GSEB Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

GSEB Solutions Class 12 Physics Chapter 8 Electromagnetic Waves

 

Question 1.Figure shows a capacitor made of two circular plates each of radius 12 cm 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 0.15
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समानांतर प्लेट संधारित्र (capacitor) को दर्शाता है जिसमें दो गोलाकार प्लेटें हैं। प्लेटों के बीच की दूरी और उनकी त्रिज्या दी गई है, और यह संधारित्र एक बाहरी स्रोत से आवेशित हो रहा है, जिससे एक स्थिर धारा प्रवाहित हो रही है।
(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 (junction rule) valid at each plate of the capacitor?
Answer:Let's find the required values for the capacitor.
Given: Current, \(I = 0.15 \text{ A}\) Radius, \(r = 12 \text{ cm} = 12 \times 10^{-2} \text{ m}\) Distance between plates, \(d = 5.0 \text{ cm} = 5 \times 10^{-2} \text{ m}\)
Area of the plates, \(A = \pi r^2 = \pi \times (12 \times 10^{-2})^2 \text{ m}^2\) Permittivity of free space, \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\)
(a) To calculate the capacitance (C) and the rate of change of potential difference (\(\frac{dV}{dt}\)): The capacitance of a parallel plate capacitor is given by: \[C = \frac{\varepsilon_0 A}{d}\] \[C = \frac{8.85 \times 10^{-12} \times \pi \times (12 \times 10^{-2})^2}{5 \times 10^{-2}}\] \[C = \frac{8.85 \times 10^{-12} \times 3.14 \times 144 \times 10^{-4}}{5 \times 10^{-2}}\] \[C \approx 80.1 \times 10^{-12} \text{ F} = 80.1 \text{ pF}\]
The charge on the capacitor is \(q = CV\). The charging current is \(I = \frac{dq}{dt}\). So, \(I = \frac{d(CV)}{dt} = C \frac{dV}{dt}\). Therefore, the rate of change of potential difference is: \[\frac{dV}{dt} = \frac{I}{C}\] \[\frac{dV}{dt} = \frac{0.15}{80.1 \times 10^{-12}}\] \[\frac{dV}{dt} \approx 1.875 \times 10^9 \text{ V/s}\]
(b) The displacement current (\(I_D\)) across the plates is equal to the conduction current (I) in the circuit for a charging capacitor. So, the displacement current \(I_D = 0.15 \text{ A}\).
(c) Kirchhoff's first rule, also known as the junction rule or current rule, states that the total current entering a junction must equal the total current leaving it. This rule is based on the conservation of charge. For a charging capacitor, conduction current flows into one plate, and displacement current effectively flows between the plates. At each plate, the conduction current arriving (or leaving) is balanced by the displacement current being generated (or collapsing). Thus, Kirchhoff's first rule is valid at each plate of the capacitor.
In simple words: We calculated how much charge the capacitor can hold and how fast the voltage changes. The displacement current inside the capacitor is the same as the current flowing in the wires. Kirchhoff's rule works here because the current entering a plate is matched by the displacement current within the capacitor.

🎯 Exam Tip: Remember to use the correct units and values for permittivity and pi. For Kirchhoff's rule, understanding the concept of displacement current bridging the gap in a capacitor is key to explaining its validity.

 

Question 2.A parallel plate capacitor (figure) made of circular plates each of radius R = 6.0 cm has capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an angular frequency of 300 rad s\(^{-1}\).
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Answer:Let's find the values for the capacitor in an AC circuit.
Given: Radius, \(R = 6.0 \text{ cm} = 6 \times 10^{-2} \text{ m}\) Capacitance, \(C = 100 \text{ pF} = 100 \times 10^{-12} \text{ F}\) RMS voltage, \(V_{\text{rms}} = 230 \text{ V}\) Angular frequency, \(\omega = 300 \text{ rad s}^{-1}\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समानांतर प्लेट संधारित्र को दर्शाता है जिसे एक AC वोल्टेज स्रोत से जोड़ा गया है। प्लेटों के बीच विद्युत और चुंबकीय क्षेत्र उत्पन्न होते हैं क्योंकि संधारित्र चार्ज और डिस्चार्ज होता है।
(a) To find the rms value of the conduction current (\(I_{\text{rms}}\)): First, calculate the capacitive reactance (\(X_C\)): \[X_C = \frac{1}{C\omega}\] Then, the rms current is: \[I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \times C\omega\] \[I_{\text{rms}} = 230 \times 100 \times 10^{-12} \times 300\] \[I_{\text{rms}} = 6.9 \times 10^{-6} \text{ A} = 6.9 \text{ μA}\]
(b) Yes, in a charging or discharging capacitor, the conduction current in the wires connecting to the plates is equal to the displacement current within the capacitor plates. This ensures charge conservation.
(c) To determine the amplitude of the magnetic field (B) at a point 3.0 cm from the axis between the plates: Given `r' = 3.0 cm = 3 \times 10^{-2} m` (distance from axis) The magnetic field (B) inside a charging capacitor, at a distance \(r'\) from the axis, is given by Ampere-Maxwell's law: \[B = \frac{\mu_0 \varepsilon_0 r'}{2R^2} \frac{d\Phi_E}{dt}\] Where \(\frac{d\Phi_E}{dt} = \frac{I}{\varepsilon_0}\). So, \[B = \frac{\mu_0 I r'}{2\pi R^2}\] We need the amplitude of B, so we use the amplitude of current \(I_0 = \sqrt{2} I_{\text{rms}}\): \[I_0 = \sqrt{2} \times 6.9 \times 10^{-6} \text{ A} \approx 9.76 \times 10^{-6} \text{ A}\] Now, substitute the values: \[B = \frac{4\pi \times 10^{-7} \times 9.76 \times 10^{-6} \times 3 \times 10^{-2}}{2 \times \pi \times (6 \times 10^{-2})^2}\] \[B = \frac{2 \times 10^{-7} \times 9.76 \times 10^{-6} \times 3 \times 10^{-2}}{(6 \times 10^{-2})^2}\] \[B = \frac{2 \times 10^{-7} \times 9.76 \times 10^{-6} \times 3 \times 10^{-2}}{36 \times 10^{-4}}\] \[B = \frac{5.856 \times 10^{-13}}{36 \times 10^{-4}}\] \[B \approx 1.626 \times 10^{-10} \text{ T}\] \[B \approx 1.63 \times 10^{-10} \text{ T}\]
In simple words: We first found the alternating current flowing through the capacitor. The current measured in the wires is the same as the displacement current between the plates. Finally, we calculated the strength of the magnetic field created inside the capacitor plates, which changes with the current.

🎯 Exam Tip: For AC circuits, remember to calculate capacitive reactance first. The equality of conduction and displacement current is a fundamental concept. When calculating the magnetic field inside a capacitor, ensure you use the correct formula derived from Ampere-Maxwell's law.

 

Question 3.What is the physical quantity the same for X-rays of wavelength \(10^{-10}\) m, the red light of wavelength 6800 Å, and radiowaves of wavelength 500 m?
Answer:X-rays, red light, and radiowaves are all types of electromagnetic waves. In a vacuum, all electromagnetic waves travel at the same speed, which is the speed of light (approximately \(3 \times 10^8\) m/s). Therefore, the physical quantity that is the same for all of them when traveling in a vacuum is their speed.
In simple words: X-rays, red light, and radio waves are all light waves. In empty space, they all travel at the same speed, which is the speed of light.

🎯 Exam Tip: This question tests a basic understanding of the electromagnetic spectrum. Always remember that all electromagnetic waves travel at the speed of light in a vacuum, regardless of their wavelength or frequency.

