PHYS 408 HOMEWORK 5 SOLUTIONS

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

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: 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...
View Full Document

Ask a homework question - tutors are online