Class 12 Physics Chapter 8 Electromagnetic Waves

Key Concepts: Advanced Field Interaction, Role of Displacement Current

d) It forms a standing wave pattern
[Solution Description] The electric field $$E$$ in an electromagnetic wave is related to the permittivity $$\epsilon$$ by Maxwell’s equations. As the permittivity varies spatially according to $$\epsilon(x) = \epsilon_0(1 + \sin kx)$$, the electric field experiences periodic modulation due to the changing permittivity. This can cause standing wave patterns or attenuation in certain regions depending on the amplitude and frequency of the variation.

Key Concepts: Equation Application

b) $$2.655 \times 10^{-14} \, \text{A}$$
[Solution Description] The displacement current can be calculated using the formula for displacement current:
$$ i_d = \epsilon_0 \frac{d\Phi_E}{dt} $$ where $$\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2$$ is the permittivity of free space. Substituting the given value of $$\frac{d\Phi_E}{dt} = 3 \times 10^{-3} \, \text{V m/s}$$:

$$ i_d = 8.85 \times 10^{-12} \left(3 \times 10^{-3}\right) = 2.655 \times 10^{-14} \, \text{A}. $$

Hence, the displacement current is $$2.655 \times 10^{-14} \, \text{A}$$.

Key Concepts: Wave Equation

c) $$2 \times 10^7\, \text{m}^{-1}$$
[Solution Description] The relationship between angular frequency $$\omega$$ and wave vector $$k$$ in a vacuum is given by:

$$
\omega = ck
$$

where $$c$$ is the speed of light in a vacuum, $$3 \times 10^8$$ m/s. Solving for $$k$$, we have:

$$
k = \frac{\omega}{c} = \frac{6 \times 10^{15}}{3 \times 10^8}
$$

Now, performing the division:

$$
k = 2 \times 10^7 \, \text{m}^{-1}
$$

Thus, the wave vector $$k$$ is $$2 \times 10^7 \, \text{m}^{-1}$$.

Key Concepts: Wave Interaction

b) $$1.68 \times 10^8$$ m/s
[Solution Description]
The velocity $$v$$ of an electromagnetic wave in a medium can be calculated using the formula:

$$
v = \frac{1}{\sqrt{\mu\epsilon}}
$$

Substituting $$\mu = 4 \pi \times 10^{-7}$$ H/m and $$\epsilon = 8.85 \times 10^{-12}$$ F/m:

$$
v = \frac{1}{\sqrt{(4 \pi \times 10^{-7})(8.85 \times 10^{-12})}}
$$

Calculating inside the square root:

$$
v = \frac{1}{\sqrt{3.54 \times 10^{-18}}}
$$

Taking the square root:

$$
v = \frac{1}{5.95 \times 10^{-9}}
$$

Finally, solving for $$v$$:

$$
v = 1.68 \times 10^{8} \text{ m/s}
$$

Key Concepts: Advanced Equations, Electromagnetic Wave Interaction

a) $$1.5 \times 10^8 \text{ m/s}$$
[Solution Description]
To find the speed of an electromagnetic wave in a medium, we use the formula $$v=\frac{c}{\sqrt{\mu_r \epsilon_r}}$$, where $$c = 3 \times 10^8 \text{ m/s}$$ is the speed of light in a vacuum. Here, $$\epsilon_r = 4$$ and $$\mu_r = 1$$. Substituting these values into the equation,

$$
v = \frac{3 \times 10^8}{\sqrt{1 \times 4}} = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \text{ m/s}.
$$

Thus, the speed of the wave in the medium is $$1.5 \times 10^8 \text{ m/s}$$.

Key Concepts: Basic Application

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion is true because infrared waves are indeed used in remote control devices for television as they can effectively transmit signals over short distances without interference from visible light. However, the reason provided is related to a different aspect of infrared waves. While it is true that infrared waves are absorbed by water molecules, this characteristic does not explain their use in remote controls. Thus, both the assertion and the reason are true individually, but the reason does not correctly explain the assertion.

Key Concepts: Field Magnitude Calculation

c) $$4 \times 10^{-8} \text{ T}$$
[Solution Description] The relationship between the magnitudes of the electric field $$(E_0)$$ and the magnetic field $$(B_0)$$ in an electromagnetic wave is given by the equation $$ B_0 = \frac{E_0}{c} $$. Substituting the given values, we have:

$$
B_0 = \frac{12 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 4 \times 10^{-8} \text{ T}
$$

Thus, the magnitude of the magnetic field is $$4 \times 10^{-8}$$ T.

