midterm2 - I 0 is incident The first polarizer is aligned...

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Spring, 2010 PH 113: Midterm Exam No. 2 Answer both problems. Each problem is worth 50 points. Only fully derived and explained solutions will receive full credit. A final answer with no derivation or justification, even if correct, will not receive credit. Problem 1. Diffraction grating . Light of wavelength λ = 600nm is incident head on (zero incidence angle) on a diffraction grating. Two adjacent principal maxima are found at sin ϑ m = 0.2 and at sin m + 1 = 0.3 (namely, these are the principal maxima of the m th and ( m +1) th orders, respectively, for some value of m ). The fourth order principal maximum (corresponding to m =4) is not seen. a) What is the separation of two adjacent rulings in the grating? b) What is the smallest possible width of a single ruling? Problem 2. Polarization . Two perfect linear wire–grid polarizers are stacked normal to a central axis (the axis is in the x -direction) along which a beam of natural light of intensity
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Unformatted text preview: I 0 is incident. The first polarizer is aligned so that the direction of the wires is in the z-direction at t =0, and the second polarizer is aligned so that the direction of the wires at t =0 is in the y-direction. The first polarizer rotates clockwise with an angular frequency ω , and the second polarizer rotates counterclockwise with the same angular frequency. 1. Write the direction of the polarization vector of each polarizer as a function of the time t . 2. What is the intensity of the light after the first polarizer but before the second? (In between the two polarizers!) 3. What is the intensity of the light after the second polarizer? 4. What is the frequency of the dark–bright–dark–… transitions on a screen placed after the second polarizer? Useful formula: sin 2 α = 1 2 1 − cos2 ( ) . Good luck !...
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midterm2 - I 0 is incident The first polarizer is aligned...

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