/*! 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 29 If a diffraction grating produce... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength \(681 \mathrm{nm},\) find (a) the number of slits per centimeter for the grating and (b) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

Short Answer

Expert verified
Grating has 4808 slits/cm; angles are 19.1° and 40.9°; no fourth-order band.

Step by step solution

01

Understanding the diffraction grating equation

The diffraction grating equation is given by \[ d \sin \theta = m \lambda \]where \(d\) is the distance between adjacent slits, \(\theta\) is the angle of diffraction, \(m\) is the order of the diffraction, and \(\lambda\) is the wavelength of the light. Given that \(\theta = 78.4^{\circ}\), \(m = 3\), and \(\lambda = 681 \text{ nm} = 681 \times 10^{-9} \text{ m}\), we can use this to find \(d\).
02

Calculate the slit spacing (d)

Rearrange the diffraction grating equation to solve for \(d\): \[ d = \frac{m \lambda}{\sin \theta} \]Substitute the given values: \[ d = \frac{3 \times 681 \times 10^{-9}}{\sin 78.4^{\circ}} \approx 2.08 \times 10^{-6} \text{ m} \]
03

Convert slit spacing to number of slits per centimeter

The number of slits per meter is given by \(N = \frac{1}{d}\). Convert this to slits per centimeter:\[ N = \frac{1}{d} \times \frac{1}{100} \]Substitute \(d\) to find \(N\):\[ N = \frac{1}{2.08 \times 10^{-6}} \] \[ N \approx 4808 \text{ slits per cm} \]
04

Find the angle of the first-order bright band

Using the same equation with \(m = 1\):\[ \sin \theta = \frac{1 \times 681 \times 10^{-9}}{2.08 \times 10^{-6}} \] \[ \sin \theta \approx 0.327 \Rightarrow \theta \approx 19.1^{\circ} \]
05

Find the angle of the second-order bright band

Using the same equation with \(m = 2\):\[ \sin \theta = \frac{2 \times 681 \times 10^{-9}}{2.08 \times 10^{-6}} \] \[ \sin \theta \approx 0.654 \Rightarrow \theta \approx 40.9^{\circ} \]
06

Check for the fourth-order bright band

To determine the existence of the fourth-order bright band, check if \(\sin \theta \leq 1\) when \(m = 4\): \[ \sin \theta = \frac{4 \times 681 \times 10^{-9}}{2.08 \times 10^{-6}} \] \[ \sin \theta \approx 1.308 \] Since \(\sin \theta > 1\), a fourth-order bright band is not possible.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffraction Order
In the context of a diffraction grating, the concept of diffraction order is a fundamental one. It is denoted by the symbol \(m\) in the diffraction grating equation. Essentially, the diffraction order refers to the series of maxima or bright bands that appear due to the constructive interference of light waves. These are labeled in ascending order starting from \(m = 0\), which is the zeroth-order or central maximum, followed by first-order, second-order, and so forth.

Each successive order represents a higher degree of path difference between light waves coming from adjacent slits in the grating. The diffraction order gives an indication of how many wavelengths fit into the path difference. Therefore, as the order increases, the angle of diffraction gets larger, following the diffraction grating equation:
  • \( d \sin \theta = m \lambda \)
The presence of higher orders is contingent on the properties of the grating and the light's wavelength. In practical terms, this means that not all orders may be visible, particularly if the angle results in \(\sin \theta > 1\). This was seen in the original exercise where the fourth-order was not allowed as the calculation led \(\sin \theta\) to exceed 1.
Wavelength
Wavelength, symbolized as \(\lambda\) in equations, is a key property of a wave. It refers to the distance between consecutive corresponding points of a wave, such as two peaks or two troughs. In the scenario of a diffraction grating, the wavelength of the light used significantly influences the diffraction pattern.

The wavelength of light is typically measured in nanometers (nm), where 1 nm = \(10^{-9}\) meters. Different colors of light have different wavelengths, with blue light being shorter and red light having a longer wavelength.

When light is passed through a diffraction grating, its wavelength determines how far apart the diffraction maxima will appear. This is explained by the diffraction grating formula:
  • \( d \sin \theta = m \lambda \)
Here, \(\lambda\) is directly proportional to \(d\sin\theta\), meaning larger wavelengths result in larger angles of diffraction. The problem introduced a wavelength of 681 nm, which is in the red region of the visible spectrum, affecting both the position and presence of different diffraction orders.
Slit Spacing
In a diffraction grating, slit spacing, denoted \(d\), is the distance between adjacent slits. It plays a crucial role in determining the angle at which light is diffracted. Essentially, it is the inverse of the number of slits per unit length. The equation:
  • \( d \sin \theta = m \lambda \)
shows that slit spacing directly affects the diffraction angle \(\theta\). Larger slit spacing implies waves from the slits constructively interfere at smaller angles, whereas smaller slit spacing results in larger angles of diffraction.

In practical terms, the number of slits per centimeter can be calculated once slit spacing is known. The original problem solved for \(d\) using known values for \(m\), \(\lambda\), and \(\theta\). The equation is rearranged to:
  • \( d = \frac{m \lambda}{\sin \theta} \)
This allows the calculation of the number of slits per cm, aiding in understanding how densely packed the grating is. For instance, the exercise calculated slit spacing as \(2.08 \times 10^{-6}\) m, which corresponds to about 4808 slits per cm.
Angle of Diffraction
The angle of diffraction, often represented as \(\theta\), defines the direction at which light waves emerge after passing through a diffraction grating. This angle is key to determining the position of the bright bands or maxima on a screen placed behind the grating.

Using the equation:
  • \( d \sin \theta = m \lambda \)
we see that the angle can be calculated for different orders \(m\) once slit spacing \(d\) and wavelength \(\lambda\) are known. Each order corresponds to a different angle, characteristic of the color or wavelength of light. For example, in the exercise, first and second order angles were found using values derived from the initial setup, leading to \(\theta\) values of 19.1° and 40.9° respectively.

This understanding allows us to predict where a particular color will appear in the spectrum and how it will change with different conditions, such as using a different wavelength or altering the grating's slit density.

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

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4\(\mu \mathrm{m} .\) The resulting diffraction pattern is observed on a screen that is 4.5 \(\mathrm{m}\) from the aperture. What is the diameter of the Airy disk on the screen?

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 \(\mathrm{cm}\) from the slits. The centers of the slits are 0.640 \(\mathrm{mm}\) apart and the width of each slit is 0.434 \(\mathrm{mm}\) . If the intensity at the center of the central maximum is \(5.00 \times\) \(10^{-4} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is 0.900 \(\mathrm{mm}\) from the center of the central maximum?

Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is \(5.93 \times 10^{8} \mathrm{km}\) away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 \(\mathrm{km}\) apart. What minimum- diameter mirror is required? Assume a wavelength of 500 \(\mathrm{nm} .\)

What is the longest wavelength that can be observed in the third order for a transmission grating having 9200 slits/cm? Assume normal incidence.

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 km. If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

See all solutions

Recommended explanations on Physics Textbooks

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.