/*! 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 34 The wavelength of yellow sodium ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The wavelength of yellow sodium light in air is \(589 \mathrm{~nm}\). (a) What is its frequency? (b) What is its wavelength in glass whose index of refraction is \(1.92 ?\) (c) From the results of (a) and (b), find its speed in this glass.

Short Answer

Expert verified
(a) Frequency: \(5.09 \times 10^{14} \text{ Hz}\); (b) Wavelength in glass: \(306.77 \text{ nm}\); (c) Speed in glass: \(1.56 \times 10^8 \text{ m/s}\).

Step by step solution

01

Identify the Known Variables

First, we identify the known variables from the problem.- Wavelength in air, \( \lambda_{\text{air}} = 589 \text{ nm} = 589 \times 10^{-9} \text{ m} \)- Speed of light in air (or vacuum), \( c = 3 \times 10^8 \text{ m/s} \)
02

Calculate the Frequency in Air

We can calculate the frequency of light using the equation:\[ f = \frac{c}{\lambda_{\text{air}}}\]Substitute the known values:\[ f = \frac{3 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} \]\[ f \approx 5.09 \times 10^{14} \text{ Hz}\]
03

Solve for the Wavelength in Glass

The wavelength in glass can be found using the formula:\[ \lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n} \]where \( n = 1.92 \) is the refractive index of glass.\[ \lambda_{\text{glass}} = \frac{589 \times 10^{-9} \text{ m}}{1.92} \] \[ \lambda_{\text{glass}} \approx 306.77 \times 10^{-9} \text{ m} = 306.77 \text{ nm}\]
04

Calculate the Speed of Light in Glass

The speed of light in glass, \( v \), can be derived using the formula:\[ v = f \times \lambda_{\text{glass}} \] Substitute the values from Step 2 and Step 3:\[ v = 5.09 \times 10^{14} \text{ Hz} \times 306.77 \times 10^{-9} \text{ m} \]\[ v \approx 1.56 \times 10^8 \text{ m/s}\]

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.

Frequency Calculation
To calculate the frequency of a light wave, you need to understand the relationship between speed, wavelength, and frequency. The formula to find the frequency (\( f \)) of light is:
  • \( f = \frac{c}{\lambda} \)

Here, \( c \) is the speed of light in a vacuum or air, which is approximately \( 3 \times 10^8 \text{ m/s} \). \( \lambda \) is the wavelength of the light.
For yellow sodium light in air, the problem gives us \( \lambda_{\text{air}} = 589 \text{ nm} \), which can be converted to meters (\( 589 \times 10^{-9} \text{ m} \)).
By substituting these values into the formula, you can calculate the frequency:
  • \( f = \frac{3 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} \approx 5.09 \times 10^{14} \text{ Hz} \)
This frequency means the number of wave cycles that pass through a point each second is approximately \( 5.09 \times 10^{14} \). Understanding this relationship helps in knowing how light behaves when traveling through different mediums.
Index of Refraction
The index of refraction, denoted by \( n \), is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. The index of refraction can be calculated using the formula:
  • \( n = \frac{c}{v} \)

Here, \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.Another way to use the index is in calculating the wavelength in a different medium. In this exercise, the wavelength of yellow sodium light in glass can be found using:
  • \( \lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n} \)

For the given problem:
  • \( n = 1.92 \)
  • \( \lambda_{\text{glass}} = \frac{589 \times 10^{-9} \text{ m}}{1.92} \approx 306.77 \times 10^{-9} \text{ m} \)
This shortened wavelength in glass is because light travels slower in glass than in the air. The index of refraction indicates the bending and slowing down of light as it travels into the glass.
Speed of Light in Medium
Light does not travel at the same speed in all substances. In a vacuum or air, it moves at \( 3 \times 10^8 \text{ m/s} \). However, in materials such as glass, it slows down. The speed of light in a medium is calculated using the frequency:
  • \( v = f \times \lambda_{\text{medium}} \)

Substituting the values obtained in previous steps of the exercise:
  • Frequency, \( f \approx 5.09 \times 10^{14} \text{ Hz} \)
  • Wavelength in glass, \( \lambda_{\text{glass}} \approx 306.77 \times 10^{-9} \text{ m} \)
The speed \( v \) is:
  • \( v = 5.09 \times 10^{14} \times 306.77 \times 10^{-9} \text{ m/s} \approx 1.56 \times 10^8 \text{ m/s} \)
This result shows that light travels roughly half as fast in glass as it does in air or a vacuum. Understanding the speed of light in different mediums explains how light behaves in everyday objects like lenses or water, affecting our perception and optics technology.

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 double-slit experiment, the distance between slits is \(5.0 \mathrm{~mm}\) and the slits are \(1.8 \mathrm{~m}\) from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength \(480 \mathrm{~nm}\), and the other due to light of wavelength \(600 \mathrm{~nm}\). What is the separation on the screen between the third-order \((m=3)\) bright fringes of the two interference patterns?

A disabled tanker leaks kerosene \((n=1.20)\) into the Persian Gulf, creating a large slick on top of the water ( \(n=1.30\) ). (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is \(380 \mathrm{~nm}\), for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

In a double-slit experiment, the fourth-order maximum for a wavelength of \(425 \mathrm{~nm}\) occurs at an angle of \(\theta=90^{\circ}\). (a) What range of wavelengths in the visible range (400 nm to \(700 \mathrm{~nm}\) ) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at \(478.8 \mathrm{~nm}\) and a minimum at \(598.5 \mathrm{~nm}\), with no minimum in between. If \(n=1.33\) for the film, what is the film thickness, assumed uniform?

Two rectangular glass plates \((n=1.60)\) are in contact along one edge and are separated along the opposite edge (Fig. 35-25). Light with a wavelength of \(700 \mathrm{~nm}\) is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the separated edges is increased by \(700 \mathrm{~nm}\), how many dark fringes will there then be across the top plate?

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.