Problem 49 - Fundamentals of Physics

Question

Assume that lasers are available whose wavelengths can be precisely "tuned" to anywhere in the visible range \(-\) that is, in the range \(450 \mathrm{~nm}<\lambda<650 \mathrm{~nm}\). If every television channel occupies a bandwidth of \(10 \mathrm{MHz}\), how many channels can be accommodated within this wavelength range?

Step-by-Step Solution

Answer

20,500,000 channels can be accommodated.

Step 1: Calculate the frequency range of given wavelengths

First, we need to convert the given wavelength range to frequency. The formula to convert wavelength \( \lambda \) to frequency \( f \) is \( f = \frac{c}{\lambda} \), where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \). We calculate the frequency corresponding to \( 450 \, \text{nm} \) and \( 650 \, \text{nm} \). Convert nanometers to meters: \( 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m} \) and \( 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \).

Step 2: Calculate frequency for 450 nm and 650 nm

Using the formula \( f = \frac{c}{\lambda} \):- For \( 450 \, \text{nm} \), \( f_{450} = \frac{3 \times 10^8}{450 \times 10^{-9}} \, \text{Hz} = 6.67 \times 10^{14} \, \text{Hz} \).- For \( 650 \, \text{nm} \), \( f_{650} = \frac{3 \times 10^8}{650 \times 10^{-9}} \, \text{Hz} = 4.62 \times 10^{14} \, \text{Hz} \).

Step 3: Determine the frequency range available for channels

The frequency range is the difference between the frequencies at \( 450 \, \text{nm} \) and \( 650 \, \text{nm} \):\[\Delta f = 6.67 \times 10^{14} \, \text{Hz} - 4.62 \times 10^{14} \, \text{Hz} = 2.05 \times 10^{14} \, \text{Hz}.\]

Step 4: Calculate the number of channels

Given that each television channel occupies \( 10 \, \text{MHz} = 10 \times 10^6 \, \text{Hz} \), the number of channels that can be accommodated is the available frequency range divided by the bandwidth of each channel:\[ \text{Number of channels} = \frac{2.05 \times 10^{14}}{10 \times 10^6} = 2.05 \times 10^7. \] This implies that \( 20,500,000 \) channels can be accommodated.

Key Concepts

Frequency Calculation

In understanding the electromagnetic spectrum, it is vital to calculate frequencies from known wavelengths. This concept is central to many practical applications, such as determining how many TV channels can fit in a specific bandwidth. To calculate frequency, we use the formula: \[ f = \frac{c}{\lambda} \]where:

• \( f \) is the frequency,
• \( c \) is the speed of light (approximately \( 3 \times 10^8 \, \text{m/s} \)), and
• \( \lambda \) is the wavelength.

For instance, with a wavelength of \( 450 \, \text{nm} \), by converting it to meters \( (450 \, \text{nm} = 450 \times 10^{-9} \, \text{m}) \) and applying it to the formula, the resulting frequency would be approximately \( 6.67 \times 10^{14} \, \text{Hz} \). This transformation from wavelength to frequency is crucial for comprehending how different electromagnetic waves behave.

Wavelength to Frequency Conversion

Converting wavelength to frequency forms a bridge to understanding the full electromagnetic spectrum. The process involves using the formula \( f = \frac{c}{\lambda} \) to transform wavelength \((\lambda)\), given in nanometers, to frequency \((f)\), in hertz.
This transformation is essential because while wavelengths are often given in terms of the visible spectrum (e.g., 450 nm to 650 nm for light), the behavior and interaction of waves with matter are more frequently understood in terms of their frequency.

For example, converting a wavelength of 650 nm, by using \( c = 3 \times 10^8 \, \text{m/s} \), results in a frequency of about \( 4.62 \times 10^{14} \, \text{Hz} \).
Such conversions not only allow for the examination of light interactions at different wavelengths but also help in practical applications like signal transmission, where precise frequency calculations are critical.

Bandwidth

Bandwidth is a pivotal factor in signal processing and communications. It defines the range of frequencies available within a particular band of the electromagnetic spectrum. In practical applications like television broadcasting, bandwidth determines how many separate channels or signals can share a particular range of frequencies.

For example, in a scenario where each TV channel requires a bandwidth of \( 10 \, \text{MHz} \), understanding the available bandwidth between frequencies derived from the limits of \( 450 \, \text{nm} \) and \( 650 \, \text{nm} \) becomes crucial.
With available frequencies between approximately \( 6.67 \times 10^{14} \) Hz and \( 4.62 \times 10^{14} \) Hz, we can find the range or bandwidth:
\[\Delta f = 6.67 \times 10^{14} \, \text{Hz} - 4.62 \times 10^{14} \, \text{Hz} = 2.05 \times 10^{14} \, \text{Hz}\]
This calculation signifies the potential for over \( 20,500,000 \) channels given their \( 10 \, \text{MHz} \) bandwidth requirement, highlighting the efficiency and capacity inherent in well-utilized bandwidth.