 

Question 4.A plane electromagnetic wave travels in a vacuum along the 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?
Answer:When a plane electromagnetic wave moves in a vacuum along the z-direction, its electric field vector (\(\overrightarrow{\mathrm{E}}\)) and magnetic field vector (\(\overrightarrow{\mathrm{B}}\)) are mutually perpendicular to each other and also perpendicular to the direction of wave propagation. This means that if the wave travels along the z-axis, then \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) must lie in the x-y plane.
To find the wavelength (\(\lambda\)): Given frequency, \(\nu = 30 \text{ MHz} = 30 \times 10^6 \text{ Hz}\) Speed of light in vacuum, \(c = 3 \times 10^8 \text{ m/s}\)
The relationship between speed, frequency, and wavelength is: \[c = \nu \lambda\] So, the wavelength is: \[\lambda = \frac{c}{\nu}\] \[\lambda = \frac{3 \times 10^8 \text{ m/s}}{30 \times 10^6 \text{ Hz}}\] \[\lambda = \frac{3 \times 10^8}{3 \times 10^7}\] \[\lambda = 10 \text{ m}\]
In simple words: If an electromagnetic wave goes forward, its electric and magnetic fields wave up and down at a right angle to each other and to the direction it's moving. For a wave with a given frequency, we can easily find its length using the speed of light.

🎯 Exam Tip: Always remember that electric and magnetic fields in an EM wave are perpendicular to each other and to the direction of propagation. The formula \(c = \nu \lambda\) is fundamental for calculating wavelength or frequency.

 

Question 5.A radio can tune into any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?
Answer:Let's find the range of wavelengths corresponding to the given frequency band.
Given: Lower frequency, \(\nu_1 = 7.5 \text{ MHz} = 7.5 \times 10^6 \text{ Hz}\) Upper frequency, \(\nu_2 = 12 \text{ MHz} = 12 \times 10^6 \text{ Hz}\) Speed of light, \(c = 3 \times 10^8 \text{ m/s}\)
The wavelength (\(\lambda\)) is related to frequency (\(\nu\)) and speed of light (c) by the formula: \[\lambda = \frac{c}{\nu}\]
For the lower frequency (\(\nu_1\)): \[\lambda_1 = \frac{c}{\nu_1} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40 \text{ m}\]
For the upper frequency (\(\nu_2\)): \[\lambda_2 = \frac{c}{\nu_2} = \frac{3 \times 10^8}{12 \times 10^6} = 25 \text{ m}\]
The corresponding wavelength band is 40 m - 25 m. (Note: Higher frequency corresponds to shorter wavelength, and lower frequency corresponds to longer wavelength).
In simple words: Radio stations operate within a specific range of frequencies. We convert these frequencies into wavelengths using the speed of light to find the corresponding range of wavelengths.

🎯 Exam Tip: Remember the inverse relationship between frequency and wavelength: higher frequency means shorter wavelength, and vice-versa. Clearly label your calculated wavelengths for each frequency extreme.

 

Question 6.A charged particle oscillates about its mean equilibrium position with a frequency of \(10^9\) Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:When a charged particle oscillates, it produces electromagnetic waves. The frequency of the electromagnetic waves generated is always equal to the frequency of oscillation of the charged particle. Given: Frequency of oscillation of the charged particle = \(10^9\) Hz.
Therefore, the frequency of the electromagnetic waves produced by the oscillator is \(10^9\) Hz.
In simple words: When a charged particle vibrates, it creates electromagnetic waves. The waves vibrate at the exact same speed as the particle itself.

🎯 Exam Tip: This is a fundamental concept: the frequency of emitted electromagnetic radiation matches the frequency of the oscillating source (charged particle).

 

Question 7.The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is \(B_0 = 510 \text{ nT}\). What is the amplitude of the electric field part of the wave?
Answer:Let's find the amplitude of the electric field.
Given: Amplitude of the magnetic field, \(B_0 = 510 \text{ nT} = 510 \times 10^{-9} \text{ T}\) Speed of light in vacuum, \(c = 3 \times 10^8 \text{ m/s}\)
The relationship between the amplitudes of the electric field (\(E_0\)) and magnetic field (\(B_0\)) in an electromagnetic wave in vacuum is: \[E_0 = c B_0\] Substitute the given values: \[E_0 = (3 \times 10^8 \text{ m/s}) \times (510 \times 10^{-9} \text{ T})\] \[E_0 = 1530 \times 10^{-1} \text{ N/C}\] \[E_0 = 153 \text{ N/C}\]
In simple words: In a light wave, the electric and magnetic fields are linked. If you know the strength of the magnetic field and the speed of light, you can easily find the strength of the electric field.

🎯 Exam Tip: The relation \(E_0 = cB_0\) is crucial for electromagnetic waves in a vacuum. Ensure proper unit conversion (nT to T) to avoid calculation errors.

 

Question 8.Suppose that the electric field amplitude of an electromagnetic wave is \(E_0 = 120 \text{ N/C}\) and that its frequency is \(\nu = 50.0 \text{ MHz}\).
(a) Determine, \(B_0, \omega, k\), and \(\lambda\).
(b) Find the expressions for E and B.
Answer:Let's calculate the various parameters of the electromagnetic wave.
Given: Electric field amplitude, \(E_0 = 120 \text{ N/C}\) Frequency, \(\nu = 50.0 \text{ MHz} = 50 \times 10^6 \text{ Hz}\) Speed of light, \(c = 3 \times 10^8 \text{ m/s}\)
(a) To determine \(B_0, \omega, k\), and \(\lambda\):
1. Magnetic field amplitude (\(B_0\)): \[B_0 = \frac{E_0}{c} = \frac{120 \text{ N/C}}{3 \times 10^8 \text{ m/s}}\] \[B_0 = 4 \times 10^{-7} \text{ T}\]
2. Angular frequency (\(\omega\)): \[\omega = 2\pi\nu = 2 \times 3.14 \times 50 \times 10^6 \text{ Hz}\] \[\omega = 314 \times 10^6 \text{ rad s}^{-1} = 3.14 \times 10^8 \text{ rad s}^{-1}\]
3. Wave number (k): \[k = \frac{\omega}{c} = \frac{3.14 \times 10^8 \text{ rad s}^{-1}}{3 \times 10^8 \text{ m/s}}\] \[k \approx 1.05 \text{ rad m}^{-1}\]
4. Wavelength (\(\lambda\)): \[\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.05 \text{ rad m}^{-1}}\] \[\lambda \approx 5.98 \text{ m}\]
(b) To find the expressions for E and B, assuming the wave propagates along the x-direction: The general expression for the electric field is \(E = E_0 \sin(kx - \omega t)\). Substituting the values: \[E = 120 \sin(1.05x - 3.14 \times 10^8 t) \text{ V/m}\]
The general expression for the magnetic field is \(B = B_0 \sin(kx - \omega t)\). Substituting the values: \[B = 4 \times 10^{-7} \sin(1.05x - 3.14 \times 10^8 t) \text{ T}\]
In simple words: We calculated the magnetic field strength, how fast the wave oscillates, its wave number (how many waves fit in a certain distance), and its length. Then, we wrote down the equations that describe how the electric and magnetic fields change as the wave travels.

🎯 Exam Tip: Remember the relationships between \(E_0, B_0, c, \omega, k, \nu,\) and \(\lambda\). Pay attention to units and ensure your final expressions for E and B follow the standard sinusoidal wave form.

