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Unformatted text preview: Ph 291, Fall 2008 Final Exam Name: SOLUT IONS 10 Dec 2008 1. [15 pts] The figure shows three pairs of polarizing sheets seen face on. The polarizing direction of each sheet is indicated by the dashed line. In each case, unpolarized light is initially incident on the 1 st polarizer (top row) and the light output from this polarizer is subsequently sent through the 2 nd polarizer (bottom row). In increasing order, rank the cases according to the fraction of the initial intensity which exits the system. The key to solving this problem is determining the angle, θ , between the preferred axes of the two polarizers. Given θ , the output of the system in terms of the fraction of initial intensity I is I out I = 1 2 cos 2 θ which is simply a restatement of Malus’ Law. The factor of 1 / 2 comes from the fact that the 1 st linear polarizer passes half of the initial unpolarized light. Analyzing each case we find: Case 1: θ = 0 ◦→ I out I = 1 2 cos 2 (0 ◦ ) = 1 2 Case 2: θ = 60 ◦→ I out I = 1 2 cos 2 (60 ◦ ) = 1 2 1 2 2 = 1 8 Case 3: θ = 90 ◦→ I out I = 1 2 cos 2 (90 ◦ ) = 0 Therefore, the Cases, in order of increasing intensity , are ranked (F) 3, 2, 1 2. [15 pts] If a pulse of electrical current is injected into a coaxial cable, it will travel down the cable much like a pulse on a string. The amplitude of the pulse decays by a factor of 1 /e within a distance λ . This distance is called the space constant of the cable and is given by the formula λ = a r R m R L where R m and R L are the transverse and longitudinal resistances of the cable, respectively, and a is the diameter of the cable. Suppose that you have made measurements of these quantities along with their associated experimental uncertainties σ R m , σ R L and σ a and you use these to compute the space constant. The uncertainty in λ is given by the expression: (D) σ λ = a r R m R L ! s σ a a 2 + σ R m 2 R m 2 + σ R L 2 R L 2 There are two approaches to calculating the error on λ : (a) direct application of the error propagation formula via computation of partial derivatives or (b) performing it in steps, each time applying a single rule ( e.g. error propagation for products, error propagation for powers, etc...). I will show you both. Direct method: We know that σ λ = s ∂λ ∂a σ a 2 + ∂λ ∂R m σ R m 2 + ∂λ ∂R L σ R L 2 . Computing the partial deriva tives, we find that: ∂λ ∂a = R 1 2 m R 1 2 L = λ a ; ∂λ ∂R m = a 1 2 R 1 2 m R 1 2 L = 1 2 λ R m ; ∂λ ∂R L = a R 1 2 m 1 2 R 3 2 L = 1 2 λ R L . Therefore, 1 Ph 291, Fall 2008 Final Exam Name: SOLUT IONS 10 Dec 2008 σ λ = s λ a σ a 2 + 1 2 λ R m σ R m 2 + 1 2 λ R L σ R L 2 = λ s σ a a 2 + 1 2 σ R m R m 2 + 1 2 σ R L R L 2 Q.E.D....
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 Spring '08
 Faculty
 Normal Distribution, Standard Deviation, Light

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