/*! 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 65 There are 5620 lines per centime... [FREE SOLUTION] | 91Ó°ÊÓ

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There are 5620 lines per centimeter in a grating that is used with light whose wavelength is \(471 \mathrm{nm}\). A flat observation screen is located at a distance of \(0.750 \mathrm{m}\) from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen?

Short Answer

Expert verified
The minimum width of the screen is 0.412 m.

Step by step solution

01

Convert Wavelength to Meters

Given the wavelength \( \lambda = 471 \mathrm{nm} \). Convert it to meters: \( \lambda = 471 \times 10^{-9} \mathrm{m} \).
02

Calculate the Grating Spacing

The grating has 5620 lines per centimeter. First, convert to meters: \( 5620 \text{ lines/cm} = 562000 \text{ lines/m} \). The spacing \( d \) between the lines is the inverse of this value: \( d = \frac{1}{562000} \, \text{m} \approx 1.78 \times 10^{-6} \mathrm{m} \).
03

Calculate the Angle for Principal Maxima

Using the diffraction grating equation \( d \sin \theta = m \lambda \), where \( m \) is the order of the maximum. For the first order maximum \( m=1 \), solve for \( \theta \):\[\sin \theta = \frac{m \lambda}{d} = \frac{1 \times 471 \times 10^{-9}}{1.78 \times 10^{-6}} = 0.264 \Rightarrow \theta = \arcsin(0.264) \approx 15.34^\circ.\]
04

Determine the Width for the First Maxima

Now, use the angle \( \theta = 15.34^\circ \) to determine the position on the screen: \[ x = L \tan(\theta) \approx 0.750 \times \tan(15.34^\circ) \approx 0.206 \mathrm{m} \].The screen must have this half-width on one side of the center for the first maximum. "},{
05

Calculate Total Minimum Width of the Screen

To capture the principal maxima on both sides of the central maximum, double the half-width calculated for one side: \[ \text{Total Width} = 2 \times 0.206 = 0.412 \mathrm{m} \].

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Key Concepts

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

Wavelength
Wavelength is a fundamental concept in understanding diffraction and other wave behaviors. It is the distance between two consecutive peaks (or troughs) of a wave. Light, which behaves as a wave, is characterized by its wavelength. The unit \'nanometer\' (nm) is commonly used to measure the wavelength of light, where 1 nm equals \(10^{-9}\) meters. In our exercise, the wavelength is given as \(471 \text{ nm}\), which is the distance the wave travels in one complete cycle. For calculations involving wavelengths, it is essential to convert all units into meters for consistency. Thus, in this exercise, the given wavelength is converted from \(471 \text{ nm}\) to \(471 \times 10^{-9} \text{ m}\). Understanding the concept of wavelength allows us to connect the properties of light to its physical manifestations, like color and diffraction patterns.
Principal Maxima
Principal maxima refer to the bright spots observed in a diffraction pattern. These spots occur due to the constructive interference of light waves passing through a diffraction grating. Each spot corresponds to a specific angle where the light waves constructively interfere. The principal maxima are determined by an integer known as the order number \(m\).
The central maximum is the brightest and located directly in line with the incoming light. Other principal maxima appear symmetrically on either side. In our exercise, determining the angles for the first-order maxima (\(m=1\)) allows us to calculate their positions on the observation screen. Recognizing and calculating the positions of these principal maxima is crucial for measuring the extent of diffraction patterns in different experimental setups.
Grating Spacing
Grating spacing is a critical factor in determining the diffraction pattern produced by a grating. It is defined as the distance between adjacent lines on a diffraction grating. In our exercise, the grating is specified to have \(5620\) lines per centimeter. By first converting this to lines per meter, we obtain \(562000\) lines per meter. Thus, the spacing \(d\) between the lines is the reciprocal of this number. Understanding grating spacing helps us comprehend how it influences the angular distribution of principal maxima. Smaller grating spacings result in wider diffraction patterns, as beams spread over a larger angular range. In calculations, always convert to standard units to ensure accuracy, as we have done by converting grating lines per centimeter to lines per meter.
Diffraction Equation
The diffraction equation is the mathematical basis for understanding how light interacts with a grating. It is expressed as \(d \sin \theta = m \lambda\), where:
  • \(d\) is the grating spacing.
  • \(\theta\) is the diffraction angle.
  • \(m\) is the order of the maximum.
  • \(\lambda\) is the wavelength of the light.
This equation connects the properties of the grating and the incoming light to the angles at which principal maxima are observed. In our exercise, using the known values, we calculated \(\theta\) for the first-order maximum \((m = 1)\), leading us to understand how the diffraction pattern spreads across the observation screen.
Mastering the diffraction equation is vital, as it is the key to predicting and analyzing the behavior of light in various optical instruments and experiments.

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Most popular questions from this chapter

A dark fringe in the diffraction pattern of a single slit is located at an angle of \(\theta_{\mathrm{A}}=34^{\circ} .\) With the same light, the same dark fringe formed with another single slit is at an angle of \(\theta_{\mathrm{B}}=56^{\circ} .\) Find the ratio \(W_{A} / W_{\mathrm{B}}\) of the widths of the two slits.

Violet light (wavelength \(=410 \mathrm{nm}\) ) and red light (wavelength \(=\) \(660 \mathrm{nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the third-order maximum. (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

Two gratings A and B have slit separations \(d_{A}\) and \(d_{\mathrm{B}},\) respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of \(\mathrm{A}\) is exactly replaced by the second-order maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

The same diffraction grating is used with two different wave-lengths of light, \(\lambda_{A}\) and \(\lambda_{\mathrm{B}} .\) The fourth-order principal maximum of light A exactly overlaps the third-order principal maximum of light B. Find the ratio \(\lambda_{N} / \lambda_{\mathrm{B}}\)

A soap film \((n=1.33)\) is \(465 \mathrm{nm}\) thick and lies on a glass plate \((n=1.52) .\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

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