub> = E
    \n\\(\\frac{\\mathrm{E}_1}{\\mathrm{E}_2}\\) = \\(\\frac{\\lambda_2}{\\lambda_1}\\)
    \n\\(\\frac{E}{\\mathrm{E}_2}\\) = \\(\\frac{2 \\lambda}{\\lambda}\\)
    \nE2<\/sub> = E\/2
    \n\u2234 The energy of photon reduces to half of its initial value.<\/p>\n

    Question 4.
    \nWhat is the principle of production of electromagnetic waves ?
    \nAnswer:
    \nIf the charge is accelerated both the magnetic field and electric field will change with space and time, then electromagnetic waves are produced.<\/p>\n

    Question 5.
    \nWhat is the ratio of speed of infrared rays and ultraviolet rays in vacuum ?
    \nAnswer:
    \nThe ratio of speed of infrared rays and ultraviolet rays in vacuum is 1 : 1.
    \nAll electromagnetic waves travel with same speed 3 \u00d7 108<\/sup> m \/s in vacuum.<\/p>\n

    Question 6.
    \nWhat is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave ?
    \nAnswer:
    \nE0<\/sub> = CB0<\/sub>
    \nWhere E0<\/sub> = Amplitude of electric field.
    \nB0<\/sub> = Amplitude of magnetic field.
    \nC = velocity of light.<\/p>\n

    \"AP<\/p>\n

    Question 7.
    \nWhat are the applications of microwaves ? (T.S. Mar. ’15)
    \nAnswer:<\/p>\n

      \n
    1. Microwaves are used in Radars.<\/li>\n
    2. Microwaves are used for cooking purposes.<\/li>\n
    3. A radar using microwave can help in detecting the speed of automobile while in motion.<\/li>\n<\/ol>\n

      Question 8.
      \nMicrowaves are used in Radars, why ? (Mar. ’14)
      \nAnswer:
      \nAs microwaves are of smaller wavelengths, hence they can be transmitted as a beam signal in a particular direction. Microwaves do not bend around the comers of any obstacle coming in their path.<\/p>\n

      Question 9.
      \nGive two uses of infrared rays. (A.P. Mar. ’19)
      \nAnswer:<\/p>\n

        \n
      1. Infrared rays are used for producing dehydrated fruits.<\/li>\n
      2. They are used in the secret writings on the ancient walls.<\/li>\n
      3. They are used in green houses to keep the plants warm.<\/li>\n<\/ol>\n

        Question 10.
        \nThe charging current for a capacitor is 0.6 A. What is the displacement current across its plates?
        \nAnswer:
        \ni = charging current for a capacitor = 0.6 A
        \ni = id = \\(\\varepsilon_0 \\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\)
        \n\u2234 i = i = id<\/sub> = 0.6A
        \n\u2234 Displacement current (id<\/sub>) = 0.6 A.<\/p>\n

        Short Answer Questions<\/span><\/p>\n

        Question 1.
        \nWhat does an electromagnetic wave consists of ? On what factors does its velocity in vacuum depend ?
        \nAnswer:
        \nMaxwell concluded that the variation in electric and magnetic field vectors perpendicular to each other leads to the production of electromagnetic waves in space. They can travel in space even without any material medium. These waves are called electromagnetic waves.<\/p>\n

        According to Maxwell, electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angles to each other as well as at right angles to the direction of wave propagation. Thus electomagnetic waves have transverse nature.<\/p>\n

        Electric field Ex<\/sub> = E0<\/sub> Sin (kz – \u03c9t)
        \nMagnetic field By<\/sub> = B0<\/sub> sin (kz – \u03c9t)
        \nWhere K is propagation constant (K = \\(\\frac{2 \\pi}{\\lambda}\\))
        \nThe velocity of electromagnetic waves C = \\(\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}\\)<\/p>\n

        Velocity of E.M waves depends on<\/p>\n

          \n
        1. Permeability in free space (\u03bc0<\/sub>).<\/li>\n
        2. Permittivity in free space (\u03b50<\/sub>).<\/li>\n<\/ol>\n

