/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Light is an electromagnetic wave... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 1

Light is an electromagnetic wave and travels at a speed of \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) The human eye is most sensitive to yellow-green light, which has a wavelength of \(5.45 \times 10^{-7} \mathrm{m} .\) What is the frequency of this light?

Short Answer

Expert verified
The frequency of the yellow-green light is approximately \( 5.50 \times 10^{14} \text{ Hz} \).

Step by step solution

01

Identify the Given Values

We know the speed of light is given as \( c = 3.00 \times 10^{8} \text{ m/s} \). The wavelength \( \lambda \) of the light is \( 5.45 \times 10^{-7} \text{ m} \).
02

Understand the Relationship Between Wavelength, Frequency, and Speed

The relationship between speed, wavelength, and frequency is given by the formula: \( c = \lambda \cdot f \), where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( f \) is the frequency.
03

Rearrange the Formula to Solve for Frequency

To find the frequency \( f \), rearrange the equation \( c = \lambda \cdot f \) to \( f = \frac{c}{\lambda} \).
04

Substitute the Given Values into the Formula

Substitute \( c = 3.00 \times 10^{8} \text{ m/s} \) and \( \lambda = 5.45 \times 10^{-7} \text{ m} \) into the formula: \[ f = \frac{3.00 \times 10^{8}}{5.45 \times 10^{-7}} \].
05

Calculate the Frequency

Perform the calculation:\[ f = \frac{3.00 \times 10^{8}}{5.45 \times 10^{-7}} \approx 5.50 \times 10^{14} \text{ Hz} \]. This gives us the frequency of the yellow-green light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Light
Light, a form of electromagnetic radiation, travels incredibly fast. One of the fundamental constants in physics, the speed of light (\(c\)), is approximately \(3.00 \times 10^{8}\) meters per second (m/s). Generally, electromagnetic waves, including visible light, travel at this speed in a vacuum. This value shows the universal speed limit, especially when we consider theoretical physics and relativity.
Here's why understanding the speed of light is essential:
  • All electromagnetic waves travel at this speed in a vacuum, allowing us to describe various light properties using it as a base.
  • It is pivotal in linking different aspects of waves, such as wavelength and frequency, through the formula \(c = \lambda \cdot f\).
Without light's speed, calculating or understanding other properties would be exceedingly more challenging.
Wavelength
Wavelength (\(\lambda\)) refers to the distance between any two corresponding points on consecutive cycles of a wave. For instance, you could measure the distance from peak to peak or trough to trough. Wavelengths of electromagnetic waves can vary widely, from very short gamma rays to the long wavelengths of radio waves.
When it comes to visible light, which human eyes can perceive, different wavelengths correspond to different colors. For example:
  • Red light has longer wavelengths than violet light.
  • The width of the visible spectrum is roughly from \(4 \times 10^{-7}\) to \(7 \times 10^{-7}\) meters.
  • Yellow-green light, highlighted in the problem, has a wavelength of \(5.45 \times 10^{-7}\) meters, which is where the human eye is most sensitive.
This numerical expression of wavelength helps connect with the speed and frequency through calculations and predictions in physics.
Frequency Calculation
Frequency (\(f\)) represents how often the wave cycles pass a certain point in one second, measured in hertz (Hz). Given that light waves travel so fast, their frequencies tend to be very high. Given the formula \(c = \lambda \cdot f\), which connects speed, wavelength, and frequency, you can solve for the frequency by rearranging it: \[f = \frac{c}{\lambda}\]Here's how you can calculate frequency:
  • Substitute the given speed of light \(c = 3.00 \times 10^{8} \text{ m/s}\) into the equation.
  • Use the given wavelength \(\lambda = 5.45 \times 10^{-7} \text{ m}\)
  • Perform the division: \[f = \frac{3.00 \times 10^{8}}{5.45 \times 10^{-7}} \approx 5.50 \times 10^{14}\text{ Hz}\]
This frequency indicates how many wave cycles occur per second for yellow-green light and is crucial for understanding the light's energetic properties and interactions.

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Most popular questions from this chapter

A wave traveling along the \(x\) axis is described mathematically by the equation \(y=0.17 \sin (8.2 \pi t+0.54 \pi x),\) where \(y\) is the displacement (in meters), \(t\) is in seconds, and \(x\) is in meters. What is the speed of the wave?

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

A loudspeaker in a parked car is producing sound whose frequency is \(20510 \mathrm{Hz} .\) A healthy young person with normal hearing is standing nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person? Why? (b) If the speed of sound is \(343 \mathrm{m} / \mathrm{s},\) what is the minimum speed of the moving car?

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does not change when the sound enters the water. The wavelength of the sound is \(2.74 \mathrm{m}\) in the air, and the temperature of both the air and the water is \(20^{\circ} \mathrm{C} .\) What is the wavelength in the water?

Two blocks are connected by a wire that has a mass per unit length of \(8.50 \times 10^{-4} \mathrm{kg} / \mathrm{m} .\) One block has a mass of \(19.0 \mathrm{kg},\) and the other has a mass of \(42.0 \mathrm{kg} .\) These blocks are being pulled across a horizontal frictionless floor by a horizontal force \(\overrightarrow{\mathbf{P}}\) that is applied to the less massive block. A transverse wave travels on the wire between the blocks with a speed of \(352 \mathrm{m} / \mathrm{s}\) (relative to the wire). The mass of the wire is negligible compared to the mass of the blocks. Find the magnitude of \(\overrightarrow{\mathbf{P}}\)
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