Solution8

# Solution8 - PHYS 434: Assignment 8 solutions Problem 7.24:...

This preview shows pages 1–3. Sign up to view the full content.

PHYS 434: Assignment 8 solutions Problem 7.24: You are asked to show that Equation 7.37 is satisfied when P ( z ) = i ln 1 + i λ z π w 0 2 ; However, we find immediately that there is a sign error in Bennett and this should be corrected to P ( z ) = i ln 1 + i z w 0 2 . Begin by taking the derivative of the given expression: P ( z ) z = i 1 + i z / w 0 2 ( ) i w 0 2 = 1 w 0 2 / ( ) 1 + i z / w 0 2 ( ) [ ] = 1 w 0 2 / + iz = i z i w 0 2 / ( ) = i z + q 0 where the last result uses Equation 7.42. Problem 7.26: When the wavelength is 633 nm , the input Rayleigh range is z R = w 0 2 = (1.00 × 10 3 m ) 2 633 × 10 9 m = 4.96 m The output beam waist location is found with Equation 7.58: 1 s 2 = 1 f 1 s 1 + z R 1 2 /( s 1 f ) = 1 0.1 m 1 0.2 m + (4.96 m ) 2 /(0.2 m 0.1 m ) s 2 = 10.0 cm The output waist size is found with Equation 7.59: w 02 = mw 01 = f s 1 f ) 2 + z R 1 2 w 01 = 0.1 m (0.1 m ) 2 + (4.96 m ) 2 (1.00 × 10 3 m ) 2.01 × 10 5 m When the wavelength is 10.6 μ m , the input Rayleigh range is z R = w 0 2 = (1.00 × 10 3 m ) 2 10.6 × 10 6 m = 0.296 m The output waist is located at 1 s 2 = 1 0.1 m 1 0.2 m + (0.296 m ) 2 /(0.2 m 0.1 m ) s 2 = 11.0 cm

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
with a waist size given by w 02 = f s 1 f ) 2 + z R 1 2 w 01 = 0.1 m (0.1 m ) 2 + (.296 m ) 2 (1.00 × 10 3 m ) 3.20 × 10 4 m Problem 7.28:
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/11/2010 for the course PHYS 434 taught by Professor Kilfoil during the Winter '09 term at McGill.

### Page1 / 5

Solution8 - PHYS 434: Assignment 8 solutions Problem 7.24:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online