PHYS 408 HOMEWORK 5 SOLUTIONS

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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...
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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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