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A physics instructor wants to project a spectrum of visiblelight colors from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 500 lines per \(\mathrm{mm},\) projecting a pattern on a screen \(2.4 \mathrm{m}\) behind the grating. a. How wide is the spectrum that corresponds to \(m=1 ?\) b. How much distance separates the end of the \(m=1\) spectrum and the start of the \(m=2\) spectrum?

Short Answer

Expert verified
The width of the spectrum is the difference between linear distances for red and violet light as calculated in step 2. The distance between \(m=1\) and \(m=2\) spectra is the difference between linear distances for violet light in the \(m=2\) spectrum and red light in the \(m=1\) spectrum as calculated in step 3.

Step by step solution

01

Angular Dispersion Calculation

First, use the formula for diffraction gratings which is \(\sin \theta = m \lambda /d\), where \(m\) is the order of maximum, \(\lambda\) is the wavelength, \(d\) is the line spacing of the grating, and \(\theta\) is the angle formed between the normal to the grating and a maximum of order \(m\). As we have to find out the width of the spectrum that corresponds to \(m=1\), put the values of \(\lambda = 400 nm\) (for violet) and \(700 nm\) (for red), \(d = 1 mm / 500 lines\) and \(m=1\) into the equation.
02

Conversion from Angular to Linear Dispersion

Next, calculate the linear distance on the screen by multiplying the tangent of the angles obtained in step 1 by the distance between the grating and the screen, which is \(2.4 m\). The difference between the distances for red and violet is the width of the \(m=1\) spectrum.
03

Distance between \(m=1\) and \(m=2\) Spectra Calculation

For the next part, again use the diffraction grating formula with \(m=2\) and \(\lambda = 400 nm\) (for violet in the \(m=2\) spectrum) and \(m=1\) and \(\lambda = 700 nm\) (for red in the \(m=1\) spectrum). After calculating the angles for these points, convert them into linear distances on the screen and take the difference to find the separation between the end of the \(m=1\) spectrum and the start of the \(m=2\) spectrum.

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Key Concepts

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

Angular Dispersion
Angular dispersion is a measure of how the angle of diffraction varies with the wavelength of light passing through a diffraction grating. It's crucial in understanding how a spectrum is formed when white light is diffracted.

Using the diffraction formula, \( \sin \theta = m \lambda / d \), where \( m \) signifies the order, \( \lambda \) the wavelength, and \( d \) the grating line spacing, one can determine the angles at which different wavelengths emerge from the grating. A higher dispersion means a greater separation between various wavelengths, making each color in the visible spectrum more distinct. This conveys how effectively a grating can spread out light into its constituent colors, an essential aspect for spectrometry and educational demonstrations.
Linear Dispersion
Linear dispersion translates the concept of angular dispersion into a measurable distance on a fixed screen. By calculating \( \tan(\theta) \) for the angles obtained through angular dispersion, and multiplying that by the grating-to-screen distance, the linear positions of each wavelength can be determined.

This aids in visually quantifying the spread of the spectrum—how broad it appears over a distance. In our example with the physics instructor's setup, the difference in linear positions for the red (\(700 nm\)) and violet (\(400 nm\)) light, projected onto a screen 2.4 meters away, would give the actual width of the spectrum for the first order (\( m=1 \) spectrum).
Spectrum Separation
Spectrum separation refers to the physical distance between two sequential orders of spectra. For instance, the separation between the end of the \( m=1 \) and the beginning of the \( m=2 \) spectrum. This becomes especially significant in determining the resolving power of a grating and is critical for ensuring distinct spectral lines in spectrometry.

As per the problem's instruction, spectrum separation can be visualized as the gap on a screen between two adjacent rainbow patterns formed due to different orders of diffracted light. These separations help students understand the concept of orders in light diffraction and their practical implications in spectral analysis.
Wavelength Calculation
Wavelength calculation in the context of a diffraction grating involves decoding the spectrum patterns back into the component wavelengths. By applying the grating equation, \( m \lambda = d \sin \theta \) with known values of grating spacing (\( d \) from the number of lines per millimeter) and diffraction angles, the wavelengths can be extracted.

The calculation of wavelengths is essential for interpreting the recorded spectra and for further scientific analysis. Spectrums generated from white light through a diffraction grating, such as that in classroom demonstrations, can be utilized to teach students about the nature of light and its interaction with different materials at a foundational level.

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Most popular questions from this chapter

\(A\) diffraction grating with 600 lines/mm is illuminated with light of wavelength 500 nm. A very wide viewing screen is \(2.0 \mathrm{m}\) behind the grating. a. What is the distance between the two \(m=1\) fringes? b. How many bright fringes can be seen on the screen?

For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer \((\lambda=680 \mathrm{nm})\) through the slit onto a screen \(5.5 \mathrm{m}\) away, a diffraction pattern appears. The bright band in the center of the pattern is \(8.0 \mathrm{cm}\) wide. What is the width of the slit?

Two narrow slits are illuminated by light of wavelength \(\lambda\). The slits are spaced 20 wavelengths apart. What is the angle, in radians, between the central maximum and the \(m=1\) bright fringe?

Solar cells are given antireflection coatings to maximize their efficiency. Consider a silicon solar cell \((n=3.50)\) coated with a layer of silicon dioxide \((n=1.45) .\) What is the minimum coating thickness that will minimize the reflection at the wavelength of \(700 \mathrm{nm},\) where solar cells are most efficient?

Figure \(P 17.10\) shows the fringes observed in a doubleslit interference experiment nated by white light. The central maximum is white because all of the colors overlap. This is not true for the other fringes. The \(m=1\) fringe clearly shows bands of color, with red appearing farther from the center of the pattern, and blue closer. If the slits that create this pattern are \(20 \mu \mathrm{m}\) apart and are located \(0.85 \mathrm{m}\) from the screen, what are the \(m=1\) distances from the central maximum for red \((700 \mathrm{nm})\) and violet \((400 \mathrm{nm})\) light?

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