          Velocity of e.m waves is 3 \u00d7 108<\/sup> m \/ s.<\/p>\n

          \"AP<\/p>\n

          Question 2.
          \nWhat is Greenhouse effect and its contribution towards the surface temperature of earth ?
          \nAnswer:
          \nGreen house effect: Temperature of the earth increases due to the radiation emitted by the earth is trapped by atmospheric gases like CO2<\/sub>, CH4<\/sub>, N2<\/sub>, Chlorofluoro carbons etc is called green house effect.<\/p>\n

            \n
          1. Radiation from the sun enters the atmosphere and heat the objects on the earth. These heated objects emit infrared rays.<\/li>\n
          2. These rays are reflected back to Earth’s surface and trapped in the Earth’s atmosphere. Due to this temperature of the earth increases.<\/li>\n
          3. The layers of carbon dioxide (CO2<\/sub>) and low lying clouds prevent infrared rays to escape Earth’s atmosphere.<\/li>\n
          4. Since day-by-day the amount of carbondioxide in the atmosphere increases, more infrared rays are entrapped in the atmosphere.<\/li>\n
          5. Hence the temperature of the Earth’s surface increases day by day.<\/li>\n<\/ol>\n

            Long Answer Questions<\/span><\/p>\n

            Question 1.
            \nGive the brief history of discovery of knowledge of electromagnetic waves.
            \nAnswer:<\/p>\n

              \n
            1. Faraday from his experimental study of electromagnetic induction magnetic field changing with time, gives rise to an electric field.<\/li>\n
            2. Maxwell in 1865 from his theoritical study concluded that, an electric field changing with time gives rise to magnetic field.<\/li>\n
            3. It is a consequence of the displacement current being a source of magnetic field.<\/li>\n
            4. It means a change in electric (or) magnetic field with time produces the other field.<\/li>\n
            5. Maxwell concluded that the variation in electric and magnetic field vectors perpendicular to each other leads to the production of electromagnetic waves in space.<\/li>\n
            6. These electromagnetic waves travel in space without any material medium.<\/li>\n
            7. Both electric and magnetic fields vary with time and space and have the same frequency.
              \n\"AP<\/li>\n
            8. The electric field vector \\(\\overrightarrow{\\mathrm{E}}\\) and magnetic field vector \\(\\overrightarrow{\\mathrm{B}}\\) are vibrating along y and z axis and propagation of electromagnetic waves along x – axis.<\/li>\n
            9. Maxwell found that the electromagnetic waves travel in vacuum with a speed is given by
              \n\"AP
              \nWhere \u03bc0<\/sub> = 4\u03c0 \u00d7 10-7<\/sup> H\/m = permeability in free space.
              \n\u03b50<\/sub> = 8.85 \u00d7 10-12<\/sup> c2<\/sup> N-1<\/sup> m-2<\/sup> = permittivity in free space.<\/li>\n
            10. The velocity of electromagnetic waves in a medium is given by v = \\(\\frac{1}{\\sqrt{\\mu \\varepsilon}}\\)<\/li>\n
            11. Maxwell also concluded that electromagnetic waves are transverse in nature.<\/li>\n
            12. In 1888 Hertz demonstrated experimentally the production and detection of E.M waves using spark oscillator.<\/li>\n
            13. In 1895 Jagadish Chandra Bose was able to produce E.M waves of wavelength 5m.m to 25 m.m.<\/li>\n
            14. 1899 Marconi was the first to establish a wireless communication at a distance of about 50 km.<\/li>\n<\/ol>\n

              \"AP<\/p>\n

              Question 2.
              \nState six characteristics of electromagnetic waves. What is Greenhouse effect ?
              \nAnswer:
              \nCharacteristics (or) properties of electromagnetic waves :<\/p>\n