 

Question 9.The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula \(E = h\nu\) (for energy of a quantum of radiation: photon) and obtain the 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?
Answer:The energy of a photon (E) is related to its frequency (\(\nu\)) by Planck's formula, \(E = h\nu\), where \(h\) is Planck's constant (\(6.63 \times 10^{-34} \text{ J s}\)). We can also use \(E = \frac{hc}{\lambda}\), where c is the speed of light and \(\lambda\) is the wavelength.
Let's calculate photon energy for a reference wavelength, say \(\lambda = 1 \text{ m}\) (radio waves): \[E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m/s}}{1 \text{ m}}\] \[E = 1.989 \times 10^{-25} \text{ J}\] To convert this energy to electron volts (eV), we divide by the charge of an electron (\(1.602 \times 10^{-19} \text{ J/eV}\)): \[E_{\text{eV}} = \frac{1.989 \times 10^{-25} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 1.24 \times 10^{-6} \text{ eV}\]
This calculated energy (\(1.24 \times 10^{-6} \text{ eV}\) for a 1 m wavelength) serves as a benchmark. We can find energies for other parts of the electromagnetic spectrum by scaling this value based on their wavelengths or frequencies.
**Relationship to Sources:** The energy of a photon produced by a source directly relates to the energy level spacings within that source. * **Radio waves** (long wavelength, low energy, e.g., \(\lambda = 10 \text{ m}\), \(E \approx 10^{-7} \text{ eV}\)): These low energies correspond to energy differences in molecular rotations and nuclear spins, typically produced by oscillating electric circuits. * **Microwaves** (e.g., \(\lambda = 10^{-2} \text{ m}\), \(E \approx 10^{-5} \text{ eV}\)): Related to molecular rotations and vibrations. * **Infrared** (e.g., \(\lambda = 10^{-5} \text{ m}\), \(E \approx 0.01 \text{ eV}\)): Associated with molecular vibrations and thermal radiation from hot bodies. * **Visible light** (e.g., \(\lambda = 5 \times 10^{-7} \text{ m}\), \(E \approx 2.5 \text{ eV}\)): These energies correspond to electron transitions within atoms and molecules. Sources include excited atoms in lamps or the outer electron shells of atoms. * **Ultraviolet** (e.g., \(\lambda = 10^{-8} \text{ m}\), \(E \approx 10 \text{ eV}\)): High enough energy to cause chemical reactions and ionize atoms. Related to electronic transitions in outer atomic shells. * **X-rays** (e.g., \(\lambda = 10^{-10} \text{ m}\), \(E \approx 1 \text{ keV}\)): These very high energies are produced by rapid acceleration of electrons (e.g., in X-ray tubes) or transitions of inner-shell electrons. * **Gamma rays** (e.g., \(\lambda = 10^{-12} \text{ m}\), \(E \approx 1 \text{ MeV}\)): The highest energies, corresponding to nuclear energy level transitions or radioactive decay processes. This indicates that nuclear energy levels are typically spaced by MeV.
In simple words: The energy of a light particle (photon) depends on its frequency or wavelength. We calculated this energy in electron volts for different types of light. This energy level tells us where that light comes from, like from electrons jumping in atoms for visible light, or from changes inside an atom's nucleus for powerful gamma rays.

🎯 Exam Tip: The formula \(E = h\nu\) or \(E = hc/\lambda\) is central. Remember to convert Joules to electron volts (\(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)). Understanding the typical energy ranges for different EM spectrum parts helps link them to their physical origins (e.g., nuclear, atomic, molecular processes).

 

Question 10.In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \(2.0 \times 10^{10} \text{ Hz}\) and amplitude \(48 \text{ V/m}\).
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field, [\(c = 3 \times 10^8 \text{ m/s}\)].
Answer:Let's find the properties of this electromagnetic wave.
Given: Frequency, \(\nu = 2.0 \times 10^{10} \text{ Hz}\) Electric field amplitude, \(E_0 = 48 \text{ V/m}\) Speed of light, \(c = 3 \times 10^8 \text{ m/s}\)
(a) To find the wavelength (\(\lambda\)): \[\lambda = \frac{c}{\nu} = \frac{3 \times 10^8 \text{ m/s}}{2.0 \times 10^{10} \text{ Hz}}\] \[\lambda = 1.5 \times 10^{-2} \text{ m}\]
(b) To find the amplitude of the oscillating magnetic field (\(B_0\)): \[B_0 = \frac{E_0}{c} = \frac{48 \text{ V/m}}{3 \times 10^8 \text{ m/s}}\] \[B_0 = 16 \times 10^{-8} \text{ T} = 1.6 \times 10^{-7} \text{ T}\]
(c) To show that the average energy density of the E field (\(U_E\)) equals the average energy density of the B field (\(U_B\)):
The average energy density of the electric field is: \[U_E = \frac{1}{2} \varepsilon_0 E_0^2\] The average energy density of the magnetic field is: \[U_B = \frac{1}{2\mu_0} B_0^2\] We know that in a vacuum, \(E_0 = cB_0\) and \(c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\), which implies \(c^2 = \frac{1}{\mu_0 \varepsilon_0}\) or \(\mu_0 \varepsilon_0 = \frac{1}{c^2}\).
Let's substitute \(B_0 = \frac{E_0}{c}\) into the expression for \(U_B\): \[U_B = \frac{1}{2\mu_0} \left(\frac{E_0}{c}\right)^2\] \[U_B = \frac{1}{2\mu_0} \frac{E_0^2}{c^2}\] Now substitute \(c^2 = \frac{1}{\mu_0 \varepsilon_0}\) into the equation for \(U_B\): \[U_B = \frac{1}{2\mu_0} E_0^2 (\mu_0 \varepsilon_0)\] \[U_B = \frac{1}{2} \varepsilon_0 E_0^2\] Since \(U_E = \frac{1}{2} \varepsilon_0 E_0^2\) and we derived \(U_B = \frac{1}{2} \varepsilon_0 E_0^2\), it proves that: \[U_E = U_B\]
In simple words: We found the length of the wave and the strength of its magnetic field. We also showed that the average energy stored in the electric part of the wave is exactly equal to the average energy stored in the magnetic part.

🎯 Exam Tip: Remember the fundamental relations \(c = \nu \lambda\) and \(E_0 = cB_0\). For part (c), the key is to use the relationship \(c = 1/\sqrt{\mu_0 \varepsilon_0}\) and substitute it into the energy density equations to prove their equality.

 

