Physical Optics First Order Maximums
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Multiple choice:
1. The middle of the first-order maximum, adjacent to the central bright fringe in the double-slit experiment, corresponds to at a point where the optical path length difference from the tow apertures is equal to: (a) h (=lander) (b) 0 (c) 1/2h (d) 1/4h (e) none of the above. Explain your choice.
2. A narrow slit of width D is illuminated by light comming from a monochromator so that the wavelength can be varied all across the visible spectrum. As h =Lander) decreases, the Faunhofer diffraction pattern, viewed in the focal plane of a lens: (a) shrinks with all fringes getting narrower (b) spreads out with all the fringes getting wider (c) remains unchanged (d) alters such that only the central maximum broadens (e) none of these. Explain your choice.
3. The effect of increasing the number of lines per centimeter of grating is to: (a) increase the number of order that can be seen (b) allow for the use of longer wavelengths (c) increase the spread of each spectral order (d) produce no change in the diffracted light (e) none of these. Explain your choice.
Problem:
Two vertical narrow slits seperated by 0.20 mm are illuminated perpendicularly by 500-nm. A fringe pattern appears on a screen 2.00 m away. How far (in mm) above and below the central axis are the first zeros of irradiance? [Hint: Horizontal slits create a vertical finge pattern composed of horizontal bands.]
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Solution Summary
The physical optics for the first order maximums are examined. The central maximum which broadens are given.
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SOLUTION.
Multiple choice:
1. The bright fringes of the diffraction pattern correspond to directions when the optical path difference between two slits is equal to an integer number of wavelengths. Only in this case we will have constructive interference between waves coming from both apertures. The central bright fringe corresponds to 0 optical path difference, the 1st bright fringes( in both directions from the central one) corresponds to 1 wavelength path difference, the 2nd - to 2 wavelength path difference and so on. So the answer to this question is A).
2. The ...
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