434: PHYS Assignment 8 solutions
Problem 7.24: You are asked to show that Equation 7.37 is satisfied when z P ( z) = i ln1 + i 2 ; w 0 However, we find immediately that there is a sign error in Bennett and this should be corrected to z P ( z) = i ln1 + i 2 . w 0 Begin by taking the derivative of the given expression: i P ( z ) i = 1 + i z / w 2 w 2 z ( 0 0 ) 1 = (w02 / ) 1 + i(z / w02 )
[
]
1 w / + iz i = 2 z i(w 0 / ) i = z + q0 where the last result uses Equation 7.42.
=
2 0
Problem 7.26: When the wavelength is 633 nm , the input Rayleigh range is 2 w 0 (1.00 103 m) 2 zR = = = 4.96 m 633 109 m The output beam waist location is found with Equation 7.58: 11 1 1 1 = = s2 = 10.0 cm 2 s2 f s1 + zR1 /( s1 f ) 0.1 m 0.2 m + ( 4.96 m) 2 /(0.2 m 0.1 m)
The output waist size is found with Equation 7.59: f w 01 w 02 = mw 01 = 2 ( s1 f ) 2 + zR1 0.1 m 3 = (0.1 m) 2 + ( 4.96 m) 2 (1.00 10 m) 5 2.01 10 m When the wavelength is 10.6 m , the input Rayleigh range is 2 w 0 (1.00 103 m) 2 zR = = = 0.296 m 10.6 106 m The output waist is located at 1 1 1 = s2 = 11.0 cm s2 0.1 m 0.2 m + (0.296 m) 2 /(0.2 m 0.1 m)
with a waist size given by
f w 02 = ( s f ) 2 + z 2 w 01 1 R1 0.1 m (1.00 103 m) 2 2 (0.1 m) + (.296 m) 3.20 104 m
=
Problem 7.28: Begin with Equations 7.58, and solve for s2 : 2 z R1 f s1 + ( s1 f ) s2 = 2 z R1 s1 + f ( s1 f ) Divide both sides by f to give 2 z R1 s1 + +ff s2 ( s1 f ) = 2 z R1 f + ( s1 f ) ( s1 f ) f = 1+ 2 z R1 + ( s1 f ) ( s1 f ) f = 1+ 2 2 zR1 + ( s1 f ) ( s1 f ) s1 f = 1+ 2 2 zR1 ( s1 f ) + f f s1 / f 1 = 1+ zR1 2 s1 2 + 1 f f
Similarly, beginning with Equations 7.59,
m=
f
(s1 f )
1
2
2 + z R1
=
(s1 f )
f2
2
+
2 z R1 f2
=
1 s1 2 zR1 2 1 + f f
Problem 7.30: Begin with the expression for the normalized waist location given in problem 7.28: s1 1 s2 f = 1+ f zR1 2 s1 2 + 1 f f s To find the extrema, take the derivative with respect to 1 and set this equal to zero f to give d s2 =0 s1 f d f s1 1 s1 1 f = 1 2 2 2 2 s1 zR1 s 2 z R1 2 f 1 + 1 1 + f f f f Thus s1 2 zR1 2 s1 2 1 + = 2 1 f f f or s1 2 zR1 2 1 = f f Take the square-root of both sides to give s1 z 1 = R1 f f For the positive solution, we obtain s1 z 1 = R1 f f Thus z R1 s2 1f f = 1 + 2 2 = 1+ f 2 z R1 z R 1 z R1 + f f With the negative result obtain we s1 z 1 = R1 f f giving z R1 s2 1f f = 1 2 2 = 11 f 2 z R1 z R 1 z R1 + f f Clearly, the first result is a maximum and the last result is a minimum.
Problem 7.44: The beam parameters within the cavity are 0.25 m g1 = 1 = 0.917 = g2 3.00 m Use Equation 7.71 to find the beam waist: 2 w0 = d (2 R d ) 2 633 109 m = (0.25)(6 0.25) = 1.2 107 m 2 2 giving w 0 = 0.348 mm . Equivalently, we may use Equation 7.81. Since the cavity is symmetric, the waist is located in the centre of the cavity. The Rayleigh range inside the cavity is 2 w 0 ( 3.48 104 m) 2 zR = = = 0.600 m . 633 109 m The beam radius at each mirror is
1
d 1 4 w1 = w 2 = 1 g 2
4 (633 109 m)(0.25 m) 1 = 2 1 (0.917) = 0.355 mm .
1
The output coupler acts like a diverging lens. Using the thin-lens formula of Equation 4.25, we find 1 1 1 1 = ( n 1) = (1.55 1) 0 = 0.183 3.00 f R1 R2 or f = 5.45 m . Use Equation 7.58 with s1 = d / 2 to find the output waist location: 1 1 1 1 = = (0.183 m1 ) = 5.458 m1 2 2 z R1 s2 f (0.600 m) s1 + 0.125 m + ( s1 f ) 0.125 m + 5.45 m or
s2 = 18.3 cm . which is 5.8 cm behind the midpoint between the mirrors. The output waist is given by Equation 7.59: f (5.45) m= = = 0.972 2 2 (0.125 + 5.45) 2 + (0.600) 2 s1 f ) + zR1 ( giving w 0,out = (0.972) w 0,in = (0.972)(0.348 mm) = 0.338 mm The output Rayleigh range is 2 w 0,out ( 3.38 104 m) 2 zR ,out = = = 0.566 m . 633 109 m
Problem 7.51: To find the output waist location, use Equation 7.58: 1 11 += 2 z R1 s2 f s1 + ( s1 f ) where 2 w 01 (1.4 104 m) 2 zR = = = 0.0972 m . 633 109 m Thus
11 = s2 f 1 s1 + z ( s1 f )
2 R1
=
1 1 = 6.61 m1 0.1 m 0.2 m + 0.0964 m
giving s2 = 15.1cm . The output beam waist is given by Equation 7.59: f 0.1 m w1 = w 2 = mw1 = (s f )2 + z 2 (0.2 m 0.1 m) 2 + (.0972 m) 2 w1 = 0.717 w1 1 R1 giving w = 0.100 mm . The far field beam divergence of the output beam is given by 2 Equation 7.24: 633 109 m = = = 2.01 103 rad . 4 w 2 (1.00 10 m)
Problem 8 : The centre region of the phase plate should be opaque. .4 Problem 8.5: The angular diameter measured from the first zero of intensity is given by = 2.44 Dob where Dob is the objective diameter. On the image plane, assumed to be at the focal plane, this corresponds to a physical size y given by f 1.00 m y = f = 2.44 = 2.44 = 24.4 Dob 0.1 m
For an illumination wavelength of 550 nm , this gives y = 24.4 (550 109 m) = 1.32 105 m .