Problem 5
Question
$$ \begin{aligned} &c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \\ &f=45.6 \mathrm{MHz} \\ &\lambda=? \end{aligned} $$
Step-by-Step Solution
Verified Answer
The wavelength is 6.58 m.
1Step 1: Identify the formula needed
To find the wavelength \( \lambda \), we use the formula: \( \lambda = \frac{c}{f} \), where \( c \) is the speed of light and \( f \) is the frequency.
2Step 2: Convert units if necessary
Check the unit of frequency. Here, the frequency \( f = 45.6 \text{ MHz} \) which is \( 45.6 \times 10^6 \text{ Hz} \). It's important to express the frequency in Hz for calculation.
3Step 3: Substitute the values into the formula
Substitute \( c = 3.00 \times 10^8 \text{ m/s} \) and \( f = 45.6 \times 10^6 \text{ Hz} \) into the formula: \( \lambda = \frac{3.00 \times 10^8}{45.6 \times 10^6} \).
4Step 4: Calculate the wavelength
Perform the division: \( \lambda = \frac{3.00 \times 10^8}{45.6 \times 10^6} = 6.58 \text{ m} \).
Key Concepts
Speed of LightFrequency ConversionUnit ConversionApplied Physics
Speed of Light
The speed of light is an incredible constant in physics, serving as a crucial bridge between various domains of science. It is represented by the symbol \( c \).
The speed of light in a vacuum is precisely \( 3.00 \times 10^8 \) meters per second (\( ext{m/s} \)). This constant helps us compute several properties of waves, especially in the electromagnetic spectrum, like radio waves, microwaves, and light itself.
The speed of light in a vacuum is precisely \( 3.00 \times 10^8 \) meters per second (\( ext{m/s} \)). This constant helps us compute several properties of waves, especially in the electromagnetic spectrum, like radio waves, microwaves, and light itself.
- Light travels amazingly fast, completing the journey from the Sun to Earth in about 8 minutes and 20 seconds.
- Importantly, \( c \) remains the same regardless of the observer’s motion or the light source.
Frequency Conversion
Frequency conversion is a necessary step when working with wave properties. It involves converting units to ensure uniformity in calculations. Frequency is typically measured in hertz (Hz), which is the number of cycles per second. However, other units like megahertz (MHz) are often used.
For example, in this exercise:
This conversion is crucial because the SI unit of frequency is Hz, allowing for accurate calculations in physics equations.
- 1 MHz means 1 million hertz or \( 10^6 \) Hz.
For example, in this exercise:
- You have to convert 45.6 MHz to Hz, which is \( 45.6 \times 10^6 \) Hz.
This conversion is crucial because the SI unit of frequency is Hz, allowing for accurate calculations in physics equations.
Unit Conversion
Unit conversion is essential in physics to maintain consistency across calculations. When dealing with quantities like wavelength, frequency, and speed, ensuring all components use compatible units is key.
Without converting units, calculations can become inaccurate or even lead to incorrect answers. Consistent units simplify the substitution into formulas and ensure that results, like wavelength here, come out as expected (e.g., in meters).
- The speed \( c \) is given in meters per second (\( ext{m/s} \)).
- The frequency was originally given in megahertz (MHz) and was converted to hertz (Hz).
Without converting units, calculations can become inaccurate or even lead to incorrect answers. Consistent units simplify the substitution into formulas and ensure that results, like wavelength here, come out as expected (e.g., in meters).
Applied Physics
Applied physics involves using scientific principles in practical scenarios, often blending theoretical analysis with experiments. In the context of waves, understanding how their characteristics, such as speed and frequency, can impact various technologies is a key aspect.
Thus, the theoretical computation of wavelength, as learned in this problem, directly influences technological advancements and applications in numerous fields like telecommunications and radar systems.
- For example, radio and microwave communications heavily rely on understanding frequency and wavelength.
- Knowing the wavelength helps design antennas and optimize signals for better communication.
Thus, the theoretical computation of wavelength, as learned in this problem, directly influences technological advancements and applications in numerous fields like telecommunications and radar systems.
Other exercises in this chapter
Problem 4
$$ \begin{aligned} &c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \\ &f=24.2 \mathrm{MHz} \\ &\lambda=? \end{aligned} $$
View solution Problem 5
$$ \text { Find the frequency of electromagnetic radiation with energy } 5.50 \times 10^{-26} \mathrm{~J} \text { . } $$
View solution Problem 6
$$ \begin{aligned} &\text { Find the energy of a photon of electromagnetic radiation with frequency }\\\ &2.50 \times 10^{12} \mathrm{~Hz} \end{aligned} $$
View solution Problem 6
$$ \begin{aligned} &c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \\ &f=415 \mathrm{~Hz} \\ &\lambda=? \end{aligned} $$
View solution