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A grating in a spectrometer is illuminated with red light \((\lambda=690 \mathrm{nm})\) and blue light \((\lambda=460 \mathrm{nm})\) simultaneously. The grating has 10,000.0 slits/cm. Sketch the pattern that would be seen on a screen \(2.0 \mathrm{m}\) from the grating. Label distances from the central maximum. Label which lines are red and which are blue.

Short Answer

Expert verified
Red lines appear at 1.91 m and blue lines at 1.03 m from the center.

Step by step solution

01

Understand the Problem

The problem involves using a diffraction grating to separate red and blue light based on their wavelengths. The grating diffracts each light into separate directions based on their wavelength, with the angle of diffraction determined using the grating equation.
02

Write Down the Grating Equation

The grating equation is given by \( d \sin \theta = m \lambda \), where \( d \) is the distance between slits \( d=1/N \) (which is the inverse of the number of lines per meter), \( \theta \) is the angle of diffraction, \( m \) is the order of the diffraction, and \( \lambda \) is the wavelength of light.
03

Calculate the Grating Spacing d

Given the grating has 10,000.0 slits per cm, convert this to slits per meter: 10,000.0 slits/cm = 1,000,000 slits/m. Thus, \( d = \frac{1}{1,000,000} = 1 \times 10^{-6} \) m.
04

Calculate Diffraction Angles for Each Wavelength

Use \( d \sin \theta = m \lambda \) for the first-order (\( m=1 \)) diffraction. For red light, \( \lambda = 690 \times 10^{-9} \) m, so \( \sin \theta_r = \frac{1 \times 690 \times 10^{-9}}{1 \times 10^{-6}} = 0.69 \). Solve for \( \theta_r \). Similarly, for blue light, \( \lambda = 460 \times 10^{-9} \), so \( \sin \theta_b = \frac{1 \times 460 \times 10^{-9}}{1 \times 10^{-6}} = 0.46 \). Solve for \( \theta_b \). \( \theta_r \approx 43.6^\circ \) and \( \theta_b \approx 27.4^\circ \).
05

Calculate DISTANCE to Screen

The screen is 2.0 m from the grating. The distance to each maxima can be found using \( x = L \tan \theta \), where \( L \) is the distance to the screen. For red light, \( x_r = 2.0 \times \tan(43.6^\circ) \approx 1.91 \text{ m} \). For blue light, \( x_b = 2.0 \times \tan(27.4^\circ) \approx 1.03 \text{ m} \).
06

Sketch and Conclude

Sketch two sets of lines on the screen starting from a central line (the 0th order where all light overlaps). The 1.91 m lines on either side are labelled as red for red light, and 1.03 m at either side are labelled as blue. This separation results from the different angles of diffraction due to the different wavelengths.

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Key Concepts

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

Wavelength
Wavelength is a critical concept in optics, representing the distance between consecutive peaks of a wave. In the context of light, it determines the color we perceive. A shorter wavelength corresponds to blue light, whereas a longer wavelength is associated with red light. For example, blue light analyzed through a diffraction grating might have a wavelength of around 460 nm, while red light typically measures around 690 nm.

Understanding wavelength helps us to predict how light will behave when interacting with various materials, including diffraction gratings. When light passes through a diffraction grating, its wavelength determines the angle at which it will be diffracted, thus affecting the position of its spectrum on a screen. This is crucial when working with spectrometers which are devices designed to separate components of light by wavelength.
Diffraction Angle
The diffraction angle is the angle at which light exits a diffraction grating, influenced by the wavelength of the light and the properties of the grating itself. The angle at which light spreads out depends on its wavelength, with the mathematical relationship given by the grating equation: \[ d \sin \theta = m \lambda \]Here, \(d\) is the distance between the grating slits, \( \theta \) is the diffraction angle, \(m\) is the order of diffraction, and \( \lambda \) is the wavelength.

When light with differing wavelengths such as red and blue hits a grating, each will exhibit unique diffraction angles due to their distinct wavelengths. This results in the red light spreading out further from the center than the blue light, forming separate spectral lines. A fundamental task in optics involves calculating these angles to predict where each color will appear on a screen, thereby mapping out the visible spectrum.
Spectrometer
A spectrometer is an essential tool in optics used to measure the properties of light over a specific portion of the electromagnetic spectrum. It works by splitting light into its component wavelengths, similar to how a prism refracts light into a spectrum.

In the case with a diffraction grating, a spectrometer helps visually display how different colors of light separate based on their wavelengths. When illuminated with mixed light, such as a combination of red and blue, the spectrometer will reveal distinct lines at different positions on the screen - determined by the diffraction angles calculated through the grating equation.
  • Spectrometers are invaluable in scientific research, allowing us to analyze chemical compositions through spectral lines.
  • They are commonly used in labs for material analysis, understanding astrophysical phenomena, and even in artistic applications, such as creating specific lighting effects.
Optics
Optics is the branch of physics devoted to studying light's properties and behavior. Key topics in optics include the study of light's interaction with lenses, mirrors, and other materials, including diffraction gratings.

A diffraction grating is an optical element with a series of equally spaced lines that diffracts light into several beams. When light interacts with a grating, it is split into various directions according to its wavelength, revealing a pattern on a screen that scientists and engineers analyze for a range of applications.

Understanding optics allows us to manipulate light for practical use, such as in lenses that correct vision or telescopes that observe distant stars. Learning about the basics in optics, such as diffraction and refraction, provides foundational knowledge for exploring complex optical systems and technologies. Embracing optics knowledge enhances one's ability to appreciate modern scientific and technological advancements.

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Most popular questions from this chapter

Light of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is \(0.530 \mathrm{cm} .\) A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of $0.640 \mathrm{cm}$ between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? [Hint: Is the small-angle approximation justified?]
Roger is in a ship offshore and listening to a baseball game on his radio. He notices that there is destructive interference when seaplanes from the nearby Coast Guard station are flying directly overhead at elevations of $780 \mathrm{m}, 975 \mathrm{m},\( and \)1170 \mathrm{m} .$ The broadcast station is \(102 \mathrm{km}\) away. Assume there is a \(180^{\circ}\) phase shift when the EM waves reflect from the seaplanes. What is the frequency of the broadcast?
The Michelson Interferometer A Michelson interferometer is adjusted so that a bright fringe appears on the screen. As one of the mirrors is moved \(25.8 \mu \mathrm{m}, 92\) bright fringes are counted on the screen. What is the wavelength of the light used in the interferometer?
Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
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