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% transmission). 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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 Spring '11
 EricSzarmes
 Physics, Work, Light, Wavelength, 5 nm, µm µm µm

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