Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 112

The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it produces two first-order maxima at \(36.093^{\circ}\) and \(36.129^{\circ}\) when projected on a 10,000 line per centimeter diffraction grating. What are the two wavelengths to an accuracy of \(0.1 \mathrm{nm}\) ?

Short Answer

Expert verified
The two wavelengths produced by the sodium vapor lamp are approximately \(591.7 \, \text{nm}\) and \(592.4 \, \text{nm}\).

Step by step solution

01

Write down the given information and the diffraction grating formula

First, list the given information: 1. Angles of the first-order maxima: \(36.093^{\circ}\) and \(36.129^{\circ}\). 2. Lines per centimeter on the diffraction grating: 10,000. Now, the diffraction grating formula is: \[d \sin{\theta} = m\lambda\] where \(d\) is the distance between the lines on the diffraction grating, \(\theta\) is the angle of the maxima, \(m\) is the order of the maxima, and \(\lambda\) is the wavelength.
02

Convert lines per centimeter to distance between lines

We need to convert the given lines per centimeter (10,000) to distance between the grating lines (d) in meters. To convert lines per centimeter to meters: \[d = \frac{1}{10,000 \cdot 100}\] Calculate the value of d: \[d = \frac{1}{1,000,000} = 10^{-6} \, \text{m}\]
03

Solve for wavelengths

Now we can use the diffraction grating formula to solve for the two wavelengths, given the angles of the maxima. Since we're dealing with first-order maxima, m=1. For the angle \(\theta_1 = 36.093^{\circ}\): \[10^{-6}\text{m} \cdot \sin{36.093^{\circ}} = 1 \cdot \lambda_1\] Calculate \(\lambda_1\): \[\lambda_1 = \frac{10^{-6}\text{m} \cdot \sin{36.093^{\circ}}}{1} = 5.917 \times 10^{-7}\text{m}\] For the angle \(\theta_2 = 36.129^{\circ}\): \[10^{-6}\text{m} \cdot \sin{36.129^{\circ}} = 1 \cdot \lambda_2\] Calculate \(\lambda_2\): \[\lambda_2 = \frac{10^{-6}\text{m} \cdot \sin{36.129^{\circ}}}{1} = 5.924 \times 10^{-7}\text{m}\]
04

Convert the wavelengths to nanometers

Finally, we need to convert the wavelengths from meters to nanometers and round them to the nearest 0.1 nm. For \(\lambda_1\): \[\lambda_1 = 5.917 \times 10^{-7}\text{m} \times \frac{10^9\text{nm}}{1\text{m}} = 591.7 \, \text{nm}\] For \(\lambda_2\): \[\lambda_2 = 5.924 \times 10^{-7}\text{m} \times \frac{10^9\text{nm}}{1\text{m}} = 592.4 \, \text{nm}\] Thus, the two wavelengths produced by the sodium vapor lamp are approximately 591.7 nm and 592.4 nm.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A water break at the entrance to a harbor consists of a rock barrier with a 50.0 -m-wide opening. Ocean waves of 20.0 -m wavelength approach the opening straight on. At what angles to the incident direction are the boats inside the harbor most protected against wave action?

Two lamps producing light of wavelength 589 nm are fixed \(1.0 \mathrm{m}\) apart on a wooden plank. What is the maximum distance an observer can be and still resolve the lamps as two separate sources of light, if the resolution is affected solely by the diffraction of light entering the eye? Assume light enters the eye through a pupil of diameter \(4.5 \mathrm{mm}\).

Two slits of width \(2 \mu \mathrm{m},\) each in an opaque material, are separated by a center-to-center distance of \(6 \mu \mathrm{m}\). A monochromatic light of wavelength \(450 \mathrm{nm}\) is incident on the double- slit. One finds a combined interference and diffraction pattern on the screen. (a) How many peaks of the interference will be observed in the central maximum of the diffraction pattem? (b) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same? (c) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, \(12 \mu \mathrm{m}\), while keeping the widths of the slits same? (d) What will happen in (a) if instead of 450-nm light another light of wavelength \(680 \mathrm{nm}\) is used? (e) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)? (f) Does this ratio depend on the wavelength of the light? (g) Does this ratio depend on the width or separation of the slits?

Compare interference and diffraction.

What is the minimum angular separation of two stars that are just-resolvable by the \(8.1-\mathrm{m}\) Gemini South telescope, if atmospheric effects do not limit resolution? Use \(550 \mathrm{nm}\) for the wavelength of the light from the stars.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Mechanics and Materials

Read Explanation

Electricity

Read Explanation

Electrostatics

Read Explanation

Engineering Physics

Read Explanation

Space Physics

Read Explanation
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.

Sign-up for free