PHYS 408 HOMEWORK 5 SOLUTIONS

# 50 cm to the right of the 2 lens negative sign

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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; %%%%%%%%%%%%%%%%%%%%%%%%%%%%...
View Full Document

## This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

Ask a homework question - tutors are online