Suppose a 2.00-cm-wide diffraction grating with 800 lines/mm is 0.500 m in front of a detector, forming a simple version of a "half- meter spectrometer." A spectrometer is used to measure the wavelengths of light by spreading them out across a detector. The quality of a spectrometer is its ability to distinguish or resolve two very similar wavelengths. A simple double slit has very poor resolution because the fringes are very wide; thus the interference patterns of two similar wavelengths would be so overlapped that you couldn't distinguish them. A diffraction grating with many slits creates much narrower interference fringes, allowing two similar wavelengths to be distinguished by where their fringes fall on the detector. What is the minimum difference between two wavelengths that this spectrometer can resolve? ▼ Part A Diffraction gratings are characterized by the number of "lines per mm." For a transmission grating, this is the number of slits per mm of width. For this grating, with 800 lines/mm, what is the position on the detector of the center of the first-order interference fringe for light with a wavelength of exactly 500 nm? The central maximum is the origin at y = 0. Express your answer in centimeters to three significant figures. ▸ View Available Hint(s) Y₁ = Submit ΑΣΦ A Previous Answers → 13 ? cm

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Suppose a 2.00-cmcm-wide diffraction grating with 800 lines/mmmm is 0.500 mm in front of a detector, forming a simple version of a "half-meter spectrometer." A spectrometer is used to measure the wavelengths of light by spreading them out across a detector. The quality of a spectrometer is its ability to distinguish or resolve two very similar wavelengths. A simple double slit has very poor resolution because the fringes are very wide; thus the interference patterns of two similar wavelengths would be so overlapped that you couldn’t distinguish them. A diffraction grating with many slits creates much narrower interference fringes, allowing two similar wavelengths to be distinguished by where their fringes fall on the detector. What is the minimum difference between two wavelengths that this spectrometer can resolve?

Part A

Diffraction gratings are characterized by the number of "lines per mmmm." For a transmission grating, this is the number of slits per mmmm of width. For this grating, with 800 lines/mmmm, what is the position on the detector of the center of the first-order interference fringe for light with a wavelength of exactly 500 nmnm? The central maximum is the origin at y� = 0.

Express your answer in centimeters to three significant figures.

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y1�1 =

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Part B

The width w� of a bright fringe is the distance from the first dark minimum on one side of the maximum to the first dark minimum on the other side. In a simple double-slit experiment, this width is simply the fringe spacing ΔyΔ�. For light with a wavelength of 500 nmnm, what is the fringe width in the interference pattern for a double slit having the value of d� you found in Part A?

Express your answer in centimeters to two significant figures.

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