Question 11.Suppose that the electric field part of an electromagnetic wave in vacuum is \(E = \{(3.1 \text{ N/C}) \cos [(1.8 \text{ rad/m})y + (5.4 \times 10^6 \text{ rad/s})t]\} \hat{i}\).
(a) What is the direction of propagation?
(b) What is the wavelength \(\lambda\)?
(c) What is the frequency?
(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.
Answer:Let's analyze the given electric field equation of an electromagnetic wave.
The general equation for a plane progressive electromagnetic wave is \(E = E_0 \cos(ky \pm \omega t)\), where the sign between ky and \(\omega t\) indicates the direction of propagation. A '+' sign means propagation in the negative direction, and a '-' sign means propagation in the positive direction.
Given equation: \(E = \{(3.1 \text{ N/C}) \cos [(1.8 \text{ rad/m})y + (5.4 \times 10^6 \text{ rad/s})t]\} \hat{i}\)
Comparing with the general form, we have: Electric field amplitude, \(E_0 = 3.1 \text{ N/C}\) Wave number, \(k = 1.8 \text{ rad/m}\) Angular frequency, \(\omega = 5.4 \times 10^6 \text{ rad/s}\)
(a) The wave propagates along the y-axis, and because the sign between `ky` and `\(\omega t\)` is positive, the direction of propagation is the negative y-direction. We can write this as \(- \hat{j}\).
(b) To find the wavelength (\(\lambda\)): \[k = \frac{2\pi}{\lambda}\] \[\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.8 \text{ rad/m}}\] \[\lambda \approx 3.49 \text{ m}\] \[\lambda \approx 3.5 \text{ m}\]
(c) To find the frequency (\(\nu\)): \[\omega = 2\pi\nu\] \[\nu = \frac{\omega}{2\pi} = \frac{5.4 \times 10^6 \text{ rad/s}}{2 \times 3.14}\] \[\nu \approx 0.859 \times 10^6 \text{ Hz}\]
(d) To find the amplitude of the magnetic field part of the wave (\(B_0\)): We use the relation \(E_0 = cB_0\), where \(c = 3 \times 10^8 \text{ m/s}\) is the speed of light in vacuum. \[B_0 = \frac{E_0}{c} = \frac{3.1 \text{ N/C}}{3 \times 10^8 \text{ m/s}}\] \[B_0 \approx 1.03 \times 10^{-8} \text{ T}\]
(e) To write an expression for the magnetic field part of the wave: Since the electric field oscillates along the \(\hat{i}\) direction and the wave propagates along the \(- \hat{j}\) direction, the magnetic field must oscillate along the \(\hat{k}\) direction (as \(\overrightarrow{E} \times \overrightarrow{B}\) must be in the direction of propagation, i.e., \(\hat{i} \times \hat{k} = -\hat{j}\)). The expression for the magnetic field part will have the same angular frequency and wave number as the electric field: \[B = \{(1.03 \times 10^{-8} \text{ T}) \cos [(1.8 \text{ rad/m})y + (5.4 \times 10^6 \text{ rad/s})t]\} \hat{k}\]
In simple words: We broke down the given electric field equation to find how the wave moves, its length, and how often it oscillates. Then, we calculated the strength of the magnetic field and wrote its equation, knowing that it's always at a right angle to both the electric field and the wave's movement.

🎯 Exam Tip: Pay close attention to the sign in the wave equation (\(ky \pm \omega t\)) to determine the direction of propagation. Remember the orthogonal relationship between \(\overrightarrow{E}\), \(\overrightarrow{B}\), and the propagation vector to correctly determine the direction of \(\overrightarrow{B}\). Ensure accurate calculation of \(\lambda, \nu,\) and \(B_0\).

 

Question 12.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 1m from the bulb
(b) at a distance of 10 m
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:Let's calculate the intensity of visible radiation at different distances.
Given: Total power of the light bulb = 100 W Percentage converted to visible radiation = 5% Visible radiation power (\(P_{\text{visible}}\)) = \(5\% \text{ of } 100 \text{ W} = 0.05 \times 100 \text{ W} = 5 \text{ W}\)
Assuming isotropic emission, the radiation spreads uniformly in all directions, forming a sphere. The area of this sphere is \(A = 4\pi r^2\). Intensity (I) is defined as power per unit area: \(I = \frac{P_{\text{visible}}}{A} = \frac{P_{\text{visible}}}{4\pi r^2}\).
(a) At a distance of \(r = 1 \text{ m}\) from the bulb: \[I = \frac{5 \text{ W}}{4 \times 3.14 \times (1 \text{ m})^2}\] \[I = \frac{5}{12.56} \text{ W/m}^2\] \[I \approx 0.398 \text{ W/m}^2\] \[I \approx 0.4 \text{ W/m}^2\]
(b) At a distance of \(r = 10 \text{ m}\) from the bulb: \[I = \frac{5 \text{ W}}{4 \times 3.14 \times (10 \text{ m})^2}\] \[I = \frac{5}{4 \times 3.14 \times 100} \text{ W/m}^2\] \[I = \frac{5}{1256} \text{ W/m}^2\] \[I \approx 0.00398 \text{ W/m}^2\] \[I \approx 0.004 \text{ W/m}^2\]
In simple words: We first found out how much light energy the bulb actually gives off as visible light. Then, assuming this light spreads out evenly in all directions, we calculated how bright it would be (intensity) at two different distances from the bulb. The brightness decreases as you move further away.

🎯 Exam Tip: Remember that intensity follows an inverse square law for isotropic sources. Calculate the actual power of visible radiation first. Use \(A = 4\pi r^2\) for the surface area of a sphere and ensure accurate calculations and units.

 

Question 13.Use the formula \(\lambda_m T = 0.29 \text{ cm K}\) to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:The formula \(\lambda_m T = 0.29 \text{ cm K}\) is Wien's displacement law, where \(\lambda_m\) is the wavelength corresponding to maximum intensity of radiation from a black body at temperature T. We can write this as \(\lambda_m T = 2.9 \times 10^{-3} \text{ m K}\). This law allows us to determine the temperature of a body based on the peak wavelength of the radiation it emits. Conversely, for a given wavelength, we can find the temperature required to emit radiation predominantly at that wavelength.
Let's calculate temperatures for different parts of the electromagnetic spectrum:
From Wien's Law, \(T = \frac{2.9 \times 10^{-3} \text{ m K}}{\lambda_m}\).
1. **Gamma Rays** (\(\lambda_m \approx 10^{-12} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10^{-12}} = 2.9 \times 10^9 \text{ K}\). This high temperature is characteristic of processes in extremely hot stellar cores or supernovae.
2. **X-rays** (\(\lambda_m \approx 10^{-10} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10^{-10}} = 2.9 \times 10^7 \text{ K}\). These temperatures are found in astrophysical phenomena like neutron stars and black hole accretion disks.
3. **Ultraviolet (UV)** (\(\lambda_m \approx 10^{-8} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10^{-8}} = 2.9 \times 10^5 \text{ K}\). Such temperatures are seen in hot, young stars.
4. **Visible Light** (\(\lambda_m \approx 5 \times 10^{-7} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{5 \times 10^{-7}} \approx 5800 \text{ K}\). This temperature range is typical for the surface of stars like our Sun, which peaks in visible light emission.
5. **Infrared (IR)** (\(\lambda_m \approx 10^{-5} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10^{-5}} = 290 \text{ K}\). This is close to room temperature or the temperature of the Earth's surface, which emits predominantly infrared radiation.
6. **Microwaves** (\(\lambda_m \approx 10^{-2} \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10^{-2}} = 0.29 \text{ K}\). This very low temperature is characteristic of the cosmic microwave background radiation, a relic of the Big Bang.
7. **Radio Waves** (\(\lambda_m \approx 10 \text{ m}\)): \(T = \frac{2.9 \times 10^{-3}}{10} = 2.9 \times 10^{-4} \text{ K}\). Such extremely low temperatures are not typically associated with direct thermal emission from macroscopic objects but rather from quantum processes or very cold regions of space.
**What these numbers tell us:** These calculated temperatures represent the characteristic temperature ranges for objects that would predominantly emit electromagnetic radiation in each specific part of the spectrum. They indicate the conditions (e.g., stellar interiors, stellar surfaces, planetary bodies, the early universe) required to produce radiation of a particular wavelength. For example, to obtain visible radiation, a source needs to be at a temperature of around 6000 K. A lower temperature source would produce radiation peaking at longer wavelengths (e.g., infrared), not visible light.
In simple words: We used a formula that links the peak color of light from a hot object to its temperature. By applying this to different types of electromagnetic waves, we found out how hot something needs to be to mostly give off that kind of light. These temperatures show us where different types of light come from, like stars for visible light or nuclear reactions for gamma rays.