Key Concepts: Frequency Range

c) 88 MHz to 108 MHz
[Solution Description] The syllabus specifies that the frequency range for FM radio is between 88 MHz and 108 MHz. This is a memorized fact from the given information.

Key Concepts: Basic Properties

d) $$10^{-8} \, \text{m}$$ to $$10^{-13} \, \text{m}$$
[Solution Description]
The syllabus states that X-rays cover wavelengths from about $$10^{-8} \, \text{m}$$ to $$10^{-13} \, \text{m}$$. This information helps in identifying the correct wavelength range.

Key Concepts: Spectrum Classification

c) Gamma rays

[Solution Description] In the electromagnetic spectrum, gamma rays have the shortest wavelength, followed by X-rays, ultraviolet, visible light, infrared, microwaves, and then radio waves having the longest wavelength.

Key Concepts: Displacement Current

c) A current due to a time-varying electric field
[Solution Description]
Displacement current, as introduced by Maxwell, is described by the formula $$i_d = \epsilon_0 \frac{d\Phi_E}{dt}$$, which is due to a time-varying electric field.

Key Concepts: Radar Application

b) Aircraft navigation
[Solution Description] Radar systems use microwaves to navigate and measure speeds, especially in fields such as aviation and automotive. This makes them suitable for applications like aircraft navigation and speed detection.

Key Concepts: Complex Application, Advanced Symmetry

c) Doubles
[Solution Description]
The displacement current $$I_d$$ is given by the equation:
$$
I_d = \varepsilon_0 \frac{d\Phi_E}{dt},
$$
where $$\Phi_E$$ is the electric flux through the surface between the plates.

For a parallel plate capacitor, the electric flux $$\Phi_E$$ can be expressed as:
$$
\Phi_E = E \cdot A,
$$
where $$E$$ is the electric field and $$A$$ is the area of the plates.

Since the electric field $$E$$ is uniform between the plates during charging, we have:
$$
I_d = \varepsilon_0 \frac{d(E \cdot A)}{dt}.
$$

If the area $$A$$ is doubled, then the new flux becomes:
$$
\Phi’_E = E \cdot 2A.
$$

The displacement current now changes to:
$$
I’_d = \varepsilon_0 \frac{d(E \cdot 2A)}{dt} = 2 \cdot \varepsilon_0 \frac{d(E \cdot A)}{dt} = 2I_d.
$$

Therefore, the displacement current $$I_d$$ doubles when the area is doubled.

Key Concepts: Advanced Calculation, Technological Application

c) 0.03 meters
[Solution Description] To find the wavelength, we use the relationship between speed, frequency, and wavelength: $$c = f \lambda$$, where $$c$$ is the speed of light, $$f$$ is the frequency, and $$\lambda$$ is the wavelength.

First, rearrange the equation to solve for $$\lambda$$:
$$ \lambda = \frac{c}{f} $$

Substitute the given values:
$$ \lambda = \frac{3 \times 10^8 \text{ m/s}}{10 \times 10^9 \text{ Hz}} $$

Calculate the result:
$$ \lambda = \frac{3 \times 10^8}{10 \times 10^9} = 3 \times 10^{-2} \text{ m} $$

Therefore, the wavelength of the microwave is 0.03 meters.

Key Concepts: UV Applications

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that UV lamps are utilized in water purifiers to eliminate germs, which is true since ultraviolet light is known for its germicidal properties. The reason discusses the effect of UV radiation on human skin, where it causes tanning by stimulating melanin production, which is also true. However, the reason provided does not explain why UV lamps are used in water purifiers; thus, both statements are correct but unconnected in their explanation.

Key Concepts: Advanced Wave Propagation, Displacement Current Impact

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that electromagnetic wave propagation in a medium is characterized by the speed $$v = \frac{1}{\sqrt{\mu \epsilon}}$$, which depends on the medium’s permeability $$(\mu)$$ and permittivity $$(\epsilon)$$. This is true and directly stems from Maxwell’s equations, demonstrating how changes in these quantities affect wave speed.

The reason provided highlights the role of displacement current in maintaining time-varying electric fields. Maxwell added this concept to Ampere’s law to account for situations where there are no actual conduction currents, ensuring that magnetic fields remain consistent when electric fields vary over time.