                \n
              1. Electromagnetic waves do not require any material medium for their propagation. They propagate in vacuum as well as in a medium.<\/li>\n
              2. Speed of E.M. waves in free space (or) vacuum is given by
                \nC = \\(\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}\\) = 3 \u00d7 108<\/sup> m\/s.<\/li>\n
              3. Speed of E.M waves in a medium is given by
                \nv = \\(\\frac{1}{\\sqrt{\\mu \\varepsilon}}\\)<\/li>\n
              4. Electromagnetic waves are transverse in nature.
                \nElectric field \\(\\overrightarrow{\\mathrm{E}}\\) and magnetic field \\(\\overrightarrow{\\mathrm{B}}\\) which constitute the E.M waves an mutually perpendicular to each other as well as perpendicular to the direction of propagation of the wave.<\/li>\n
              5. Electromagnetic waves are self sustaining electric and magnetic field oscillations in space.<\/li>\n
              6. Electromagnetic waves transport energy.
                \nPoynting vector (\\(\\overrightarrow{\\mathrm{P}}\\)) = \\(\\frac{1}{\\mu_0}(\\overrightarrow{\\mathrm{E}} \\times \\overrightarrow{\\mathrm{B}})\\)<\/li>\n
              7. Relation between electric field vector E and magnetic field vector g in vacuum is given by
                \nC = \\(\\frac{\\mathrm{E}_0}{\\mathrm{~B}_0}\\)<\/li>\n
              8. Electromagnetic waves are not deflected by magnetic and electric fields.<\/li>\n
              9. Electromagnetic waves can be reflected, refracted, interferenced, diffracted and polarised.<\/li>\n
              10. Electromagnetic wave follow the superposition principle.<\/li>\n
              11. Average electric energy density of E.M wave is given by
                \nUav<\/sub> = UE<\/sub> + UB<\/sub>
                \nUav<\/sub> = \\(\\frac{1}{2} \\varepsilon_0 \\mathrm{E}^2\\) + \\(\\frac{1}{2} \\cdot \\frac{\\mathrm{B}^2}{\\mu_0}\\)
                \nUav<\/sub> = 2UE<\/sub> = 2UB<\/sub><\/li>\n
              12. Intensity of an E.M waves depends on its average energy density.
                \nI = \\(\\frac{1}{2} \\varepsilon_0 \\mathrm{C} \\mathrm{E}_0^2\\)<\/li>\n
              13. E.M. waves carry momentum and exert radiation pressure is given by
                \nP = \\(\\frac{\\mathrm{F}}{\\mathrm{A}}\\) = \\(\\frac{1}{\\mathrm{~A}} \\frac{\\mathrm{dp}}{\\mathrm{dt}}\\) = \\(\\frac{\\text { Intensity (I) }}{C}\\)<\/li>\n<\/ol>\n

                Green house effect:
                \nThe temperature of the Earth increases due to radiation emitted by the Earth.is trapped by atmospheric gases like CO2<\/sub>, CH4<\/sub>, N2<\/sub>O, chlorofluoro carbons etc is called green house effect.<\/p>\n

                Textual Exercises<\/span><\/p>\n

                Question 1.
                \nThe figure shows a capacitor made of two circular plates each of radius 12cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
                \na) Calculate the capacitance and the rate of charge of potential difference between the plates.
                \n\"AP
                \nb) Obtain the displacement current across the plates.
                \nc) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor ? Explain.
                \nAnswer:
                \nGiven \u03b50<\/sub> = 8.85 \u00d7 10-12<\/sup> C2<\/sup> N-1<\/sup> m-2<\/sup>
                \nHere, R = 12cm = 0.12m, d = 5.0mm = 5 \u00d7 10-3<\/sup>m, I = 0.15A
                \nArea, A = \u03c0R2<\/sup> = 3.14 \u00d7 (0.12)2<\/sup>m2<\/sup><\/p>\n

                a) We know that capacity of a parallel plate capacitor is given by
                \n\"AP
                \nb) Displacement current is equal to conduction current i.e., 0.15 A.
                \nc) Yes, Kirchhoffs first rule is valid at each plate of the capacitor provided. We take the current to be the sum of the conduction and displacement currents.<\/p>\n