🎯 Exam Tip: Clearly state Wien's displacement law. When calculating temperature ranges, use typical wavelengths for each part of the EM spectrum. Explain that the calculated temperatures indicate the conditions under which a body would primarily emit radiation at that specific wavelength, linking energy scales to physical phenomena.

 

Question 14.Given 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.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 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].
(d) 5890 Å - 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in \(^{57}\text{Fe}\) nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
Answer:
(a) A wavelength of 21 cm: This corresponds to **radio waves**. Specifically, it's a famous spectral line emitted by neutral hydrogen atoms in space, used in radio astronomy.
(b) A frequency of 1057 MHz: This also corresponds to **radio waves** (specifically, in the microwave part of the radio spectrum due to its high frequency). The Lamb shift refers to a small energy difference in hydrogen's energy levels.
(c) A temperature of 2.7 K: This temperature is associated with **microwaves**. It's the temperature of the cosmic microwave background (CMB) radiation, which is low-energy microwave radiation pervading the universe, a remnant from the early universe.
(d) Double lines of sodium [5890 Å - 5896 Å]: These wavelengths fall within the **visible light** spectrum, specifically in the yellow region. These are characteristic emission lines of sodium vapor, often seen in streetlights.
(e) An energy of 14.4 keV: This energy corresponds to **X-rays** (specifically, soft X-rays or low-energy X-rays). Such high energies are typical for transitions involving inner electrons in heavier atoms or nuclear transitions, as seen in Mössbauer spectroscopy involving the \(^{57}\text{Fe}\) nucleus.
In simple words: We identified which type of electromagnetic wave each given number represents. These numbers describe different light forms, from long radio waves from hydrogen in space to high-energy X-rays from atomic nuclei, each telling us something unique about its origin.

🎯 Exam Tip: Be familiar with the approximate wavelength, frequency, energy, and temperature ranges for each major part of the electromagnetic spectrum. Connect specific physical phenomena (e.g., hydrogen emission, CMB, sodium lamps, nuclear transitions) to their corresponding EM radiation type.

 

Question 15.Answer the following questions:
(a) Long-distance radio broadcasts use short-wave bands. Why?
(b) It is necessary to use satellites for long-distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) 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?
Answer:
(a) Long-distance radio broadcasts use short-wave bands because shortwaves (wavelengths less than 200m or frequencies greater than 1500 kHz) are effectively reflected by the F-layer of the ionosphere. This reflection allows the short waves to travel long distances by bouncing between the Earth's surface and the ionosphere, reaching receivers far from the transmitter.
(b) Satellites are necessary for long-distance TV transmission because television signals, which are typically in the very high frequency (VHF) or ultra-high frequency (UHF) range, have short wavelengths and are not reflected by the Earth's ionosphere. Instead, they pass through it. To transmit these signals over long distances, artificial satellites in orbit receive the signals from an earth station and then re-transmit them back to different locations on Earth.
(c) Optical and radio telescopes can be built on the ground because the Earth's atmosphere is transparent to most visible light and radio waves, allowing these radiations to reach the surface. However, X-rays are largely absorbed by the Earth's atmosphere. Therefore, X-ray astronomy requires telescopes to be placed on satellites orbiting above the atmosphere to detect X-rays from space.
(d) The small ozone layer in the stratosphere is crucial for human survival because it absorbs most of the harmful ultraviolet (UV) radiation from the Sun. UV radiation can cause skin cancer, cataracts, and suppress the immune system in humans, and can damage plant life. By absorbing these harmful rays, the ozone layer protects life on Earth.
(e) If the Earth did not have an atmosphere, its average surface temperature would be much lower than what it is now. The atmosphere, particularly gases like carbon dioxide and water vapor, traps infrared radiation emitted by the Earth's surface, a phenomenon known as the greenhouse effect. This natural greenhouse effect keeps the Earth warm enough to support life. Without an atmosphere to trap this heat, heat would radiate away into space much more quickly, leading to significantly colder surface temperatures.
(f) The prediction of a ‘nuclear winter' after a global nuclear war is based on the idea that massive amounts of dust, soot, and smoke would be injected into the atmosphere from widespread fires caused by nuclear explosions. This thick layer of particles would block out a significant portion of sunlight from reaching the Earth's surface. The reduction in solar radiation would lead to a drastic drop in global temperatures, prolonged darkness, and disruptions to photosynthesis, severely impacting ecosystems and human food production, leading to a "winter-like" environment.
In simple words: This question covers several key aspects of how electromagnetic waves interact with Earth. It explains why some radio waves bounce off the atmosphere, why TV signals need satellites, why X-ray telescopes must be in space, how the ozone layer protects us, how the atmosphere keeps Earth warm, and why a nuclear war could cause a "nuclear winter" by blocking sunlight.

🎯 Exam Tip: For each part, clearly identify the specific type of electromagnetic radiation and its interaction with the atmosphere or ionosphere. Connect the properties of waves (reflection, absorption, penetration) to practical applications (radio, TV, astronomy) and environmental impacts (ozone layer, greenhouse effect, nuclear winter).

 

GSEB Class 12 Physics Electromagnetic Waves Additional Important Questions and Answers

 

Question 1.What is an electromagnetic wave?
Answer:An electromagnetic wave is a wave that consists of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction in which the wave travels. These fields periodically change in strength as the wave propagates through space, carrying energy and momentum.
In simple words: An electromagnetic wave is a special type of wave made of vibrating electric and magnetic fields. These fields move together, are at right angles to each other, and point perpendicular to the direction the wave is traveling.

🎯 Exam Tip: Define an electromagnetic wave by specifying its components (electric and magnetic fields), their orientation (mutually perpendicular and perpendicular to propagation), and their nature (oscillating and propagating).

 

Question 2.What are the contributions of Faraday and Maxwell in this field?
Answer:The contributions of Faraday and Maxwell were fundamental to understanding electromagnetic waves:
* **Faraday's Contribution:** Michael Faraday discovered that a changing magnetic field produces an electric field. This principle is known as Faraday's Law of Induction and is crucial for the generation of electricity and understanding electromagnetic phenomena.
* **Maxwell's Contribution:** James Clerk Maxwell extended Faraday's work by proposing that a changing electric field also produces a magnetic field (known as displacement current). By unifying electricity, magnetism, and optics, Maxwell formulated his famous equations, which predicted the existence of electromagnetic waves and showed that light itself is an electromagnetic wave.
In simple words: Faraday showed that changing magnetism creates electricity. Maxwell built on this, proving that changing electricity also creates magnetism, and combined these ideas to predict that light is an electromagnetic wave.

🎯 Exam Tip: Clearly distinguish between Faraday's discovery (changing magnetic field creates electric field) and Maxwell's key insight (changing electric field creates magnetic field via displacement current), and how Maxwell unified these concepts to predict EM waves.

 

Question 3.Match the following.

A	B	C
i. y ray	Sunburn	Photon emission by fast-moving electrons
iii. UV ray	Diagnosis	Oscillating current
iv. Microwave	Radioactivity	Nucleus

Answer:

A	B	C
i. \(\gamma\)-ray	Radioactivity	Nucleus
ii. X-ray	Diagnosis	Photon emission by fast-moving electrons
iii. UV ray	Sunburn	Electronic de-excitation
iv. Microwave	Remote sensing	Oscillating current

In simple words: We matched different types of electromagnetic waves (like gamma rays or X-rays) to their common uses, their sources, and where they typically originate from.

🎯 Exam Tip: Familiarize yourself with the applications, sources, and origins of different parts of the electromagnetic spectrum. Pay close attention to distinguishing characteristics for each type of radiation (e.g., gamma rays from nuclear processes, X-rays for medical imaging).

