Problem 3 - University Physics with Modern Physics

Question

For an electromagnetic wave propagating in air, determine the frequency of a wave with a wavelength of (a) \(5.0 \mathrm{km} ;\) (b) 5.0 \(\mathrm{m}\) (c) \(5.0 \mu \mathrm{m} ;\) (d) 5.0 \(\mathrm{nm}\)

Step-by-Step Solution

Answer

The frequencies are 60 kHz, 60 MHz, 60 THz, and 60 PHz respectively.

Step 1: Understanding the Relationship

The frequency \( f \) of an electromagnetic wave is related to its wavelength \( \lambda \) by the formula \( f = \frac{c}{\lambda} \), where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \) meters per second.

Step 2: Convert the Wavelength to Meters

Ensure that each given wavelength is converted into meters as it is necessary for consistency in using the speed of light expressed in meters per second. For example:- For \( 5.0 \text{ km} \), \( \lambda = 5000 \text{ meters} \).- For \( 5.0 \text{ m} \), \( \lambda = 5 \text{ meters} \).- For \( 5.0 \mu \text{m} \), \( \lambda = 5.0 \times 10^{-6} \text{ meters} \).- For \( 5.0 \text{ nm} \), \( \lambda = 5.0 \times 10^{-9} \text{ meters} \).

Step 3: Calculate the Frequency for Each Wavelength

Use the formula \( f = \frac{c}{\lambda} \) to determine the frequency for each wavelength:- For \( 5.0 \text{ km} \): \( f = \frac{3 \times 10^8}{5000} = 6 \times 10^4 \text{ Hz} \).- For \( 5.0 \text{ m} \): \( f = \frac{3 \times 10^8}{5} = 6 \times 10^7 \text{ Hz} \).- For \( 5.0 \mu \text{m} \): \( f = \frac{3 \times 10^8}{5.0 \times 10^{-6}} = 6 \times 10^{13} \text{ Hz} \).- For \( 5.0 \text{ nm} \): \( f = \frac{3 \times 10^8}{5.0 \times 10^{-9}} = 6 \times 10^{16} \text{ Hz} \).

Key Concepts

Frequency Calculation

Frequency calculation is a foundational concept in understanding electromagnetic waves. The frequency of a wave refers to the number of cycles that occur in one second. It is measured in hertz (Hz). To compute frequency, we use the formula:

• \( f = \frac{c}{\lambda} \)

Here, \( f \) stands for frequency, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the wave.

The speed of light \( c \) is a constant, approximately \( 3 \times 10^8 \) meters per second. By rearranging the formula \( f = \frac{c}{\lambda} \), it becomes clear that as the wavelength decreases, the frequency increases. This inverse relationship explains why shorter wavelengths result in higher frequencies, which can lead to very different properties and behaviors of waves in applications like radio, microwaves, and light.

Wavelength Conversion

Wavelength conversion is crucial for simplifying the frequency calculation process to maintain unit consistency. It's important to have the wavelength in meters, especially because the speed of light is expressed in meters per second under the International System of Units (SI).

To convert different units to meters:

• 1 km = 1,000 meters, so 5 km = 5,000 meters.
• 1 µm (micrometer) = \( 10^{-6} \) meters, so 5 µm = \( 5 \times 10^{-6} \) meters.
• 1 nm (nanometer) = \( 10^{-9} \) meters, so 5 nm = \( 5 \times 10^{-9} \) meters.

Always check the unit of the given wavelength to ensure you are using the formula correctly. This seamless conversion is necessary to perform accurate calculations of frequency.

Speed of Light

The speed of light is a critical constant in physics, denoted by \( c \). It is defined as precisely \( 299,792,458 \) meters per second but often approximated as \( 3 \times 10^8 \) meters per second for simplicity in calculations. This speed is remarkable because it remains constant in a vacuum, unaffected by the observer's frame of reference.

The speed of light's constancy facilitates calculations in electromagnetic theory and is fundamental when relating wavelength to frequency. Its unchanging nature in a vacuum is what allows us to rely on simple equations like \( f = \frac{c}{\lambda} \) without worrying about more complex variables, such as medium variability. This plays a pivotal role in technologies ranging from GPS systems to fiber optics.

Wave Propagation in Air

Wave propagation describes how electromagnetic waves travel from one point to another through a medium. In the case of air, wave propagation is similar to that in a vacuum, as air has a minimal effect on the speed of light. Hence, the speed is still approximately \( 3 \times 10^8 \) meters per second.

When waves propagate through air, they can be absorbed, reflected, refracted, or scattered. However, these effects typically only slightly alter the speed, making air a practical medium for everyday wave transmission, like radio waves and light waves. Understanding these behaviors helps in the optimization of signals in wireless communications and the study of phenomena like refraction and diffraction in weather prediction and astronomy.