                Question 2.
                \nA parallel plate capacitor in the figure 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 a (angular) frequency of 300 rad s-1<\/sup>.
                \na) What is the rms value of the conduction current ?
                \nb) Is the conduction current equal to the displacement current ?
                \nc) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
                \n\"AP
                \nAnswer:
                \na) Irms<\/sub> = \\(\\frac{E_{\\mathrm{fms}}}{X_c}\\) = \\(\\frac{E_{\\mathrm{rms}}}{\\frac{1}{\\omega C}}\\) = Erms<\/sub> \u00d7 \u03c9C
                \n\u2234 Irms<\/sub> = 230 \u00d7 300 \u00d7 100 \u00d7 10-12<\/sup> = 6.9 \u00d7 10-6<\/sup>A = 6.9\u00b5A<\/p>\n

                b) Yes, I = Id<\/sub> where I is steady d.c or a.c. This is shown below
                \n\"AP<\/p>\n

                c) We know, B = \\(\\frac{\\mu_0}{2 \\pi} \\times \\frac{\\pi}{R^2} \\times I_d\\)
                \nThe formula is valid even if Id<\/sub> is oscillating. As Id<\/sub> = I, therefore
                \nB = \\(\\frac{\\mu_0 r \\mathrm{I}}{2 \\pi \\mathrm{R}^2}\\)
                \nIf I = I0<\/sub>, the maximum value of current, then
                \nAmplitude of B = max. value of B = \\(\\frac{\\mu_0 \\mathrm{rI}_0}{2 \\pi \\mathrm{R}^2}\\) = \\(\\frac{\\mu_0 \\mathrm{r} \\sqrt{2} \\mathrm{I}_{\\mathrm{rms}}}{2 \\pi \\mathrm{R}^2}\\)
                \n= \\(\\frac{4 \\pi \\times 10^{-7} \\times 0.03 \\times \\sqrt{2} \\times 6.9 \\times 10^{-6}}{2 \\times 3.14 \\times(0.06)^2}\\) = 1.63 \u00d7 10-11<\/sup>T.<\/p>\n

                \"AP<\/p>\n

                Question 3.
                \nWhat physical quantity is the same for X-rays of wavelength 10-10<\/sup>m, red light of wavelength 6800 A and radiowaves of wavelength 500m?
                \nAnswer:
                \nThe speed in vacuum is same for all the given wavelengths, which is 3 \u00d7 108<\/sup> m\/s.<\/p>\n

                Question 4.
                \nA plane electromagnetic wave travels In vacuum alo\u00f1g z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz, what is its wavelength?
                \nAnswer:
                \nIn electromagnetic wave, the electric field vector \\(\\overrightarrow{\\mathrm{E}}\\) and magnetic field vector \\(\\overrightarrow{\\mathrm{B}}\\) show their variations perpendicular to the direction of propagation of wave as well as perpendicular to each other. As the electromagnetic wave is travelling along z – direction, hence \\(\\overrightarrow{\\mathrm{E}}\\) and \\(\\overrightarrow{\\mathrm{B}}\\) show their variation in x – y plane.
                \nwave length \u03bb = \\(\\frac{\\mathrm{c}}{\\mathrm{v}}\\) = \\(\\frac{3 \\times 10^8 \\mathrm{~m} \/ \\mathrm{s}}{30 \\times 10^6 \\mathrm{~s}^{-1}}\\) = 10m<\/p>\n

                Question 5.
                \nA radio can tune into any station in the 7.5 MHz to 12MHz band. What is-the corresponding wavelength band?
                \nAnswer:
                \n\u03bb1<\/sub> = \\(\\frac{3 \\times 10^8}{7.5 \\times 10^6}\\) = 40m
                \n\u03bb2<\/sub> = \\(\\frac{3 \\times 10^8}{12 \\times 10^6}\\) = 25m
                \nThus wavelength band is 40m to 25m.<\/p>\n

                Question 6.
                \nA charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator ?
                \nAnswer:
                \nThe frequency of electromagnetic wave is same as that of oscillating cliarged particle about its equilibrium position; which is 109<\/sup>Hz.<\/p>\n

