PHYS 408 HOMEWORK 5 SOLUTIONS

50 cm to the right of the 2 lens negative sign

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Unformatted text preview: /λ was used. The position of the minimum waist of the transmitted beam is 7.50 cm to the right of the 2 lens (negative sign indicates the lens is to the left of the minimum waist). Using z0 = π w0 /λ again, we can calculate the new minimum waist to be 0.0127 cm = 12.7 microns. 3.2 lens of finite thickness Lengths in centimeters, to-be-determined thickness d. M = ￿ 1 ￿ 1−3/2 − (1)(−15) 0 3/2 1 1 0 d 1 ￿ 1 ￿ 3/2−1 − (3/2)(+5) 0 1 3/2 1 30 0 1 = 1 − d/15 −2/15 + d/450 30 − 4d/3 −3 + 2d/45 (22) We are intersted in the value of d that yields a z that is 10 percent different than the z computed above. The introduction of more material of index n will “pull” the waist closer to the lens, so we expect a meaningful solution only for 0.90 ∗ (−7.5) (i.e. the first solution for d when increasing d from zero). The value of z is the real part of q : ￿ (1 − d/15)(−iz0 ) + (30 − 4d/3) z = Real[q ] = Real (−2/15 + d/450)(−iz0 ) + (1 − d/45) ￿ The Matlab script below can be run (twice) to calculate the desired root. The answer is 1.94 cm. clear all %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % calculate total matrix, in terms of variable thickness x %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% syms x n = 3/2; 4 (23) a b c d = = = = [ [ [ [ 1 1 1 1 30 0 x 0 ; ; ; ; 0 1]; -(n-1)/(n*5) 1/n]; 0 1]; -(1-n)/(-15) n]; M = d*c*b*a; %%%%%%%%%%%%%%%%%%%%%%%%%%%%...
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This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

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