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Unformatted text preview: Ph 214, Spring 2008 Exam # 1 Question #1 SOLUTIONS 1. [20 pts] Unpolarized or circularly polarized beam? The only components from the kit that we need to use are the λ 4-plate and a single linear polarizer. Qualitatively: The order of elements in the optical path are (1) light source, (2) λ 4-plate, (3) linear polarizer and (4) light meter. We first pass the light through a λ 4-plate (orientation doesn’t matter). If the input is unpolarized, the output will be unpolarized. If the input is circularly polarized, the output will be linearly polarized. If we use a light meter at this stage, the intensity in both cases will equal the input intensity from the source (assuming perfect polarizing filters) so we cannot distinguish between the two. We now send the output of the λ 4-plate through the linear polarizer. In both cases, the final output will be linearly polarized. However, the intensity of the output will possibly depend on the relative orientation of the preferred axis of the linear polarizer and the fast axis of the λ 4-plate. Unpolarized light will exit a (perfect) linear polarizer with an intensity which is half of the input intensity and independent of orientation angle whereas linearly polarized light will exit a (perfect) linear polarizer with an intensity which is dependent on orientation (Malus’ Law). Therefore our apparatus will register the following: a final intensity which is half of the initial intensity and independent of the orientation of the linear polarizer (unpolarized input) -or- a variable intensity as a function of the orientation of the linear polarizer (circularly polarized input). In summary: Case 1. Unpolarized light ( I )-→ λ 4-plate-→ unpolarized light ( I = I )-→ Linear polarizer-→ linearly polarized light ( I = 0 . 5 I ), independent of θ . Case 2. Circularly polarized light ( I )-→ λ 4-plate-→ linearly polarized light ( I = I )-→ Linear polarizer-→ linearly polarized light ( I = I cos 2 θ ), Malus’ Law. [ For full credit, you must explain both how your apparatus transforms the input light and how you can use this output to distinguish between the two cases ] [ 5 pts deduction for any incorrect statement of how light is transformed (e.g. passing circularly polarized light through a λ 2-plate produces linearly polarized light)] [ 5 pts deduction for not describing what your apparatus does to unpolarized light ( existence and uniqueness )] 1 Ph 214, Spring 2008 Exam # 1 Question #2 SOLUTIONS 2. [30 pts] Synchrotron Radiation a. The electric field radiated by an acceler- ated charge is given by vector E rad ( vector r, t ) = − qvectora ⊥ ( t ′ ) 4 π ǫ c 2 r where t ′ = t − r/c . Therefore, our first step should be to compute the acceleration of the charged particle. In the problem, you are given the position of the particle as a func- tion of time vector r ( t ) 1 . The acceleration is just given by two time derivatives of this function....
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This note was uploaded on 02/27/2012 for the course CHEMISTRY/ CH/ECE/PH/ taught by Professor Faculty during the Spring '08 term at Cooper Union.
- Spring '08