                Question 7.
                \nThe amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0<\/sub> = 510 nT. What is the amplitude of the electric field part of the wave ?
                \nAnswer:
                \nHere, B0<\/sub> = 510nT = 510 \u00d7 10-9<\/sup>T
                \nE0<\/sub> = CB0<\/sub> = 3 \u00d7 108<\/sup> \u00d7 510 \u00d7 10-9<\/sup> = 153 NC-1<\/sup>.<\/p>\n

                Question 8.
                \nSuppose that the electric field amplitude of an electromagnetic wave is E0<\/sub> = 120 N\/C and that its frequency is v = 50.0 MHz.
                \n(a) Determine. B0<\/sub>, \u03c9, k, and \u03bb.
                \n(b) Find expressions for E and B.
                \nAnswer:
                \na)
                \nB0<\/sub> = \\(\\frac{E_0}{\\mathrm{c}}\\) = \\(\\frac{120}{3 \\times 10^8}\\) = 4 \u00d7 10-7<\/sup> T
                \n\u03c9 = 2\u03c0v = 2 \u00d7 3.14 \u00d7 (50 \u00d7 106<\/sup>) = 3.14 \u00d7 108<\/sup> rad\/s
                \nK = \\(\\frac{\\omega}{\\mathrm{C}}\\) = \\(\\frac{3.14 \\times 10^8}{3 \\times 10^8}\\) = 1.05 rad\/m
                \n\u03bb = \\(\\frac{C}{V}\\) = \\(\\frac{3 \\times 10^8}{50 \\times 10^6}\\) = 6.00 m<\/p>\n

                b) Expression for \\(\\overrightarrow{\\mathrm{E}}\\) is E = E0<\/sub> sin (kx – \u03c9t)
                \n= (120 N\/c) Sin [(1.05 rad\/m) x – (3.14 \u00d7 108 rad \/s)t] \\(\\hat{\\mathrm{j}}\\)
                \nExpression for \\(\\overrightarrow{\\mathrm{B}}\\) is B = B0<\/sub> sin (kx – \u03c9t)
                \n= (4 \u00d7 10-7<\/sup> T) sin [(1.05 rad\/m) x – (3.14 \u00d7 108<\/sup> rad\/s)t] \\(\\hat{\\mathrm{k}}\\).<\/p>\n

                \"AP<\/p>\n

                Question 9.
                \nThe terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum, in what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation ?
                \nAnswer:
                \nEnergy of a photon of frequency v is given by E = hv joules = \\(\\frac{\\mathrm{hv}}{1.6 \\times 10^{-19} \\mathrm{ev}}\\)
                \nWhere h = 6.6 \u00d7 10-34<\/sup> J. The energy of photon of different parts of electromagnetic spectrum in joules and eV are shown in table below, along with their sources of origin.
                \n\"AP
                \n\"AP<\/p>\n

                Question 10.
                \nIn a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 \u00d7 1010<\/sup> Hz and amplitude 48 V m-1<\/sup>.
                \na) What is the wavelength of the wave ?
                \nb) What is the amplitude of the oscillating magnetic field ?
                \nc) Show that the average energy density of the E field equals the average energy density of the B field. [C = 3 \u00d7 108<\/sup> ms-1<\/sup>].
                \nAnswer:
                \nHere, v = 2.0 \u00d7 1010<\/sup>Hz, E0<\/sub> = 48 Vm-1<\/sup>, C = 3 \u00d7 108<\/sup> m\/s<\/p>\n

                a) wavelength of the wave, \u03bb = \\(\\frac{C}{v}\\) = \\(\\frac{3 \\times 10^8}{2.0 \\times 10^{10}}\\) = 1.5 \u00d7 10-2<\/sup> m<\/p>\n

                b) Amplitude of oscillating magnetic field,
                \nB0<\/sub> = \\(\\frac{E_0}{C}\\) = \\(\\frac{48}{3 \\times 10^8}\\) = 1.6 \u00d7 10-7<\/sup>T<\/p>\n

