PHYS 408 HOMEWORK 5 SOLUTIONS

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: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 2 3 3 3 3 4 4 4 1 beam focused into aquarium 1.1 spot size in air θ0 ≈ w0 z0 w0 ≡ (λz0 /π )1/2 → w0 = λ πθ0 (1) Minimum spot size: 2w0 = 2 λ = 4 × 10−4 m πθ0 (2) Spot size at 12 cm to left of minimum spot size location, noting z0 = w0 /θ0 = 0.2 m: 1.2 ￿ ￿ 1 /2 2w(z = 12 cm) = 4 × 10−4 × 1 + (0.12/0.2)2 m = 4.66 × 10−4 m (3) spot size and location in aquarium Just before the aquarium, noting z is negative b/c it is to the left of the minimum waist: qa = −0.12 − 0.2i (4) The eﬀective ABCD matrix for entering into to the glass, then propagating through the glass, then entering into the water is: A B C D = 1 0 0 nglass /nwater 1 dglass 0 1 1 0 0 nair /nglass The q at just to the right of the glass (inside the aquarium) is: qb = (1)(−0.12 − 0.2i) + (0.02) = −0.1333 − 0.2667i (1)(−0.12 − 0.2i) + 0.75 = 1 0.02...
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