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Chapter 15: Problem 2124

Speed of electromagnetic wave is the same (A) for all wavelengths (B) in all media (C) for all intensities (D) for all frequencies

Short Answer

Expert verified
The speed of electromagnetic waves is the same for all intensities (C), as it is not directly dependent on the amount of energy the wave carries. This holds true as long as the wavelength and medium do not change.

Step by step solution

01

Analyzing Option A: For all wavelengths

Electromagnetic waves can have different wavelengths, but, in a vacuum, their speed remains constant, and the frequency will adjust accordingly to maintain that speed. However, when they enter different media, the speed may change, so this statement is not always true.
02

Analyzing Option B: In all media

The speed of electromagnetic waves depends on the medium through which they travel. In a vacuum, the speed remains constant at \(3 \times 10^8 m/s\), but in other media such as air, water, or glass, the speed will generally reduce. This is due to the interaction between the waves and the atoms in the medium. Therefore, this statement is not true.
03

Analyzing Option C: For all intensities

The intensity of an electromagnetic wave refers to the amount of energy it carries. The speed of the wave is not directly dependent on its intensity. A higher intensity wave may have more energy, but it will still travel with the same speed if the wavelength and medium do not change. Therefore, this statement is true.
04

Analyzing Option D: For all frequencies

The speed of electromagnetic waves in a vacuum is constant, and the wavelength changes according to the frequency to maintain that constant speed. However, in different media, the speed may vary with frequency. So, this statement is not always true. Given the analysis above, the correct answer is: (C) for all intensities

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Most popular questions from this chapter

A plane electromagnetic wave of frequency \(25 \mathrm{MHz}\) travels in free space along the \(\mathrm{x}\) direction. At a particular point in space and time \(\mathrm{E}^{-}=6.3 \mathrm{j} \wedge \mathrm{Vm}^{-1}\) then \(\mathrm{B}^{-}\) at this point is (A) \(2.1 \times 10^{-8}\) i \(\mathrm{T}\) (B) \(2.1 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (C) \(1.89 \times 10^{9} \mathrm{k} \wedge \mathrm{T}\) (D) \(2.52 \times 10^{-7} \mathrm{k} \wedge \mathrm{T}\)

A point source of electromagnetic radiation has an average output power of \(800 \mathrm{~W}\). The maximum value of electric field at a distance of $4.0 \mathrm{~m}$ from the source is (A) \(64.7 \mathrm{Vm}^{-1}\) (B) \(57.8 \mathrm{Vm}^{-1}\) (C) \(56.72 \mathrm{Vm}^{-1}\) (D) \(54.77 \mathrm{Vm}^{-1}\)

Unit of \(\mu_{0} \mathrm{c}\) is same as that of (A) current (B) resistance (C) electric charge (D) velocity

The frequencies of \(\mathrm{x}\) rays, \(\gamma\) rays and ultraviolet rays are respectively \(\mathrm{p}, \mathrm{q}\) and \(\mathrm{r}\) then (A) \(pr\) (B) \(p>q, q>r\) (C) \(p>q, q

Astronomers have found that electromagnetic waves of wavelength $21 \mathrm{~cm}$ are continuously reaching the Earth's surface. Calculate the frequency of this radiation. $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ (A) \(14.28 \mathrm{GHz}\) (B) \(1.428 \mathrm{kHz}\) (C) \(1.428 \mathrm{MHz}\) (D) \(1.428 \mathrm{GHz}\)
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