                c) For average energy density
                \nUE<\/sub> = \\(\\frac{1}{2} \\varepsilon_0 \\mathrm{E}_0^2\\) …… (1)
                \nWe know that \\(\\frac{E_0}{B_0}\\) = C
                \nPutting in Eq (1)
                \nUE<\/sub> = \\(\\frac{1}{4} \\varepsilon_0 \\cdot C^2 \\mathrm{~B}_0^2\\) …. (2)
                \nSpeed of Electro magnetic waves, C = \\(\\frac{1}{\\sqrt{\\mu_0 \\mathrm{E}_0}}\\)
                \nPutting in Eq (2) We get.
                \nUE<\/sub> = \\(\\frac{1}{4} \\varepsilon_0 B_0^2 \\cdot \\frac{1}{\\mu_0 \\varepsilon_0}\\)
                \nUE<\/sub> = \\(\\frac{1}{4} \\cdot \\frac{\\mathrm{B}_{\\mathrm{O}}^2}{\\mu_0}=\\frac{\\mathrm{Bo}^2}{2 \\mu_0}\\) = \u03bcB<\/sub>
                \nThus, the average energy density of E field equals the average energy density of B field.<\/p>\n

                Additional Exercises<\/span><\/p>\n

                Question 1.
                \nSuppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N\/C) cos [(1.8 rad\/m) y + {5.4 \u00d7 106<\/sup> rad \/s} t]} i.
                \na) What is the direction of propagation ?
                \nb) What is the wavelength \u03bb ?
                \nc) What is the frequency v ?
                \nd) What is the amplitude of the magnetic field part of the wave ?
                \ne) Write an expression for the magnetic field part of the wave.
                \nAnswer:
                \na) From the given question it is clear that direction of motion of e.m. wave is along negative y direction i.e along – \\(\\hat{\\mathrm{j}}\\)<\/p>\n

                b) Comparing the given question with equation E = E0<\/sub> cos (ky + \u03c9t).
                \nWe have, K = 1.8 rad\/m, \u03c9 = 5.4 \u00d7 108<\/sup> rad\/s, E0<\/sub> = 3.1 N\/C
                \n\u03bb = \\(\\frac{2 \\pi}{\\mathrm{k}}\\) = \\(\\frac{2 \\times(22 \/ 7)}{1.8}\\) = 3.492 m \u2248 3.5m<\/p>\n

                c) V = \\(\\frac{\\omega}{2 \\pi}\\) = \\(\\frac{5.4 \\times 10^8}{2 \\times\\left(\\frac{22}{7}\\right)}\\) = 85.9 \u00d7 106<\/sup> \u2248 86MHz<\/p>\n

                d) B0<\/sub> = \\(\\frac{\\mathrm{E}_0}{\\mathrm{C}}\\) = \\(\\frac{3.1}{3 \\times 10^8}\\) = 1.03 \u00d7 10-8<\/sup>T \u2248 10.3nT<\/p>\n

                e) B = B0 cos (ky + \u03c9t) \\(\\hat{\\mathbf{k}}\\) = (10.3nT) cos [(1.8 rad\/m\/y + (5.4 \u00d7 108<\/sup> rad \/ s)t] \\(\\hat{\\mathbf{k}}\\)<\/p>\n

                Question 2.
                \nAbout 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
                \na) at a distance of lm from the bulb ?
                \nb) at a distance of 10 m ?
                \nAssume that the radiation is emitted isotropically and neglect reflection.
                \nAnswer:
                \na) Intensity,
                \n\"AP
                \nb) I = \\(\\frac{100 \\times(5 \/ 100)}{4 \\pi(10)^2}\\) = 4 \u00d7 10-3<\/sup> W\/m2<\/sup><\/p>\n

                Question 3.
                \nUse the formula \u03bbm<\/sub> T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you ?
                \nAnswer:
                \nWe know, every body at given temperature T1<\/sub> emits radiations of all wavelengths in certain range. For a black body, the wavelength corresponding to maximum intensity of radiation at a given temperature.
                \n\u03bbm<\/sub> T = 0.29cmk or T = \\(\\frac{0.29}{\\lambda_{\\mathrm{m}}}\\)
                \nFor \u03bbm<\/sub> = 10-6<\/sup>m = 10-4<\/sup>cm, T = \\(\\frac{0.29}{10^{-4}}\\) = 2900 k.<\/p>\n