While both statements are true individually, the reason does not directly justify or explain the specific formula for wave speed in a medium as given in the assertion. Therefore, the correct response is that both are true, but the reason is not the correct explanation of the assertion.

Key Concepts: Intermediate Calculations

c) $$70.5$$ degrees
[Solution Description] According to Snell’s Law, $$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$$ where $$n_1 = 1.33$$, $$n_2 = 1.00$$. The angle of incidence $$\theta_1 = 45^\circ$$. So we have:

$$
1.33 \times \sin(45^\circ) = 1.00 \times \sin(\theta_2)
$$

Using the value $$\sin(45^\circ) = \frac{\sqrt{2}}{2} = 0.707$$, we get:
$$
1.33 \times 0.707 = \sin(\theta_2)
$$

Thus,
$$
\sin(\theta_2) = 0.9403
$$

Finding the inverse sine:
$$
\theta_2 \approx \arcsin(0.9403) \approx 70.5^\circ
$$

So, the angle of refraction is approximately $$70.5$$ degrees.

Key Concepts: Formula Application

b) $$1 \times 10^{-6}$$ T
[Solution Description]
Use the relationship between electric and magnetic fields: $$B_0 = \frac{E_0}{c}$$. Given $$E_0 = 300$$ V/m and $$c = 3 \times 10^8$$ m/s:
$$
B_0 = \frac{300}{3 \times 10^8} = 1 \times 10^{-6} \text{ T}
$$
Thus, the amplitude of the magnetic field is $$1 \times 10^{-6}$$ T.

Key Concepts: Wave Equation Derivation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that the wave equation for electric fields can be derived from Maxwell’s equations in a vacuum, which involves utilizing the complete set of Maxwell’s equations including Gauss’s Law for electricity, Gauss’s Law for magnetism, Faraday’s Law, and Ampere-Maxwell Law. These equations are used to establish the relationships between electric and magnetic fields in electromagnetic waves.

The Reason mentions Gauss’s Law for electricity: $$\oint E \cdot dA = Q/\varepsilon_0$$ as essential for deriving the wave equation. However, while Gauss’s law is an important part of Maxwell’s equations, it alone does not directly lead to the wave equation for electric fields. Instead, it is the combination of all four Maxwell’s equations, particularly Faraday’s Law and Ampere-Maxwell Law, that leads to the derivation of the wave equation.

Thus, both the Assertion and the Reason are true, but the Reason is NOT the correct explanation of the Assertion.

Key Concepts: Ampere-Maxwell Law

a) It substitutes for conduction current in non-conductive regions.
[Solution Description]

Displacement current plays a crucial role in the generalized Ampere-Maxwell law by providing a link between time-varying electric fields and magnetic fields, especially in cases where no physical current is present. The law can be written as:

$$
\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \left( i_c + \epsilon_0 \frac{d\Phi_E}{dt} \right)
$$

In this equation, $$\epsilon_0 \frac{d\Phi_E}{dt}$$ represents the displacement current density component that arises due to time-varying electric fields ($$ \frac{d\Phi_E}{dt} $$) even in the absence of actual charge flow (conduction current). This allows Maxwell’s equations to predict electromagnetic waves and ensure the consistency of electromagnetism theory.

Key Concepts: Application Recognition

b) Physical therapy

[Solution Description] Infrared lamps are commonly used in physical therapy. They provide therapeutic warmth that can penetrate tissues, promoting circulation and easing muscle stiffness.

Key Concepts: Frequency Range

c) 530 kHz to 1710 kHz
[Solution Description] The AM radio band operates within a specific frequency range to transmit audio signals effectively. The standard frequency range for AM radio broadcasting is from 530 kHz to 1710 kHz.

Key Concepts: Visible Spectrum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion correctly states the wavelength range for the visible spectrum, which comprises wavelengths between 700 nm and 400 nm. The reason accurately explains that this spectrum is within the range detectable by the human eye. Both statements are true, and the reason provides a valid explanation for the assertion.

Key Concepts: Synthesis of Concepts, Complex Problem Solving

c) $$30 \, \text{V/m}$$

[Solution Description]

To find the amplitude $$E_0$$ of the electric field, we use the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave:

$$\frac{E_0}{B_0} = c$$

Given:

\[
\begin{align*}
B_0 &= 1 \times 10^{-7} \, \text{T} \\
c &= 3 \times 10^8 \, \text{m/s}
\end{align*}
\]

Plugging these values into the formula:

$$E_0 = c \cdot B_0$$

Calculating:

$$E_0 = (3 \times 10^8 \, \text{m/s}) \cdot (1 \times 10^{-7} \, \text{T}) = 30 \, \text{V/m}$$

Therefore, the amplitude $$E_0$$ is $$30 \, \text{V/m}$$.

