PH291_2009FALL_EXAM1_PROFSOLN_[0]

PH291_2009FALL_EXAM1_PROFSOLN_[0] - Ph 291, Fall 2009 Final...

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Final Exam SOLUTIONS 1. [14 pts] Suppose that an object positioned 10 inches to the left of a positive (converging) lens is imaged 30 inches to the right of the lens. Where will the image appear if the object is now moved so that it is 5 inches from the lens? The solution simply requires the application of the thin-lens equation. For the initial configuration, we are given an image ( s i = 30 00 ) and an object distance ( s o = 10 00 ) which we use to determine the focal length of the lens. f = s o s i s o + s i = 300 40 = 7 . 5 00 We are now told that the object is moved such that s o = 5 00 . We immediately notice that s o < f which implies 1 We again use the thin-lens equation, this time to compute the final image location given s o f 1 f - 1 s o = 1 s i = 2 15 - 3 15 = - 1 15 -→ s i = - 15 00 (A) +6 inches (B) - 6 inches (C) +10 inches (D) - 10 inches (E) +15 inches (F) - 15 inches 2. [14 pts] The nervous system transmits information from cell to cell using stereotypical electrical pulses. The energy for these discharges is stored in the form of electric charges held on opposite sides of a semi-permeable membrane. An ionic species with a di erence in concentration between the inside and the outside of the cell leads to the establishment of an electric potential di erence given by: V = RT F log ± X Y ² where R , T and F are constants, log log e and X and Y are the external and internal concentrations, re- spectively. Suppose that you have made measurements of X and Y along with their associated experimental uncertainties σ X and σ Y and you use these to predict the expected membrane potential. The uncertainty in V is given by the expression: ( N.B. assume all measurements are independent. ) Before performing any computations, we can immediately reject four of the six possible answers simply on the basis of dimensional analysis 2 (our answer has to have final units of [Volts]). Answers (A) and (D) have units of [ Volts ] × [ concentration ] while Answers (C) and (F) have units of [ Volts ] × [ concentration ] 2 which leaves us with a choice between (B) and (E). We simply perform error propagation using the one-shot method: V = RT F log ± X Y ² = RT F [log X - log Y ] ∂V ∂X = RT F 1 X ; ∂V ∂Y = - RT F 1 Y σ V = s ∂V ∂X σ X ! 2 + ∂V ∂Y σ Y ! 2 = s ± RT F 1 X σ X ² 2 + ± - RT F 1 Y σ Y ² 2 = RT F r ± σ X X ² 2 + ± σ Y Y ² 2 1 if you chose one of these, that would be bad news. 2
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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.

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PH291_2009FALL_EXAM1_PROFSOLN_[0] - Ph 291, Fall 2009 Final...

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