 

Question 4.(a) If the fundamental source of a sound wave is a vibrating object, what is the fundamental source of an electromagnetic wave?
(b) Give the expression for the velocity of E.M waves in a vacuum.
(c) On what factors does its velocity in vacuum depend?
Answer:
(a) The fundamental source of an electromagnetic wave is an oscillating or accelerating charged particle. When a charged particle vibrates or changes its speed, it creates disturbances in the electric and magnetic fields around it, which then propagate as electromagnetic waves.
(b) The expression for the velocity (c) of electromagnetic waves in a vacuum is given by: \[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\] where \(\mu_0\) is the permeability of free space and \(\varepsilon_0\) is the permittivity of free space.
(c) The velocity of electromagnetic waves in a vacuum depends on two fundamental constants: the **permittivity of free space (\(\varepsilon_0\))** and the **permeability of free space (\(\mu_0\))**. These constants describe how electric and magnetic fields behave in a vacuum.
In simple words: Vibrating electric charges create electromagnetic waves. The speed of these waves in empty space depends only on how easily electric and magnetic fields can form in that space.

🎯 Exam Tip: Remember that accelerating charges generate EM waves. The formula \(c = 1/\sqrt{\mu_0 \varepsilon_0}\) is a key Maxwell's equation result. Clearly state that the speed in vacuum depends only on the fundamental constants \(\mu_0\) and \(\varepsilon_0\).

 

Question 5.Maxwell noticed that Ampere's circuital law is inconsistent where the electric current changes with time. He showed that consistency requires an additional source of the magnetic field.
(a) What is the current called and which is responsible for the magnetic field?
(b) Give the expression for this type of current.
(c) How it differs from conduction current?
Answer:
(a) The additional current introduced by Maxwell to resolve the inconsistency in Ampere's circuital law is called **displacement current**. This current is responsible for producing the magnetic field in regions where the electric field is changing, such as between the plates of a charging capacitor.
(b) The expression for displacement current (\(I_D\)) is: \[I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\] where \(\varepsilon_0\) is the permittivity of free space and \(\frac{d\Phi_E}{dt}\) is the rate of change of electric flux through a surface.
(c) Displacement current differs from conduction current in several ways: * **Nature:** Conduction current is due to the actual flow of free charges (like electrons) in a conductor. Displacement current, on the other hand, is not a flow of charges but rather arises from a time-varying electric field. * **Medium:** Conduction current requires a conducting medium to flow. Displacement current can exist even in a vacuum or insulating materials where no free charges are present. * **Conditions:** Conduction current exists under steady conditions (DC circuits) as well as varying conditions. Displacement current exists only when the electric field is changing with time. Under steady conditions, the electric field is constant, so the displacement current is zero.
In simple words: Maxwell added a new type of current, called displacement current, to explain how magnetic fields can exist even where no wires or charges flow, like between capacitor plates. This current is caused by a changing electric field, unlike regular conduction current which is the flow of actual charges.

🎯 Exam Tip: Understand that displacement current completes Ampere's law, allowing for consistent description of EM waves. Remember its definition \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\) and the key differences from conduction current (no charge flow, exists in vacuum, depends on changing electric field).

 

Question 6.(a) What is an electromagnetic spectrum?
(b) How the electromagnetic waves are arranged?
(c) Mention some of the properties of E.M waves.
Answer:
(a) The electromagnetic spectrum is the entire range of all possible electromagnetic radiation. It extends from very long wavelength radio waves to very short wavelength gamma rays, encompassing all forms of light, including visible light, that exist in the universe. It is a continuous distribution of electromagnetic waves, ordered by wavelength or frequency.
(b) Electromagnetic waves are arranged in the electromagnetic spectrum based on their wavelength and frequency. They are typically ordered in terms of increasing wavelength (and decreasing frequency), or vice-versa. The common order from longest wavelength (lowest frequency) to shortest wavelength (highest frequency) is:
Radio waves, Microwaves, Infrared radiation, Visible light, Ultraviolet radiation, X-rays, Gamma rays.
(c) Some properties of electromagnetic (EM) waves include:
(i) They are produced by accelerated or oscillating electric charges.
(ii) They do not require any material medium for their propagation; they can travel through a vacuum.
(iii) The velocity of electromagnetic waves in free space is constant and equal to the speed of light, \(c = 3 \times 10^8 \text{ m/s}\).
(iv) They obey the principle of superposition, meaning multiple waves can pass through the same point in space without affecting each other.
(v) They carry both energy and momentum as they propagate through space.
(vi) They are transverse in nature, meaning the oscillating electric and magnetic fields are perpendicular to the direction of wave propagation.
In simple words: The electromagnetic spectrum is a complete list of all types of light, arranged by how long their waves are or how fast they vibrate. These waves, like radio waves or X-rays, don't need a medium to travel, move at the speed of light, carry energy, and their electric and magnetic parts vibrate at right angles to their movement.

🎯 Exam Tip: Define the EM spectrum as the full range of EM radiation. Memorize the order of waves in the spectrum (e.g., in terms of increasing wavelength). List 4-5 key properties, such as being transverse, not requiring a medium, constant speed in vacuum, and carrying energy/momentum.

 

Question 7.The electric field vector of an electromagnetic wave is \(E = E_0 \sin [2\pi(\frac{x}{\lambda} - \frac{t}{T})]\hat{j}\).
(a) Write the magnetic field vector of this wave.
(b) What is the direction of propagation?
(c) With what velocity the wave propagates?
Answer:Let's analyze the given electric field equation of an electromagnetic wave.
Given electric field vector: \(E = E_0 \sin [2\pi(\frac{x}{\lambda} - \frac{t}{T})]\hat{j}\)
(a) To write the magnetic field vector of this wave: In an electromagnetic wave, the electric field \(\overrightarrow{E}\), magnetic field \(\overrightarrow{B}\), and the direction of propagation are mutually perpendicular. Also, \(\overrightarrow{E} \times \overrightarrow{B}\) gives the direction of propagation.
Here, the electric field is along the \(\hat{j}\) (y-direction), and the wave propagates along the \(\hat{i}\) (x-direction) (due to \(x/\lambda\) term and '-' sign for positive direction). For \(\hat{j} \times \overrightarrow{B}\) to be in the \(\hat{i}\) direction, \(\overrightarrow{B}\) must be in the \(\hat{k}\) (z-direction). (Since \(\hat{j} \times \hat{k} = \hat{i}\)).
The magnetic field amplitude \(B_0 = E_0/c\). The magnetic field also oscillates in phase with the electric field.
So, the magnetic field vector is: \[\vec{B} = B_0 \sin [2\pi(\frac{x}{\lambda} - \frac{t}{T})]\hat{k}\] or \[\vec{B} = \frac{E_0}{c} \sin [2\pi(\frac{x}{\lambda} - \frac{t}{T})]\hat{k}\]
(b) The direction of propagation is determined by the variable in the sinusoidal argument (x in this case) and the sign before the `t` term. Since the argument contains \((\frac{x}{\lambda} - \frac{t}{T})\), the wave is propagating in the positive x-direction (or along \(\hat{i}\)).
(c) The velocity with which the wave propagates in a vacuum is the speed of light (c). It is given by the expression: \[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\] In a medium, the velocity (v) is given by: \[v = \frac{1}{\sqrt{\mu \varepsilon}}\]
In simple words: We took the equation for the electric field of a light wave to figure out its magnetic field. We found that the wave is moving forward along the x-axis, and its magnetic field vibrates perpendicular to both the electric field and the wave's path. We also recalled that light waves in empty space travel at a constant speed, which depends on the fundamental properties of space itself.

