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? Submit Part C Correct Previous Answers It's shown in more advanced optics courses that the width of the fringes of a diffraction grating are Wgrating Wdouble/N, where wdouble is the fringe width of a double slit with the same slit spacing d and N (which is assumed to be >> 1) is the number of slits illuminated. What is the fringe with of the half-meter spectrometer of Part A? Express your answer in centimeters to two significant figures.

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? Submit Part C Correct Previous Answers It's shown in more advanced optics courses that the width of the fringes of a diffraction grating are Wgrating Wdouble/N, where wdouble is the fringe width of a double slit with the same slit spacing d and N (which is assumed to be >> 1) is the number of slits illuminated. What is the fringe with of the half-meter spectrometer of Part A? Express your answer in centimeters to two significant figures.