Key Concepts: Basic Concept

d) Guglielmo Marconi
[Solution Description]
Guglielmo Marconi is credited with the first successful long-distance transmission of electromagnetic waves, which marked the beginning of modern wireless communication. His work laid the foundation for the development of radio.

Key Concepts: Real-World Application, Complex Wave Equations

b) $$\frac{\lambda_0}{2}$$
[Solution Description]
To solve this problem, we use the relationship between speed, frequency, and wavelength: $$v = \nu \lambda$$. In a vacuum, the wave speed $$c$$ is given by $$c = \nu_0 \lambda_0$$, where $$\nu_0 = f_0$$ is the original frequency, and $$\lambda_0$$ is the original wavelength. When the wave enters the medium, the speed becomes $$v = \frac{c}{2}$$. The frequency remains unchanged $$(\nu = \nu_0)$$, so we set up the equation for the medium:

$$
v = \nu \lambda_{\text{medium}}
$$

Substituting the known values:

$$
\frac{c}{2} = \nu_0 \lambda_{\text{medium}}
$$

Solving for $$\lambda_{\text{medium}}$$ gives us:

$$
\lambda_{\text{medium}} = \frac{c}{2\nu_0} = \frac{\lambda_0}{2}
$$

Hence, the new wavelength in the medium is half of the original wavelength.

Key Concepts: Real-World Application

b) Solar panel
[Solution Description] The Poynting vector represents energy flow in electromagnetic fields. Among the given options, a solar panel directly converts electromagnetic energy from sunlight into electrical energy using the principles of energy transfer described by the Poynting vector.

Key Concepts: Capacitor Analysis

c) Displacement current allows continuity in the magnetic field even without conduction current between the plates.

[Solution Description]

In a capacitor circuit with an AC source, the conduction current flows in the wire leading to one plate of the capacitor but not between the plates because there is no direct conductive path. However, the changing electric field between the capacitor plates gives rise to a displacement current, allowing the continuation of the magnetic effects within the loop as described by the Ampere-Maxwell Law.

Hence, the correct statement is that the displacement current allows continuity in the magnetic field even without conduction current between the plates.

Key Concepts: Conceptual Linkage

d) It decreases by a factor of $$\sqrt{8}$$
[Solution Description]
The speed of light $$v$$ in a medium can be determined using the formula:

$$
v = \frac{1}{\sqrt{\mu \epsilon}}
$$

Given that $$\epsilon = 4\epsilon_0$$ and $$\mu = 2\mu_0$$, substitute these values into the equation:

$$
v = \frac{1}{\sqrt{(2\mu_0)(4\epsilon_0)}} = \frac{1}{\sqrt{8\mu_0\epsilon_0}}
$$

Since $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$, we have:

$$
v = \frac{c}{\sqrt{8}} = \frac{3 \times 10^8}{2 \sqrt{2}} \approx 1.06 \times 10^8 \, \text{m/s}
$$

Key Concepts: Frequency Calculation

a) 9.55 $$\times 10^{13}$$ Hz
[Solution Description]

The frequency $$\nu$$ can be calculated from the angular frequency $$\omega$$ using the formula:

$$
\nu = \frac{\omega}{2\pi}
$$

Substituting the given value of $$\omega = 6 \times 10^{14}$$ rad/s:

$$
\nu = \frac{6 \times 10^{14}}{2\pi} \approx \frac{6 \times 10^{14}}{6.283} \approx 9.55 \times 10^{13} \text{ Hz}
$$

Therefore, the frequency of the wave is approximately $$9.55 \times 10^{13}$$ Hz.

Key Concepts: Oscillating Fields

c) Assertion is true, but Reason is false.

[Solution Description]

Assertion: “In electromagnetic waves, electric and magnetic fields oscillate in phase.” is true because Maxwell’s equations predict that both fields reach their maximum values simultaneously.

Reason: “The oscillations of electric and magnetic fields are independent of each other.” is false because the oscillations are not independent; they are coupled as described by Maxwell’s equations and occur due to changes in one another.

