PHYS 408 HOMEWORK 5 SOLUTIONS

02 1 075 5 6 so the new minimum waist is located

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Unformatted text preview: 1 0.75 (5) (6) So the “new” minimum waist is located 0.133 m to the right of point b, equivalently 0.163 to the right of the outside edge of the aquarium. The new minimum spot size is 2w0 = 2((2π · 10−7 /1.3333)0.2667/π )1/2 = 4 × 10−4 m. 2 2 gaussian aperture 2.1 divergence angle and minimum waist location Let position a be at the position of initial minimum waist, position b just before the thin film, and position c be just after the thin film. 2 z0,a = π w0,a /λ = 4 m (7) qb = qa + 0.8 m Aqb + B qc = Cqb + D 1 1 iλ = + 2 qc Rc π Wc 1 1 iλ = + 2 qb Rb π Wb (8) q-expressions: (9) (10) (11) The thin film introduces no ρ-dependent phase shift, so R from b to c is unchanged (equating real parts of q expressions): R b = Rc (12) 1 1 1 2 + 2 × 10−6 m2 = W 2 Wb c (13) and, equating imaginary parts of qb and qc expressions, And, equating both real and imaginary parts of each q : 1 1 iλ = + qc qb 2π · 10−6 m qc = 0.0873 m − (i)(1.345 m) (14) The minimum waist position...
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