/*! 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 52 The human eye is most sensitive ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The human eye is most sensitive to light with a frequency of about \(5.5 \times 10^{14} \mathrm{Hz},\) which is in the yellow-green region of the electromagnetic spectrum. How many wavelengths of this light can fit across the width of your thumb, a distance of about \(2.0 \mathrm{cm} ?\)

Short Answer

Expert verified
About 36,789 wavelengths of yellow-green light can fit across a 2.0 cm width.

Step by step solution

01

Understand the Relationship between Frequency and Wavelength

We know that the speed of light, \( c \), is related to the frequency, \( f \), and the wavelength, \( \lambda \), of light by the equation:\[ c = f \times \lambda \]In this problem, we want to find the wavelength, so we rearrange the formula to:\[ \lambda = \frac{c}{f} \]where \( c \approx 3.0 \times 10^8 \text{ m/s} \) is the speed of light.
02

Calculate the Wavelength of the Light

Given that the frequency, \( f \), is \( 5.5 \times 10^{14} \text{ Hz} \), we substitute this and the speed of light into the formula for wavelength:\[ \lambda = \frac{3.0 \times 10^8 \text{ m/s}}{5.5 \times 10^{14} \text{ Hz}} \]Calculate \( \lambda \):\[ \lambda \approx 5.45 \times 10^{-7} \text{ m} \]
03

Convert the Width of the Thumb to Meters

The width of your thumb is given as \( 2.0 \text{ cm} \). Convert this into meters, since the wavelength is in meters:\[ 2.0 \text{ cm} = 0.02 \text{ m} \]
04

Determine How Many Wavelengths Fit Across the Width of the Thumb

Now, divide the thumb width by the wavelength to find how many wavelengths can fit:\[ \text{Number of wavelengths} = \frac{0.02 \text{ m}}{5.45 \times 10^{-7} \text{ m}} \]Calculate the number of wavelengths:\[ \approx 36789 \text{ wavelengths} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation. This includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each type of radiation varies in frequency and wavelength.
Radiation types in the electromagnetic spectrum are ordered by frequency and wavelength.
  • Low-frequency radiation like radio waves have long wavelengths.
  • High-frequency radiation like gamma rays have short wavelengths.
Visible light, which is a small part of the electromagnetic spectrum, can be perceived by the human eye. It ranges from violet (which has a shorter wavelength and higher frequency) to red (which has a longer wavelength and lower frequency). The problem mentioned involves yellow-green light, which the human eye is most sensitive to, and falls somewhere in the middle of the visible spectrum.
Frequency
Frequency refers to the number of waves that pass a given point in one second. It is measured in Hertz (Hz). In the context of the electromagnetic spectrum, frequency can vary greatly—from a few Hz to many GHz.In the exercise,
  • The frequency is given as \(5.5 \times 10^{14} \text{ Hz}\).
  • This is characteristic of visible light.
Frequency is fundamentally important to understanding light waves because it relates to the energy of the light. Higher frequency means higher energy. Since the frequency is known, we can calculate the wavelength using the speed of light, making frequency a core part of solving such problems.
Speed of Light
The speed of light in a vacuum is a constant and crucial to calculations not just in this problem, but in all of physics. The value is approximately \(3.0 \times 10^8 \text{ m/s}\). The relationship between speed, frequency, and wavelength is expressed in the equation:
  • \(c = f \times \lambda \)
This equation shows us that the speed of light \(c\) is the product of the frequency \(f\) and the wavelength \(\lambda\). To calculate a known wavelength or frequency, this equation must be rearranged. Here, it has been used to determine the wavelength of light using its frequency. This knowledge is critical when measuring the physical dimensions light can occupy, such as in the exercise.
Unit Conversion
Correct unit conversion is pivotal in physics to ensure calculations are accurate. In this problem, we need to convert the physical width of a thumb from centimeters to meters.
  • Given: \(2.0 \text{ cm} = 0.02 \text{ m}\)
In the International System of Units (SI), wavelengths are usually expressed in meters. So, converting thumb width to meters ensures consistency and correctness in the calculation.Converting units ensures dimensions are in the same system, allowing calculations to proceed unhampered. Mistakes in unit conversion might lead to significant errors in computations. Understanding unit conversion bridges the gap between problem measurements and proper solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a certain UHF radio wave, the shortest distance between positions at which the electric and magnetic fields are zero is \(0.34 \mathrm{m} .\) Determine the frequency of this UHF radio wave.

Suppose that a police car is moving to the right at \(27 \mathrm{m} / \mathrm{s},\) while a speeder is coming up from behind at a speed of \(39 \mathrm{m} / \mathrm{s}\), both speeds being with respect to the ground. Assume that the electromagnetic wave emitted by the police car's radar gun has a frequency of \(8.0 \times 10^{9} \mathrm{Hz}\). Find the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the original frequency emitted by the police car.

Magnetic resonance imaging, or MRI (see Section 21.7 ), and positron emission tomography, or PET scanning (see Section 32.6 ), are two medical diagnostic techniques. Both employ electromagnetic waves. For these waves, find the ratio of the MRI wavelength (frequency \(=6.38 \times 10^{7} \mathrm{Hz}\) ) to the PET scanning wavelength (frequency \(=1.23 \times 10^{20} \mathrm{Hz}\)).

A truck driver is broadcasting at a frequency of 26.965 MHz with a CB (citizen's band) radio. Determine the wavelength of the electromagnetic wave being used. The speed of light is \(c=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\).

A laser emits a narrow beam of light. The radius of the beam is \(1.0 \times\) \(10^{-3} \mathrm{m},\) and the power is \(1.2 \times 10^{-3} \mathrm{W} .\) What is the intensity of the laser beam?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.