Hence, the correct answer is option c).

Key Concepts: Wave Equation Application

c) $$503 \text{ nm}$$
[Solution Description]
To find the wavelength $$\lambda$$, we use the relation between the wavenumber $$k$$ and wavelength:
$$
k = \frac{2 \pi}{\lambda}
$$
Rearranging for $$\lambda$$ gives:
$$
\lambda = \frac{2 \pi}{k}
$$
Substituting $$k = 1.25 \times 10^7 \text{ m}^{-1}$$:
$$
\lambda = \frac{2 \times \pi}{1.25 \times 10^7}
$$
$$
\lambda \approx 5.03 \times 10^{-7} \text{ m}
$$
Hence, the wavelength is approximately $$503 \text{ nm}$$.

Key Concepts: Conceptual Understanding

c) It is doubled
[Solution Description] The relationship between the electric field $$E_0$$ and the magnetic field $$B_0$$ in an electromagnetic wave is given by $$B_0 = \frac{E_0}{c}$$. If $$E_0$$ is doubled, the equation becomes:
$$
B_0′ = \frac{2E_0}{c} = 2 \times \frac{E_0}{c} = 2B_0
$$
Thus, when the electric field is doubled, the magnetic field also doubles.

Key Concepts: Advanced Calculations, Conceptual Understanding

c) $$7.07 \times 10^7 \mathrm{\ m/s}$$
[Solution Description] To find the speed of an electromagnetic wave in a medium, we use the formula $$v = 1/\sqrt{\mu \epsilon}$$. Here, $$\mu = 5 \times 10^{-7} \mathrm{\ H/m}$$ and $$\epsilon = 4 \times 10^{-11} \mathrm{\ F/m}$$.

First, calculate the product $$\mu \epsilon$$:
$$
\mu \epsilon = (5 \times 10^{-7}) (4 \times 10^{-11}) = 20 \times 10^{-18}
$$

Then, find the square root of $$\mu \epsilon$$:
$$
\sqrt{\mu \epsilon} = \sqrt{20 \times 10^{-18}} = \sqrt{2 \times 10^{-17}} \approx 1.414 \times 10^{-8.5}
$$

Finally, calculate the speed $$v$$:
$$
v = \frac{1}{\sqrt{\mu \epsilon}} \approx \frac{1}{1.414 \times 10^{-8.5}} = 7.07 \times 10^7 \mathrm{\ m/s}
$$

Key Concepts: Energy Transfer

b) They can transfer energy even in a vacuum
[Solution Description] Electromagnetic waves can propagate through a vacuum without requiring a medium, as they carry energy from one place to another via oscillating electric and magnetic fields. They do not lose energy rapidly in space unless interacting with matter.

Key Concepts: Advanced Wave Properties, Real-world Problem Solving

b) $$1 \times 10^{-6} \, \text{T}$$
[Solution Description]
To find the magnetic field amplitude $$B_0$$, we use the relationship between the electric field and magnetic field amplitudes:
$$
E_0 = c \times B_0.
$$
Rearranging for $$B_0$$, we have:
$$
B_0 = \frac{E_0}{c}.
$$
Substituting $$E_0 = 300 \, \text{V/m}$$ and $$c = 3 \times 10^8 \, \text{m/s}$$, we get:
$$
B_0 = \frac{300}{3 \times 10^8} = 1 \times 10^{-6} \, \text{T}.
$$
Therefore, the magnetic field amplitude $$B_0$$ is $$1 \times 10^{-6} \, \text{T}$$.

Key Concepts: Law Application

d) Faraday’s Law
[Solution Description] According to Faraday’s Law, a time-varying magnetic field induces an electromotive force and hence an electric field. Faraday’s Law is represented by the equation:
$$
\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}
$$
where $$\nabla \times \mathbf{E}$$ is the curl of the electric field, and $$-\frac{\partial \mathbf{B}}{\partial t}$$ represents the rate of change of the magnetic field.

Key Concepts: Field Relationship

b) Doubled
[Solution Description]
The relationship between the electric field amplitude $$E_0$$ and the magnetic field amplitude $$B_0$$ in an electromagnetic wave is given by:

$$
B_0 = \frac{E_0}{c}
$$

If the electric field $$E_0$$ is doubled, let’s denote the new electric field as $$2E_0$$. The new magnetic field $$B’_0$$ becomes:

$$
B’_0 = \frac{2E_0}{c}
$$

Comparing with the original magnetic field $$B_0$$:

$$
\frac{2E_0}{c} = 2 \left(\frac{E_0}{c}\right) = 2B_0
$$

Thus, the magnetic field amplitude $$B_0$$ also doubles.