🎯 Exam Tip: To determine the direction of \(\vec{B}\), use the cross product rule: \(\vec{E} \times \vec{B}\) points in the direction of propagation. The direction of propagation is given by the spatial variable's direction (e.g., x) and the sign with the time variable (negative sign means positive direction). The speed of light in vacuum is a constant defined by \(\mu_0\) and \(\varepsilon_0\).

 

Question 8.(a) The small ozone layer on the top of the stratosphere is crucial for human survival Why?
(b) Where is the position of UV radiation in the electromagnetic spectrum?
(c) Can you convert UV to visible radiation? Explain.
(d) Which radiation is adjacent to UV to the higher frequency side of the electromagnetic spectrum?
Answer:
(a) The small ozone layer in the stratosphere is vital for human survival because it effectively absorbs the majority of harmful ultraviolet (UV) radiation coming from the sun. Exposure to high levels of UV radiation can cause severe damage to living cells and DNA, leading to skin cancer, eye damage (like cataracts), and weakening of the immune system. The ozone layer acts as a protective shield, filtering out these dangerous rays.
(b) In the electromagnetic spectrum, ultraviolet (UV) radiation is located between X-rays (higher energy/frequency) and visible light (lower energy/frequency). So, it falls after visible light and before X-rays.
(c) Yes, it is possible to convert UV radiation into visible radiation. This process is utilized in fluorescent lights and some phosphorescent materials. In these devices, UV radiation excites atoms or molecules, causing their electrons to jump to higher energy levels. As these electrons return to lower energy levels, they emit photons of visible light. For example, in a mercury tube light, electrical energy excites mercury vapor, producing UV light, which then strikes a fluorescent coating to emit visible light.
(d) The radiation adjacent to UV on the higher frequency (and thus higher energy, shorter wavelength) side of the electromagnetic spectrum is **X-rays**.
In simple words: The ozone layer protects us from harmful sun rays. UV light sits between visible light and X-rays in the light spectrum. We can turn UV light into visible light, like in fluorescent lamps. The type of light next to UV, but with higher energy, is X-rays.

🎯 Exam Tip: Understand the ozone layer's role (UV absorption). Know the relative positions of different EM radiations (UV is between visible and X-rays). Be aware of practical applications like fluorescence for UV-to-visible conversion. Remember the order of increasing frequency: Visible -> UV -> X-rays.

 

Question 9.Both radio waves and gamma rays are transverse in nature and moving at the same speed in free space. Then in what aspect are they different?
Answer:While both radio waves and gamma rays are electromagnetic waves, are transverse, and travel at the speed of light in free space, they differ significantly in their origin, wavelength, frequency, and energy:
(i) **Origin:** Radio waves are typically produced by the oscillation of electrons in electrical circuits (atomic or molecular transitions involving outer electrons), such as in antennas. Gamma rays, on the other hand, originate from nuclear processes, such as radioactive decay or nuclear reactions, involving changes within the atomic nucleus.
(ii) **Energy, Frequency, and Wavelength:** Gamma rays have extremely high frequencies, very short wavelengths, and consequently, very high photon energies. Radio waves have very low frequencies, very long wavelengths, and very low photon energies. For example, a typical radio wave might have a wavelength of several meters and an energy of \(\sim 10^{-7} \text{ eV}\), while a gamma ray might have a wavelength of \(\sim 10^{-12} \text{ m}\) and an energy of \(\sim 1 \text{ MeV}\).
(iii) **Penetrating Power:** Due to their very high energy, gamma rays have extremely high penetrating power and can pass through thick materials. Radio waves, with their much lower energy, have comparatively low penetrating power and can be easily blocked or absorbed by many materials.
In simple words: Even though both radio waves and gamma rays are light waves, they are very different. Radio waves come from electrons wiggling in circuits, are long and low-energy. Gamma rays come from changes inside an atom's center, are super short and super high-energy, meaning they can pass through almost anything.

🎯 Exam Tip: Focus on the key differences: origin (atomic vs. nuclear), energy/frequency/wavelength (low vs. high), and penetrating power (low vs. high). These distinctions highlight their different applications and effects.

 

Question 10.Prof. J. Prakash asked Viji whether the average energy density of the electric field and magnetic field are equal or not? Viji said "yes". Justify the answer.
Answer:Yes, Viji's answer is correct. In an electromagnetic wave propagating in a vacuum, the average energy density associated with the electric field is indeed equal to the average energy density associated with the magnetic field. This can be justified mathematically.
The average energy density of the electric field (\(U_E\)) is given by: \[U_E = \frac{1}{2} \varepsilon_0 E_0^2\] where \(\varepsilon_0\) is the permittivity of free space and \(E_0\) is the amplitude of the electric field.
The average energy density of the magnetic field (\(U_B\)) is given by: \[U_B = \frac{1}{2\mu_0} B_0^2\] where \(\mu_0\) is the permeability of free space and \(B_0\) is the amplitude of the magnetic field.
In an electromagnetic wave in vacuum, the amplitudes of the electric and magnetic fields are related by: \[E_0 = c B_0 \implies B_0 = \frac{E_0}{c}\] Also, the speed of light (c) in vacuum is related to \(\mu_0\) and \(\varepsilon_0\) by: \[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies c^2 = \frac{1}{\mu_0 \varepsilon_0} \implies \mu_0 \varepsilon_0 = \frac{1}{c^2}\]
Now, let's substitute \(B_0 = \frac{E_0}{c}\) into the expression for \(U_B\): \[U_B = \frac{1}{2\mu_0} \left(\frac{E_0}{c}\right)^2\] \[U_B = \frac{1}{2\mu_0} \frac{E_0^2}{c^2}\] Next, substitute \(c^2 = \frac{1}{\mu_0 \varepsilon_0}\) into this equation: \[U_B = \frac{1}{2\mu_0} E_0^2 (\mu_0 \varepsilon_0)\] \[U_B = \frac{1}{2} \varepsilon_0 E_0^2\] Since \(U_E = \frac{1}{2} \varepsilon_0 E_0^2\) and we have shown that \(U_B = \frac{1}{2} \varepsilon_0 E_0^2\), it proves that: \[U_E = U_B\] Thus, the average energy densities of the electric and magnetic fields in an electromagnetic wave in vacuum are equal.
In simple words: Yes, the average energy stored in the electric part of a light wave is exactly the same as the average energy stored in its magnetic part. This is shown by equations that link electric and magnetic field strengths to the speed of light and fundamental properties of empty space.

🎯 Exam Tip: This is a standard derivation. Clearly state the formulas for \(U_E\) and \(U_B\), then use the relations \(E_0 = cB_0\) and \(c = 1/\sqrt{\mu_0 \varepsilon_0}\) to transform one energy density expression into the other. Ensure all steps are logical and algebraically correct.

