/*! 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 61 With a particular grating the so... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

With a particular grating the sodium doublet \((589.00 \mathrm{nm}\) and \(589.59 \mathrm{nm}\) ) is viewed in the third order at \(10^{\circ}\) to the normal and is barely resolved. Find (a) the grating spacing and (b) the total width of the rulings.

Short Answer

Expert verified
The grating spacing \(d\) is approximately \(1.02 \times 10^{-5} \, \text{m}\), and the total width of the rulings is approximately \(0.34 \, \text{m}\).

Step by step solution

01

Understand the Problem

We have to calculate the grating spacing given the observation of the sodium doublet in the third order diffraction at an angle of \(10^\circ\). Additionally, we will calculate the total width of the rulings needed to resolve this doublet.
02

Use the Diffraction Grating Equation

The diffraction grating equation is given by \(d \sin \theta = m \lambda\), where \(d\) is the grating spacing, \(\theta\) is the angle, \(m\) is the order of diffraction, and \(\lambda\) is the wavelength. As the sodium doublet consists of two wavelengths \(\lambda_1 = 589.00 \, \text{nm}\) and \(\lambda_2 = 589.59 \, \text{nm}\), we'll use both to find the average grating spacing.
03

Calculate Grating Spacing `d`

To calculate the grating spacing, we'll use the average wavelength of the sodium doublet. Average wavelength: \(\lambda = \frac{\lambda_1 + \lambda_2}{2} = \frac{589.00 \, \text{nm} + 589.59 \, \text{nm}}{2} = 589.295 \, \text{nm}\). Convert nm to meters by multiplying by \(10^{-9}\). So, \(\lambda = 589.295 \times 10^{-9} \, \text{m}\). Now, substitute this in the equation: \(d \sin(10^\circ) = 3 \times 589.295 \times 10^{-9} \, \text{m}\). Calculate \(d = \frac{3 \times 589.295 \times 10^{-9}}{\sin(10^\circ)} \, \text{m}\).
04

Calculate the Total Width of the Rulings

The resolving power \(R\) of a grating is given by \(R = mN = \frac{\lambda}{\Delta \lambda}\), where \(\Delta \lambda = \lambda_2 - \lambda_1\) is the separation between the wavelengths. Given \(m = 3\) and \(\Delta \lambda = 589.59 \, \text{nm} - 589.00 \, \text{nm} = 0.59 \, \text{nm}\). \(R = \frac{589.295}{0.59} \approx 998.8\). Thus, \(N = \frac{R}{m} \approx \frac{998.8}{3}\). Calculate \(N\) and use it to find the total ruled width: \(L = Nd\). Substitute \(N\) and the previously found \(d\) to calculate \(L\) in meters.
05

Final Calculations and Rounding

Calculate numerical values from previous steps rounding if necessary to two decimal places for practical purposes. This includes determining the value of \(d\) and \(L\) to obtain the final results for grating spacing and total ruled width in meters.

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.

Sodium Doublet
The sodium doublet is one of the most famous examples in spectroscopy. It refers to two closely spaced spectral lines from sodium, with wavelengths of 589.00 nm and 589.59 nm. These wavelengths result from transitions between certain energy levels in sodium atoms. When examining such small differences, scientific instruments must be finely tuned to resolve or distinguish between these two lines, especially in the presence of a diffraction grating. This level of precision is crucial when studying the properties of light or analyzing materials using spectroscopy. The sodium doublet serves as an excellent example of how sensitive and precise optical measurements can capture minute differences in light.
Grating Spacing
Grating spacing, represented by the symbol \(d\), is a critical factor in a diffraction grating's ability to resolve details in light. It refers to the distance between adjacent lines or slits on the grating. This spacing influences the angles at which different wavelengths are diffracted and is pivotal in determining the grating's performance.
When light strikes the grating, each of these slits acts as a source of diffracted waves which interfere with each other. The condition for the maximum intensity or constructive interference is given by the grating equation: \(d \sin \theta = m \lambda\), where \(\theta\) is the angle of diffraction, \(m\) is the order of diffraction, and \(\lambda\) is the wavelength of light. Solving this equation for \(d\) helps understand how well a grating can separate spectral lines like the sodium doublet.
Resolving Power
The resolving power of a diffraction grating is a measure of its ability to separate closely spaced spectral lines. It's denoted by \(R\) and is defined as \(R = \frac{\lambda}{\Delta \lambda}\), where \(\lambda\) is the average wavelength, and \(\Delta \lambda\) is the difference between the two wavelengths being resolved. A higher resolving power indicates a greater ability to distinguish between very close wavelengths, which is essential for detailed spectroscopic studies.
For practical applications, resolving power can also be expressed in terms of the diffraction order \(m\) and the total number of slits \(N\) or ruling lines, by the formula \(R = mN\). Here, resolving the sodium doublet in the third order with close lines demands a high resolving power, which can be obtained by having more lines or finer grating spacing.
Diffraction Order
Diffraction order \(m\) in the context of diffraction gratings refers to the series of angles at which light of a particular wavelength will constructively interfere and create an observable pattern. The first order (\(m=1\)) corresponds to the first instance where the path difference equals one wavelength, the second order (\(m=2\)) corresponds to two wavelengths, and so on.
Higher diffraction orders imply multiples of the wavelength fitting the path difference, allowing light to be viewed at sharper angles. These higher orders often require more precise instrumentation to observe because each incremental order amplifies the complexities involved in separating similar wavelengths - as seen with the sodium doublet viewed in the third order at \(10^{\circ}\). This fascination with diffraction orders in practical experiments exemplifies how different light properties are explored using optical components.

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 spy satellite orbiting at \(160 \mathrm{~km}\) above Earth's surface has a lens with a focal length of \(3.6 \mathrm{~m}\) and can resolve objects on the ground as small as \(30 \mathrm{~cm}\). For example, it can easily measure the size of an aircraft's air intake port. What is the effective diameter of the lens as determined by diffraction consideration alone? Assume \(\lambda=550 \mathrm{nm}\).

If first-order reflection occurs in a crystal at Bragg angle \(3.4^{\circ}\) at what Bragg angle does second-order reflection occur from the same family of reflecting planes?

(A) A circular diaphragm \(60 \mathrm{~cm}\) in diameter oscillates at a frequency of \(25 \mathrm{kHz}\) as an underwater source of sound used for submarine detection. Far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be \(1450 \mathrm{~m} / \mathrm{s}\) and find the angle between the normal to the diaphragm and a line from the diaphragm to the first minimum. (b) Is there such minimum for a source having an (audible) freguency of \(1.0 \mathrm{kHz}\) ?

A diffraction grating has 8900 slits across \(1.20 \mathrm{~cm}\). If light with a wavelength of \(500 \mathrm{nm}\) is sent through it, how many orders (maxima) lie to one side of the central maximum?

An \(x\) -ray beam of wavelength \(A\) undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is \(23^{\circ}\), and an x-ray beam of wavelength 97 pm undergoes third-order reflection when its angle of incidence to that face is \(60^{\circ} .\) Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength \(A\).

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.