/*! 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 39 A grating has 5000.0 slits/cm. H... [FREE SOLUTION] | 91影视

91影视

Log out

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.
Light from a helium-neon laser \((630 \mathrm{nm})\) is incident on a pair of slits. In the interference pattern on a screen \(1.5 \mathrm{m}\) from the slits, the bright fringes are separated by \(1.35 \mathrm{cm} .\) What is the slit separation? [Hint: Is the small angle approximation justified?]
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 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.

The first two dark fringes on one side of the central maximum in a single-slit diffraction pattern are \(1.0 \mathrm{mm}\) apart. The wavelength of the light is \(610 \mathrm{nm}\) and the screen is \(1.0 \mathrm{m}\) from the slit. What is the slit width?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Engineering Physics

Read Explanation

Work Energy and Power

Read Explanation

Quantum Physics

Read Explanation

Linear Momentum

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.