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Daily_011.SOLN

# Daily_011.SOLN - by a factor of two This would otherwise...

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Phys 274 – c Eric B. Szarmes 11 February 2011 Daily Homework 11 SOLUTIONS Q5T.2 The table is filled out with the corresponding patterns as follows: Slit Slit Case Wavelength λ Width a Separation d Detectors? Pattern Sample 5 nm 3 μ m 10 μ m No Sample (a) 5 nm 6 μ m 10 μ m No C (b) 5 nm 6 μ m 20 μ m No E (c) 10 nm 6 μ m 20 μ m No B (d) 10 nm 6 μ m 20 μ m Yes D (e) 10 nm 6 μ m 10 μ m Yes D Here are the steps: (a) Starting from the sample pattern, we now double the slit width . This halves the envelope (however, the fringes remain the same because we didn’t change λ or d ): C (b) Starting from pattern C, we now double the slit separation . This halves the fringe spacing (however, the envelope remains the same because we didn’t change λ or a ): E (c) Starting from pattern E, we now double the wavelength . This doubles both the envelope and the fringe spacing (because both increase as λ increases): B (d) Starting from pattern B, the only thing we do is add proximity detectors; this destroys the interference pattern (i.e the fringes disappear ): D (e) Starting from pattern D, we retain the detectors but decrease the slit
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Unformatted text preview: by a factor of two. This would otherwise have the eﬀect of doubling the fringe spacing while keeping the envelope the same; but the fringes are washed out anyway by the detectors, so the pattern remains the same: D Q5B.3 Let the initial intensity be I , and let the transmittance factor of each ﬁlter be T = 0 . 15 (i.e. 15% transmis-sion). Then, after one ﬁlter, the transmitted intensity is I 1 = TI ; after two ﬁlters, the transmitted intensity is I 2 = T 2 I ; and after N ﬁlters, the transmitted intensity is I N = T N I . (1) We want to choose N so that I N = I / (3 × 10 10 ), or I 3 × 10 10 = T N I . (2) Divide both sides by I , and then take the logarithm of both sides to get ln ( 1 3 × 10 10 ) = ln ( T N )-ln(3 × 10 10 ) = N ln( T ) and so N =-ln(3 × 10 10 ) ln( T ) =-ln(3 × 10 10 ) ln(0 . 15) = 12 . 7 . (3) Evidently, 13 ﬁlters would be suﬃcient. 1...
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