Problem 9 - College Physics

Question

\(\bullet\) Visible light. The wavelength of visible light ranges from 400 nm to 700 nm. Find the corresponding ranges of this light's (a) frequency, (b) angular frequency, (c) wave number.

Step-by-Step Solution

Answer

Frequency: \(4.29 \times 10^{14}\) Hz to \(7.5 \times 10^{14}\) Hz, Angular Frequency: \(2.696 \times 10^{15}\) rad/s to \(4.712 \times 10^{15}\) rad/s, Wave Number: \(8.98 \times 10^6\) m\(^{-1}\) to \(1.57 \times 10^7\) m\(^{-1}\).

Step 1: Understand the Relationship Between Wavelength and Frequency

The speed of light in a vacuum, denoted as \( c \), is approximately \( 3 \times 10^8 \) m/s. The relationship between the wavelength \( \lambda \) (in meters) and the frequency \( f \) (in Hertz) is given by the equation \( c = \lambda f \). We must convert the wavelength from nanometers (nm) to meters (m) before using this formula.

Step 2: Convert Wavelength from Nanometers to Meters

Given that 1 nm = \( 1 \times 10^{-9} \) m, we convert the wavelengths: \( 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4 \times 10^{-7} \text{ m} \) and \( 700 \text{ nm} = 700 \times 10^{-9} \text{ m} = 7 \times 10^{-7} \text{ m} \).

Step 3: Calculate the Frequency Range

Using \( f = \frac{c}{\lambda} \), calculate the frequency for both wavelength limits:- For \( 4 \times 10^{-7} \text{ m} \):\[ f = \frac{3 \times 10^8 \text{ m/s}}{4 \times 10^{-7} \text{ m}} = 7.5 \times 10^{14} \text{ Hz} \]- For \( 7 \times 10^{-7} \text{ m} \):\[ f = \frac{3 \times 10^8 \text{ m/s}}{7 \times 10^{-7} \text{ m}} = 4.29 \times 10^{14} \text{ Hz} \].Thus, the frequency range is \( 4.29 \times 10^{14} \text{ Hz} \) to \( 7.5 \times 10^{14} \text{ Hz} \).

Step 4: Calculate the Angular Frequency Range

The angular frequency \( \omega \) is related to frequency by \( \omega = 2\pi f \). Calculate for both:- \( \omega_{min} = 2\pi \times 4.29 \times 10^{14} \text{ Hz} \approx 2.696 \times 10^{15} \text{ rad/s} \)- \( \omega_{max} = 2\pi \times 7.5 \times 10^{14} \text{ Hz} \approx 4.712 \times 10^{15} \text{ rad/s} \).Thus, the angular frequency range is approximately from \( 2.696 \times 10^{15} \text{ rad/s} \) to \( 4.712 \times 10^{15} \text{ rad/s} \).

Step 5: Calculate the Wave Number Range

The wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \). Compute for both wavelengths:- \( k_{min} = \frac{2\pi}{7 \times 10^{-7} \text{ m}} \approx 8.98 \times 10^6 \text{ m}^{-1} \)- \( k_{max} = \frac{2\pi}{4 \times 10^{-7} \text{ m}} \approx 1.57 \times 10^7 \text{ m}^{-1} \).So, the wave number range is \( 8.98 \times 10^6 \text{ m}^{-1} \) to \( 1.57 \times 10^7 \text{ m}^{-1} \).

Key Concepts

Wavelength

The concept of wavelength is fundamental in understanding the nature of light. Wavelength, usually denoted by the symbol \( \lambda \), is the distance between consecutive peaks (or troughs) of a wave. It is a crucial parameter in determining the color of visible light. In the context of visible light, wavelength ranges from 400 nm (nanometers) to 700 nm. This range corresponds to the spectrum of light perceivable by the human eye.

• Wavelength affects how we perceive color.
• Shorter wavelengths (around 400 nm) correspond to blue/violet light.
• Longer wavelengths (around 700 nm) correspond to red light.

To convert these wavelengths into meters for calculations, remember that 1 nm equals \( 1 \times 10^{-9} \) meters. Therefore, 400 nm converts to \( 4 \times 10^{-7} \) meters, and 700 nm converts to \( 7 \times 10^{-7} \) meters.

Frequency

Frequency \( f \) represents how often a wave peak passes a fixed point in one second, measured in Hertz (Hz). For visible light, it is calculated using the speed of light \( c \) and wavelength \( \lambda \) with the formula \( c = \lambda f \).

• In a vacuum, the speed of light \( c \approx 3 \times 10^8 \) m/s.
• Inverse relation: As wavelength decreases, frequency increases.

For the visible light range:

• Using the shortest wavelength \( 4 \times 10^{-7} \) m: \( f = \frac{3 \times 10^8}{4 \times 10^{-7}} = 7.5 \times 10^{14} \) Hz.
• Using the longest wavelength \( 7 \times 10^{-7} \) m: \( f = \frac{3 \times 10^8}{7 \times 10^{-7}} = 4.29 \times 10^{14} \) Hz.

This means that frequency for visible light spans from approximately \( 4.29 \times 10^{14} \) Hz to \( 7.5 \times 10^{14} \) Hz, with each frequency corresponding to a different color in the spectrum.

Angular Frequency

Angular frequency, denoted \( \omega \), extends our understanding of frequency. It defines how fast the wave oscillates within a circle, measured in radians per second (rad/s). It is related to the regular frequency by the equation \( \omega = 2\pi f \).

• \( 2\pi \) in the formula accounts for the angular measure in a full wave cycle.

To find the angular frequency range for visible light:

• Minimum angular frequency \( \omega_{min} = 2\pi \times 4.29 \times 10^{14} \approx 2.696 \times 10^{15} \) rad/s.
• Maximum angular frequency \( \omega_{max} = 2\pi \times 7.5 \times 10^{14} \approx 4.712 \times 10^{15} \) rad/s.

This shows how angular frequency provides another perspective on wave oscillations, highlighting its occurrence in circular motion.

Wave Number

Wave number, represented as \( k \), describes the spatial frequency of a wave, or how many wave cycles fit into a unit distance. It relates to the wavelength as \( k = \frac{2\pi}{\lambda} \), expressing the number of wave cycles per meter.

• The wave number links directly with wavelength: as wavelength increases, wave number decreases.

For the visible light range:

• Minimum wave number \( k_{min} = \frac{2\pi}{7 \times 10^{-7}} \approx 8.98 \times 10^6 \) m\(^{-1}\).
• Maximum wave number \( k_{max} = \frac{2\pi}{4 \times 10^{-7}} \approx 1.57 \times 10^7 \) m\(^{-1}\).

Wave number is critical in fields like spectroscopy and quantum mechanics, where understanding wave behavior in terms of space is essential.