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Chapter 22: Problem 57

A star is moving away from Earth at a speed of $2.4 \times 10^{8} \mathrm{m} / \mathrm{s} .\( Light of wavelength \)480 \mathrm{nm}$ is emitted by the star. What is the wavelength as measured by an Earth observer?

Short Answer

Expert verified
Answer: The redshifted wavelength of the light as observed from Earth is approximately 864 nm.

Step by step solution

01

Understand the redshift formula

The redshift formula relates the initial wavelength (λ₀) of light emitted by a moving object and the observed wavelength (λ), taking into account the relative speed (v) of the observer and the object. The redshift formula is given by: \(λ = λ₀\left(1+\frac{v}{c}\right)\) where c is the speed of light in vacuum \(c = 3 × 10^8 m/s \). In this exercise, we are given the initial wavelength of light (λ₀) and the speed of the star (v) moving away from Earth. We need to find the redshifted wavelength (λ) as seen by an observer on Earth.
02

List the given values

We are given the following values: Initial Wavelength (λ₀): \(480 nm = 480 × 10^{-9} m\) Speed of the Star (v): \(2.4 × 10^8 m/s\) Speed of Light (c): \(3 × 10^8 m/s\) We need to find the observed wavelength (λ).
03

Calculate the redshifted wavelength

Plug in the given values into the redshift formula: \(λ = λ₀\left(1+\frac{v}{c}\right) = (480 × 10^{-9})\left(1+\frac{2.4 × 10^8}{3 × 10^8}\right)\)
04

Solve for the observed wavelength

Calculate the observed wavelength by solving the equation: \(λ = (480 × 10^{-9})\left(1+\frac{2.4 × 10^8}{3 × 10^8}\right) = (480 × 10^{-9})(1 + 0.8) ≈ (480 × 10^{-9})(1.8) = 864 × 10^{-9} m\)
05

Convert the result to nanometers

Convert the wavelength from meters back to nanometers: \(λ = 864 × 10^{-9} m = 864 nm\) The wavelength of the light as measured by an Earth observer is approximately 864 nm.

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Most popular questions from this chapter

An unpolarized beam of light (intensity \(I_{0}\) ) is moving in the \(x\) -direction. The light passes through three ideal polarizers whose transmission axes are (in order) at angles \(0.0^{\circ}, 45.0^{\circ},\) and \(30.0^{\circ}\) counterclockwise from the \(y\) -axis in the \(y z\) -plane. (a) What is the intensity and polarization of the light that is transmitted by the last polarizer? (b) If the polarizer in the middle is removed, what is the intensity and polarization of the light transmitted by the last polarizer?
Vertically polarized light with intensity \(I_{0}\) is normally incident on an ideal polarizer. As the polarizer is rotated about a horizontal axis, the intensity \(I\) of light transmitted through the polarizer varies with the orientation of the polarizer \((\theta),\) where \(\theta=0\) corresponds to a vertical transmission axis. Sketch a graph of \(I\) as a function of \(\theta\) for one complete rotation of the polarizer $\left(0 \leq \theta \leq 360^{\circ}\right)$.(W) tutorial: polarized light).
Polarized light of intensity \(I_{0}\) is incident on a pair of polarizing sheets. Let \(\theta_{1}\) and \(\theta_{2}\) be the angles between the direction of polarization of the incident light and the transmission axes of the first and second sheets, respectively. Show that the intensity of the transmitted light is $I=I_{0} \cos ^{2} \theta_{1} \cos ^{2}\left(\theta_{1}-\theta_{2}\right)$.
How far does a beam of light travel in 1 ns?
A 60.0 -m \(\mathrm{W}\) pulsed laser produces a pulse of EM radia tion with wavelength \(1060 \mathrm{nm}\) (in air) that lasts for 20.0 ps (picoseconds). (a) In what part of the EM spectrum is this pulse? (b) How long (in centimeters) is : single pulse in air? (c) How long is it in water \((n=1.33) \equiv\) (d) How many wavelengths fit in one pulse? (e) What is the total electromagnetic energy in one pulse?
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