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.

A double-slit experiment has slit spacing 0.12 mm. What should be the slit-to-screen distance L if the bright fringes are to be 5.0 mm apart when the slits are illuminated with 633-nm laser light?

You are evaluating a method for studying liquid crystals. Two identical crystals are mounted over the two slits of a double-slit apparatus. The slits are illuminated with monochromatic coherent light and the interference pattern is observed on a screen. An electric voltage is then applied to one of the two crystals, causing its index of refraction to increase and the interference pattern to shift to one side. Your boss has asked you to calculate the change in the index of refraction of the electrified crystal in terms of distance between the two slits, the distance from the slits to the observing screen, the thickness of the crystals, the original index of refraction of the crystals (when not electrified), the wavelength of the light used, and the distance that the central maximum of the pattern is observed to shift when the voltage is applied (you might not use all of these quantities in your answer). two crystals double slits Light observing screen

In the single-slit diffraction experiment of the figure, let the wavelength of the light be 541 nm, the slit width be 6.11 pm, and the viewing screen be at distance D = 3.41 m. Let a y axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let Ip represent the intensity of the diffracted light at point P at y = 15.8 cm. (a) What is the ratio of Ip the intensity Im at the center of the pattern? Determine where point P is in the diffraction pattern by giving the orders of nearest (b) maximum (0 is the central maximum) and (c) minimum between which it lies. Totally destructive interference- P1 12 Po Central axis a/2 Viewing screen Incident wave Units (a) Number 2.98e-4 This answer has no units Units (b) Number This answer has no units Version 4.24.1 2.