Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 96

A grating with 4000 lines per centimeter is used to diffract light that contains all wavelengths between 400 and \(650 \mathrm{nm} .\) How wide is the first-order spectrum on a screen \(3.0 \mathrm{m}\) from the grating?

Short Answer

Expert verified
The width of the first-order spectrum on a screen 3.0 m from the grating is the difference of positions of light of wavelengths 400 nm and 650 nm, calculated using the grating equation and trigonometry.

Step by step solution

01

Identify Given Values

First, recognize the given values: the number of lines per cm in the grating (4000), the range of wavelengths between \(400 \) and \(650 \) nm, and the distance from the grating to the screen (\(3.0 \) m). The goal is to find the width of the first order spectrum on the screen.
02

Convert Units

Convert the given values into appropriate units for this problem. The number of lines per centimeter needs to be converted to lines per meter by multiplying by 10000 (since there are 10000 millimeters in 1 meter), which gives \(4.0 \times 10^7 \) lines/m. Similarly, the range of wavelengths must be converted from nanometers to meters by multiplying by \(1 \times 10^{-9} \)\, yielding a range of wavelengths from \(4.0 \times 10^{-7} \) m to \(6.5 \times 10^{-7} \) m.
03

Implement Grating Equation

Use the grating equation to calculate the angle the light is deflected for each wavelength, for the first order (\(m = 1\)): \(\sin(\theta) = m \times \(\lambda / d)\) where \(\theta \) is the angle of deflection, m is the diffraction order, \(\lambda\) is the wavelength, and d is the separation between the lines in the grating. Solve this for each end of the spectrum to determine two angles – \(\theta1 \) and \(\theta2 \).
04

Calculate Screen Positions

Next, determine the position of each end of the spectrum on the screen by using the right triangle formed, where the distance from the grating to the screen is the adjacent side and the position on the screen is the opposite side. Using \(\tan(\theta) = \text{opposite} / \text{adjacent}\), the position can be found by multiplying the tangents of the angles by the distance to the screen.
05

Determine Spectrum Width

The width of the spectrum is the difference between the two positions. Subtract the smaller position from the larger one to obtain the desired width.

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

Microwaves of wavelength 10.0 mm fall normally on a metal plate that contains a slit \(25 \mathrm{mm}\) wide. (a) Where are the first minima of the diffraction pattern? (b) Would there be minima if the wavelength were \(30.0 \mathrm{mm}\) ?

A spy satellite orbits Earth at a height of \(180 \mathrm{km}\). What is the minimum diameter of the objective lens in a telescope that must be used to resolve columns of troops marching \(2.0 \mathrm{m}\) apart? Assume \(\lambda=550 \mathrm{nm}\)

Find the radius of a star's image on the retina of an eye if its pupil is open to \(0.65 \mathrm{cm}\) and the distance from the pupil to the retina is \(2.8 \mathrm{cm}\). Assume \(\lambda=550 \mathrm{nm}\).

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.

Red light (wavelength 632.8 nm in air) from a HeliumNeon laser is incident on a single slit of width \(0.05 \mathrm{mm}\). The entire apparatus is immersed in water of refractive index \(1.333 .\) Determine the angular width of the central peak.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Energy Physics

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Astrophysics

Read Explanation

Work Energy and Power

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