Key Concepts: Experimental Validation, Wavelength Independence, Unification of Forces

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that experiments have shown that the speed of electromagnetic waves in a vacuum does not depend on their wavelength. According to the syllabus, this is true, as the speed $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ remains constant regardless of the wavelength.

The reason states that this constancy supports the unification of electric and magnetic forces through Maxwell’s equations. This is also true because Maxwell’s equations demonstrate how changing electric fields produce magnetic fields and vice versa, leading to self-propagating electromagnetic waves with constant speed $$c$$.

Therefore, both the assertion and the reason are true, and the reason correctly explains why the speed being independent of wavelength supports the unification of the forces.

Key Concepts: Wave Propagation

c) The electric and magnetic fields are perpendicular to each other and to the direction of wave propagation.

[Solution Description] Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of wave propagation. This characteristic allows electromagnetic waves to travel through the vacuum of space.

Key Concepts: Advanced Problem Solving, Real-world Scenario

b) 0.0833 m
[Solution Description]
The speed of an electromagnetic wave in a medium is given by $$v = \frac{1}{\sqrt{\mu \epsilon}}$$. For this problem, we have:

$$
v = \frac{1}{\sqrt{(4 \mu_0)(9 \epsilon_0)}} = \frac{1}{6}c = \frac{3 \times 10^8 \text{ m/s}}{6} = 5 \times 10^7 \text{ m/s}
$$

Using the relationship between speed, frequency, and wavelength $$(v = \nu \lambda)$$, we can find the wavelength:

$$
\lambda = \frac{v}{\nu} = \frac{5 \times 10^7 \text{ m/s}}{60 \times 10^6 \text{ Hz}} = \frac{5}{60} = \frac{1}{12} \text{ m} \approx 0.0833 \text{ m}
$$

Key Concepts: UV Light Sources

c) The Sun
[Solution Description]
The Sun is a significant natural source of ultraviolet light. It emits a broad spectrum of UV radiation that reaches the Earth.

Key Concepts: Frequency Relation

b) Oscillating charge
[Solution Description] The frequency of an electromagnetic wave equals the frequency of oscillation of the charge that produces it. Thus, when charges oscillate at a certain frequency, they emit electromagnetic waves at that same frequency.

Key Concepts: Application of Laws

c) Total internal reflection occurs
[Solution Description] Using Snell’s Law, we have $$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$$ where $$\theta_1$$ is the angle of incidence and $$\theta_2$$ is the angle of refraction. Substituting values: $$1.5 \times \sin{45^\circ} = 1.0 \times \sin{\theta_2}$$. Solving for $$\theta_2$$, $$\sin{\theta_2} = 1.5 \times 0.707 = 1.061$$. Since $$\sin{\theta_2} > 1$$, total internal reflection occurs, and no refraction happens.

Key Concepts: Propagation Direction

a) x-direction
[Solution Description]
In electromagnetic waves, the direction of wave propagation is perpendicular to both the electric and magnetic fields. According to the right-hand rule, if the electric field points in the y-direction and the magnetic field points in the z-direction, then the wave propagates in the x-direction.

Key Concepts: Common Uses

b) Microwaves
[Solution Description] Microwaves are often used in radar systems to detect objects at a distance. They are transmitted towards an object, and the reflected waves are analyzed to determine the object’s location and speed.

Key Concepts: Real-world Application

d) Infrared waves
[Solution Description] Infrared waves are known to be absorbed by water molecules. They are produced by hot bodies and molecules and find applications in physical therapy and remote controls due to this property.

Key Concepts: Comparative Analysis, Real World Application

b) Photographic cameras use visible rays to function.

[Solution Description]

Statement analysis:

– Option A suggests visible rays have higher frequencies than gamma rays, which is incorrect as gamma rays possess far higher frequencies.

– Option B claims photographic cameras use visible light, which is true because they capture images using light within the visible spectrum.

– Option C says visible rays have longer wavelengths than radio waves, which is false since radio waves have significantly longer wavelengths.

– Option D posits that visible rays are used in radios, but radios operate at much lower frequencies outside the visible spectrum.

Thus, option B is the correct choice.