PHYS 408 HOMEWORK 5 SOLUTIONS

PHYS 408 HOMEWORK 5 SOLUTIONS - solution set 5 Contents 1 2...

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solution set 5 November 10, 2009 Contents 1 beam focused into aquarium ........................................... 2 1.1 spot size in air ............................................ 2 1.2 spot size and location in aquarium ................................. 2 2 gaussian aperture ................................................. 3 2.1 divergence angle and minimum waist location ........................... 3 2.2 equivalent ABCD matrix ...................................... 3 2.3 creation of Flm 3 3 lens of Fnite thickness .............................................. 4 3.1 lens of negligible thickness 4 3.2 lens of Fnite thickness ........................................ 4 1
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1 beam focused into aquarium 1.1 spot size in air θ 0 w 0 z 0 w 0 ( λz 0 ) 1 / 2 w 0 = λ πθ 0 (1) Minimum spot size: 2 w 0 =2 λ 0 =4 × 10 4 m (2) Spot size at 12 cm to left of minimum spot size location, noting z 0 = w 0 0 =0 . 2m : 2 w ( z = 12 cm) = 4 × 10 4 × ° 1+(0 . 12 / 0 . 2) 2 ± 1 / 2 m=4 . 66 × 10 4 m (3) 1.2 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: q a = 0 . 12 0 . 2 i (4) The eFective ABCD matrix for entering into to the glass, then propagating through the glass, then entering into the water is: AB CD = 10 0 n glass /n water 1 d glass 01 0 n air /n glass = . 02 . 75 (5) The q at just to the right of the glass (inside the aquarium) is: q b = (1)( 0 . 12 0 . 2 i )+(0 . 02) (1)( 0 . 12 0 . 2 i )+0 . 75 = 0 . 1333 0 . 2667 i (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 2 w 0 = 2((2 π · 10 7 / 1 . 3333)0 . 2667 ) 1 / 2 × 10 4 m.
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PHYS 408 HOMEWORK 5 SOLUTIONS - solution set 5 Contents 1 2...

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