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A transmission grating has 5900 lines/cm. Light in the range from \(400 \mathrm{nm}\) to \(720 \mathrm{nm}\) impinges perpendicularly on the grating. How big is the angular width of the first-order spectrum?

Short Answer

Expert verified
The angular width of the first-order spectrum is approximately 12.2 degrees.

Step by step solution

01

Identify given values

The transmission grating has 5900 lines/cm. Light wavelengths range from \(\lambda_1 = 400\, \text{nm}\) to \(\lambda_2 = 720\, \text{nm}\). We need to find the angular width of the first-order spectrum.
02

Convert grating lines to grating spacing

Convert the grating lines per cm to grating spacing \(d\). Given 5900 lines/cm, the spacing \(d\) in meters is:\[d = \frac{1}{5900 \, \text{lines/cm}} = \frac{1}{590000} \, \text{m}\]
03

Determine the diffraction angle formula

The angle \(\theta\) for each wavelength is given by the diffraction equation for the first order:\[d \sin \theta = m\lambda\]where \(m = 1\) for the first-order spectrum.
04

Calculate diffraction angles for the range

Using the diffraction equation, calculate \(\theta_1\) and \(\theta_2\) for the wavelengths \(\lambda_1\) and \(\lambda_2\):\[\sin \theta_1 = \frac{1 \times 400 \times 10^{-9}}{1/590000}\]\[\sin \theta_2 = \frac{1 \times 720 \times 10^{-9}}{1/590000}\]
05

Solve for each diffraction angle

Calculate \(\sin \theta_1\) and \(\sin \theta_2\):\[\sin \theta_1 = 0.236\]\[\sin \theta_2 = 0.424\]Use \(\sin^{-1}\) to find \(\theta_1\) and \(\theta_2\):\[\theta_1 = \sin^{-1}(0.236)\]\[\theta_2 = \sin^{-1}(0.424)\]
06

Calculate the angular width

Compute the angular width \(\Delta \theta\) by subtracting \(\theta_1\) from \(\theta_2\):\[\Delta \theta = \theta_2 - \theta_1\]

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Key Concepts

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

Angular Spectrum
The angular spectrum refers to the spread of colors emitted at different angles when light passes through a diffraction grating, a device with many closely spaced lines. This spectrum is a result of the diffraction process, which involves bending of light waves and creating patterns of light and dark bands. Each wavelength, or color of light, emerges at a slightly different angle, contributing to the angular spread.

This spread is what we refer to as the angular spectrum. When light of varying wavelengths, such as those between 400 nm to 720 nm, hits the grating, each component spreads out into a spectrum of colors. This separation allows for detailed studies and applications, such as identifying spectral lines of elements.
Grating Spacing
Grating spacing, denoted by the symbol \(d\), is a crucial parameter in diffraction grating experiments. It represents the distance between adjacent lines or slits on the grating. Knowing the grating spacing allows us to calculate the angles at which light of specific wavelengths will diffract.

To determine \(d\), we need to know the number of lines per centimeter, which is often provided, such as in our example with 5900 lines/cm. Conversion is then performed to get the spacing in meters. This spacing is calculated using the formula:
  • \[d = \frac{1}{ ext{number of lines/cm}}\]
Understanding grating spacing is fundamental as it directly influences the diffraction pattern and the angles where different wavelengths appear.
First-Order Spectrum
The first-order spectrum is one of the primary outcomes in diffraction experiments. It refers to the first set of angles at which light of various wavelengths is diffracted by the grating. This first-order is given by \(m = 1\) in the diffraction formula. Each higher order corresponds to higher values of \(m\).

For instance, selecting \(m = 1\) in the equation \(d \sin \theta = m \lambda\), where \(\theta\) is the diffraction angle and \(\lambda\) is the wavelength, describes the first-order diffraction. This means, the entire visible spectrum is spread out once by the grating, providing essential information for analysis in a single, clear pattern.
Wavelength Range
Wavelength range in optical experiments, like with a diffraction grating, defines the variety of wavelengths, or colors, that are being analyzed. In this specific example, the wavelength range is from 400 nm to 720 nm, which covers a large portion of the visible spectrum.

Visible light comprises this range, starting from violet (400 nm) to red (720 nm). By knowing this range, it is possible to understand exactly which wavelengths will be diffracted and measured. These wavelengths determine where each color will appear in the diffraction pattern, allowing for precise measurement of the angular spectrum.
Diffraction Angle
The diffraction angle \(\theta\) is a key concept in the study of light diffraction. This angle represents the direction at which a particular wavelength of light deviates as it passes through a grating. Each wavelength has a unique diffraction angle that depends on the order of the spectrum and the grating spacing.

Using the formula \(d \sin \theta = m \lambda\), where \(d\) is the grating spacing, \(m\) the order of the spectrum, and \(\lambda\) the wavelength, we can calculate this angle. Assessing diffraction angles for various wavelengths within a range (400 nm to 720 nm, in the exercise) allows us to determine how the spectrum is spread. This distribution of angles, when calculated for a first-order spectrum, leads to the notion of the angular width of the spectrum, showing how extensive the spread of wavelengths is over their respective angles.

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Most popular questions from this chapter

A collimated beam from a ruby laser \((694.3 \mathrm{nm})\) having an irradiance of \(10 \mathrm{W} / \mathrm{m}^{2}\) is incident perpendicularly on an opaque screen containing a square hole \(5.0 \mathrm{mm}\) on a side. Compute the irradiance at a point on the central axis \(250 \mathrm{cm}\) from the aperture. Check that this is near-field diffraction.

The circular hole in an opaque screen is \(6.00 \mathrm{mm}\) in diameter. It is perpendicularly illuminated by collimated light of wavelength \(500 \mathrm{nm} .\) How many Fresnel zones will be "seen" from a point- \(P\) on the central axis \(6.00 \mathrm{m}\) from the screen? Will that point be bright or dark? Roughly, what will the diffraction pattern look like on a vertical plane containing \(P ?\)

Collimated light from a krypton ion laser at \(568.19 \mathrm{nm}\) impinges normally on a circular aperture. When viewed axially from a distance of \(1.00 \mathrm{m}\), the hole uncovers the first half-period Fresnel zone. Determine its diameter.

We intend to observe two distant equal-brightness stars whose angular separation is \(50.0 \times 10^{-7}\) rad. Assuming a mean wavelength of \(550 \mathrm{nm},\) what is the smallest-diameter objective lens that will resolve the stars (according to Rayleigh's criterion)?

A narrow single slit (in air) in an opaque screen is illuminated by infrared from a He-Ne laser at \(1152.2 \mathrm{nm}\), and it is found that the center of the tenth dark band in the Fraunhofer pattern lies at an angle of \(6.2^{\circ}\) off the central axis. Determine the width of the slit. At what angle will the tenth minimum appear if the entire arrangement is immersed in water \(\left(n_{w}=1.33\right)\) rather than air \(\left(n_{a}=1.00029\right) ?\)

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