Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 25: Problem 39

A grating has 5000.0 slits/cm. How many orders of violet light of wavelength \(412 \mathrm{nm}\) can be observed with this grating? (tutorial: grating).

Short Answer

Expert verified
Answer: There can be 4 orders of violet light observed with a grating of 5000.0 slits/cm.

Step by step solution

01

Write down the known values

We are given: - Number of slits per centimeter: \(n = 5000.0\) slits/cm - Wavelength of violet light: \(\lambda = 412 \mathrm{nm}\)
02

Convert the given values to compatible units

As we are given the number of slits per centimeter, it is better to convert the wavelength into centimeters: \(\lambda = 412 \mathrm{nm} \times \frac{1 \mathrm{cm}}{10^7 \mathrm{nm}} = 4.12 \times 10^{-5} \mathrm{cm}\)
03

Rewrite the grating equation

The grating equation is given by: \(\sin \theta_{m} = \frac{m \lambda}{d}\) where: - \(\theta_m\) is the angle of diffraction for the \(m\)th order - \(m\) is the order of the spectrum (a positive integer) - \(\lambda\) is the wavelength of light - \(d\) is the distance between the slits, which is the inverse of the number of slits per length
04

Calculate the distance between slits

Given the number of slits per centimeter (\(n\)), we can find the distance between the slits (\(d\)) by: \(d = \frac{1}{n} = \frac{1}{5000.0 \thinspace \mathrm{slits/cm}} = 2 \times 10^{-4} \thinspace \mathrm{cm}\)
05

Calculate the maximum order \(m\)

The maximum order, \(m\), occurs when \(\sin \theta_{m} = 1\). So we can rewrite the grating equation as: \(1 = \frac{m \lambda}{d}\) Now let's solve for \(m\): \(m = \frac{d}{\lambda} = \frac{2 \times 10^{-4}\thinspace \mathrm{cm}}{4.12 \times 10^{-5}\thinspace \mathrm{cm}} \approx 4.85\). Since \(m\) is a positive integer, we need to round down 4.85 to the nearest integer, which gives us \(m = 4\).
06

Answer

There can be 4 orders of violet light (\(\lambda = 412 \thinspace \mathrm{nm}\)) observed with a grating of 5000.0 slits/cm.

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 grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.
If you shine a laser (wavelength \(0.60 \mu \mathrm{m}\) ) with a small aperture at the Moon, diffraction makes the beam spread out and the spot on the Moon is large. Making the aperture smaller only makes the spot on the Moon larger. On the other hand, shining a wide searchlight at the Moon can't make a tiny spot- the spot on the Moon is at least as wide as the searchlight. What is the radius of the smallest possible spot you can make on the Moon by shining a light from Earth? Assume the light is perfectly parallel before passing through a circular aperture.
Find the height \(h\) of the pits on a CD (Fig. \(25.6 \mathrm{a}\) ). When the laser beam reflects partly from a pit and partly from land (the flat aluminum surface) on either side of the "pit," the two reflected beams interfere destructively; \(h\) is chosen to be the smallest possible height that causes destructive interference. The wavelength of the laser is \(780 \mathrm{nm}\) and the index of refraction of the poly carbonate plastic is \(n=1.55.\)
Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?
A mica sheet \(1.00 \mu \mathrm{m}\) thick is suspended in air. In reflected light, there are gaps in the visible spectrum at \(450,525,\) and $630 \mathrm{nm} .$ Calculate the index of refraction of the mica sheet.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Physics of Motion

Read Explanation

Force

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Medical 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