 

Question 11.What is the greenhouse effect? Explain.
Answer:The greenhouse effect is a natural process that warms the Earth's surface and keeps its temperature suitable for life. It occurs when certain gases in Earth's atmosphere trap heat that would otherwise escape into space.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी पर ग्रीनहाउस प्रभाव को दर्शाता है। सूर्य से आने वाली विकिरण पृथ्वी की सतह द्वारा अवशोषित की जाती है, जिससे वह गर्म हो जाती है। पृथ्वी फिर अवरक्त विकिरण (infrared radiation) उत्सर्जित करती है, जिसे वायुमंडल में कुछ गैसें रोक लेती हैं, जिससे पृथ्वी का तापमान बढ़ जाता है।
Here's how it works:
1. **Solar Radiation Absorption:** The Earth's surface absorbs short-wavelength solar radiation (like visible light) from the Sun, which causes the surface to heat up.
2. **Infrared Emission:** Like any warm body, the heated Earth's surface then emits longer-wavelength infrared (IR) radiation back towards space.
3. **Atmospheric Trapping:** Certain gases in the Earth's lower atmosphere, known as greenhouse gases (such as carbon dioxide, methane, water vapor, and nitrous oxide), are transparent to incoming solar radiation but absorb and re-emit this outgoing infrared radiation. These gases effectively trap the heat.
4. **Warming Effect:** This trapped heat warms the lower atmosphere and the Earth's surface, much like the glass panes of a greenhouse trap heat inside. Low-lying clouds also contribute to this effect by reflecting IR radiation back to Earth, especially at night.
Without this natural greenhouse effect, the Earth's average temperature would be much colder, around -18°C (0°F), making it largely uninhabitable.
In simple words: The greenhouse effect is like a natural blanket around Earth. Sunlight warms the ground, and then the Earth sends out heat. Certain gases in the air catch this heat, stopping it from escaping into space, which keeps our planet warm enough for life.

🎯 Exam Tip: Explain the greenhouse effect as a two-step process: absorption of solar radiation by Earth, followed by emission of infrared radiation, which is then trapped by greenhouse gases. Mention key greenhouse gases and the natural role of this effect in maintaining Earth's temperature.

 

Question 12.Radio waves diffract around the building, while light waves do not. Why?
Answer:The phenomenon of diffraction (bending of waves around obstacles) is significant when the wavelength of the wave is comparable to or larger than the size of the obstacle. If the wavelength is much smaller than the obstacle, diffraction effects are negligible, and the wave appears to travel in straight lines.
* **Radio waves:** These waves have relatively long wavelengths, typically ranging from a few centimeters to several kilometers. Buildings and other common obstacles usually have sizes comparable to or smaller than these wavelengths. Therefore, radio waves easily diffract around buildings and hills, allowing radio signals to be received even when the transmitter is not in a direct line of sight.
* **Visible light waves:** Visible light has extremely short wavelengths, typically in the range of 400 nm to 700 nm (i.e., \(4 \times 10^{-7}\) m to \(7 \times 10^{-7}\) m). Buildings and most everyday objects are vastly larger than these wavelengths. Consequently, visible light does not significantly diffract around these obstacles, and we observe sharp shadows, meaning light travels in straight lines from our perspective.
In simple words: Radio waves bend around buildings because their wavelengths are long, similar to the size of the buildings. Light waves, however, have very short wavelengths, much smaller than buildings, so they travel in straight lines and don't bend much around obstacles, creating shadows.

🎯 Exam Tip: The key to explaining diffraction is the comparison between the wavelength of the wave and the size of the obstacle. Emphasize that significant diffraction occurs when these are comparable. Clearly state the typical wavelength ranges for radio waves and visible light.

 

Question 13.Why do we use short waves for long-distance radio broadcasts?
Answer:Shortwave radio bands are used for long-distance radio broadcasts because they interact favorably with the Earth's ionosphere. The ionosphere is a region of the upper atmosphere containing ionized gases. Shortwave signals, with wavelengths typically in the range of 10 to 100 meters, are reflected by the ionosphere back towards the Earth's surface. This reflection allows the waves to "hop" across vast distances, making long-range communication possible. Unlike very long waves which are absorbed by the F-layer or very short waves (like TV signals) which pass straight through, shortwaves are ideally suited for this sky-wave propagation.
In simple words: We use short radio waves for talking over long distances because they bounce off a special layer in Earth's upper atmosphere called the ionosphere. This bouncing lets the waves travel far around the globe.

🎯 Exam Tip: The crucial point here is the interaction with the ionosphere. Explain that short waves are reflected by the ionosphere, allowing for "sky-wave" propagation over long distances. Avoid confusing them with TV signals that pass through the ionosphere.

 

Question 14.Why are microwaves used in RADAR?
Answer:Microwaves are used in RADAR (Radio Detection and Ranging) systems for several important reasons:
1. **Small Wavelength:** Microwaves have relatively short wavelengths (typically from 1 mm to 1 meter). This allows radar systems to use small antennas to produce highly directional beams. A narrow beam improves angular resolution, making it easier to pinpoint the exact location of objects.
2. **Minimal Attenuation:** Microwaves can travel long distances through the atmosphere with very little attenuation (loss of signal strength) due to absorption or scattering by rain, fog, or clouds. This makes them effective for detecting distant targets in various weather conditions.
3. **Penetration:** They can penetrate non-metallic objects (like clouds, light foliage, or building materials) to some extent, which is useful for weather forecasting and certain military applications.
4. **Doppler Effect:** Microwaves are excellent for utilizing the Doppler effect to measure the speed of moving targets, as small frequency shifts can be accurately detected.
In simple words: Microwaves are good for RADAR because their waves are short, allowing for focused beams that can precisely locate objects. They also travel far without weakening too much, even through bad weather, and can be used to measure how fast things are moving.

🎯 Exam Tip: Focus on the properties that make microwaves suitable for RADAR: short wavelength (for high directionality and resolution), low atmospheric attenuation (for long-distance detection), and the ability to utilize the Doppler effect for speed measurement.

 

Question 15.It is necessary to use satellites for long TV transmission. Why?
Answer:Satellites are essential for long-distance TV transmission because television signals operate at very high frequencies (VHF and UHF bands), which correspond to very short wavelengths. Unlike shortwave radio signals, these high-frequency TV signals are not reflected by the Earth's ionosphere; instead, they pass straight through it into space. Therefore, to transmit TV signals over distances beyond the line of sight (the curvature of the Earth), artificial satellites are used. These satellites receive the signals from a transmitting station on Earth and then re-broadcast them to a wide area on the ground, effectively bypassing the limitations of the Earth's curvature and atmospheric reflection properties.
In simple words: TV signals have short wavelengths and do not bounce off the atmosphere like some radio waves do; they go straight through. So, for TV broadcasts to reach far places beyond the curve of the Earth, satellites in space are needed to catch and re-send these signals.

🎯 Exam Tip: The key reason is that TV signals (VHF/UHF) are not reflected by the ionosphere; they penetrate it. Therefore, satellites are used as relays to overcome the Earth's curvature and enable line-of-sight communication over global distances.

 

Question 16.How are electromagnetic waves emitted?
Answer:Electromagnetic waves are fundamentally emitted by charged particles (like electrons or protons) when they undergo acceleration. This acceleration can take various forms:
1. **Oscillation:** A charged particle oscillating back and forth (like an electron in an antenna) continuously changes its velocity, thus accelerating, and emits electromagnetic waves.
2. **Deceleration/Acceleration:** When charged particles are abruptly accelerated (e.g., in an X-ray tube where electrons are suddenly stopped) or decelerated, they release electromagnetic radiation.
3. **Atomic/Nuclear Transitions:** Electromagnetic waves (photons) are also emitted when electrons in atoms jump from higher energy levels to lower ones (e.g., visible light, UV, X-rays), or when changes occur within the nucleus of an atom (e.g., gamma rays during radioactive decay). These transitions involve the acceleration of charges within the atom or nucleus.
In simple words: Electromagnetic waves are created when charged particles, like electrons, speed up, slow down, or vibrate. This acceleration makes them release energy in the form of light waves.

🎯 Exam Tip: The core concept is that accelerated or oscillating charged particles are the source of electromagnetic waves. Provide examples like oscillating electrons in antennas, decelerating electrons for X-rays, and atomic/nuclear transitions for light and gamma rays.

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