                Temperature for other wavelengths can be similarly found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of e.m spectrum. Thus to obtain visible radiation, say, \u03bbm<\/sub> = 5 \u00d7 10-5<\/sup>cm, the source should have a temperature
                \nT = \\(\\frac{0.29}{5 \\times 10^{-5}}\\) = 6000 k
                \nIt is to be noted that, a body at lower temperature will also produce this wavelength but not with maximum intensity.<\/p>\n

                Question 4.
                \nGiven below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
                \na) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
                \nb) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
                \nc) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
                \nd) 5890 A – 5896 A [double lines of sodium]
                \ne) 14.4 keV [energy of a particular transition in 57<\/sup>Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy).
                \nAnswer:
                \na) This wavelength corresponds to radiowaves.<\/p>\n

                b) This frequency also corresponds to radiowaves.<\/p>\n

                c) Given T = 2.7 K As \u03bbm<\/sub> T = 0.29cm \u00b0k
                \n\u2234 \u03bbm<\/sub> = \\(\\frac{0.29}{\\mathrm{~T}}\\) = \\(\\frac{0.29}{2.7}\\) \u2248 0.11cm
                \nThis wavelength corresponds to microwave region of the electromagnetic spectrum.<\/p>\n

                d) This wavelength lies in the visible region of the electromagnetic spectrum.<\/p>\n

                e) Here, Energy E = 14.4KeV = 14.4 \u00d7 103<\/sup> \u00d7 1.6 \u00d7 10-19<\/sup>J
                \nAs E = hv
                \n\u2234 v = \\(\\frac{E}{h}\\) = \\(\\frac{14.4 \\times 10^3 \\times 1.6 \\times 10^{-19}}{6.6 \\times 10^{-34}}\\) \u2248 3 \u00d7 1011<\/sup> MHz
                \nThis frequency lies in the X-ray region of electromagnetic spectrum.<\/p>\n

                \"AP<\/p>\n

                Question 5.
                \nAnswer the following questions :
                \na) Long distance radio broadcasts use short-wave bands. Why ?
                \nb) It is neccessary to use satellites for long distance TV transmission. Why ?
                \nc) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the earth. Why ?
                \nd) The small ozone layer on top of the stratosphere is crucial for human survival. Why ?
                \ne) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now ?
                \nf) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction ?
                \nAnswer:
                \na) It is so because ionosphere reflects the waves in these bands.<\/p>\n

                b) Yes, television signals being of high frequency are not reflected by ionosphere, Therefore to reflect them satellites are needed. That is why, satellites are used for long distance TV transmission.<\/p>\n

                c) Optical and radiowaves can penetrate the atmosphere where as X-rays being of much smaller wavelength are absorbed by the atmosphere. That is why we can work with optical and radio telescopes on earth’s surface but X-ray astronomical telescopes must be used on the satellite orbiting above the earth’s atmosphere.<\/p>\n

                d) The small ozone layer present on the top of the stratosphere absorbs most of ultraviolet radiations from the sun which are dangerous and cause genetic damage to living cells, prevents them from reaching the earth’s surface and thus helps in the survival of life.<\/p>\n

                e) The temperature of earth would be lower because the green house effect of atmosphere would be absent.<\/p>\n

                f) The clouds by a global nuclear war would perhaps cover most parts of sky preventing solar light from reaching many parts of globe. This would cause a winter.<\/p>\n","protected":false},"excerpt":{"rendered":"

                Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves Textbook Questions and Answers. AP Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves Very Short Answer Questions Question 1. What is the average wavelength of X-rays? (A.P. Mar. ’16 ) Answer: The wavelength range of X-rays is from 10-8m(10nm) … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22666"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22666"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22666\/revisions"}],"predecessor-version":[{"id":22667,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22666\/revisions\/22667"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22666"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22666"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22666"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}