PROBLEM 31 & 32 May 05 2010

PROBLEM 31 & 32 May 05 2010 - 1 mm 1 1 5 1 2 nm 500...

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Solving problems 31 and 32 May 5, 2010 Problem 31 (like 27.19): An optical reflector for back-reflecting a small portion of a normally incident laser beam is made with a wedge shape to deflect the unwanted beams reflected from the rear surface of the window. The index of refraction of the window is n= 1.5. What should be the angle of the wedge so that the interference fringes observed in the forward direction are spaced 0.1 mm apart using green laser light of wavelength 500 nm? x a) 237 mrad; b) 0.15 mrad; c) 500 mrad; d) 15 mrad; e) 1.67 mrad. What do you know? Twice the path length through the glass 2 t changes by one wavelength in the glass for every fringe. Thus n t 2 between fringes. The wedge angle is thus
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Unformatted text preview: radians 10 67 . 1 mm 1 . 1 5 . 1 2 nm 500 1 2 3 x n x t . Problem 32 (like 27.22): What is the separation of the 2 second order diffraction peaks (to either side of the principle diffraction maximum) on a screen 5 m away when green laser light of wavelength 500 nm passes through a slit with a width of 25 m? a) 100 m ; b) 10 cm ; c) 10 m ; d) 1 m ; e) 100 nm What do you know? From the graph, the angular separation of the two 2 nd order diffraction peaks will be a / 3 3 . We calculate a / = 500 nm / 25 m = 20 x 10-3 radians. At 5 m away this becomes 5 m x 20 x 10-3 radians = 0.1 m. 1...
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