PHYS 408, Optics Problem Set 4 - Spring 06 Posted: Fri, March 4, 06 Due: 5pm Thu, March 7, 06. Refraction at a Spherical Boundary. Derive the M matrix of.4-6 in the textbook. You may use Snell s Law directly.. Exercise.4-5. Imaging with a thick lens. 3. Exercise 3.-. Validity of the Paraxial Approximation for a Gaussian beam. 4. Exercise 3.-5. Identification of a Beam with Known Curvatures at Two Points. 5. Steck-Probelm 6.6 (PHYS 408 website edition). 6. Two concave mirrors with radii of curvature R and R, are placed at a distance d from each other, as shown in figure, forming a cavity. Figure Use the approach of Problem 5 to prove that radii of curvature of Gaussian mode wavefronts must match those of cavity mirrors. Note that R and R are both negative for concave mirrors. 7. Cavity modes. A defect in a periodic stacks may form a cavity. (a) A structure placed in vaccum is shown as below, where n =.5, n = 3, n 3 = 4, d = 300nm, d = 50nm, d 3 = 00nm, and 0 layers on both sides. Plot the the transmission spectrum versus wavenumber /λ = 0 0000cm. Is this structure going to work as a good cavity? Why?
Figure (b) Same parameters as in part (a), but with a symmetric sequence on either side of the defect. Plot the the transmission spectrum versus wavenumber /λ = 0 0000cm. Identify all modes, find resonant freqencies and plot the mode profiles ( E ) for each resonant frequency. Figure 3
Solutions:. As Figure 4: Figure 4 Suppose the ray right before and after the surface are characterized by (y, θ ) and (y, θ ), where we know that the position of the ray is not changed at the surface and that θ is negative in our chosen coordinate system. We want to know the transfer matrix in the euqation: [ ] [ ] [ ] y A B y =, C D θ which coorsponds to two equations: θ y = Ay + Bθ () θ = Cy + Dθ. () It is obvious from equation that A = and B = 0 have to be true for arbitrary y and θ to satisfy this equation. Since the lense is spherical, by geometry, we know that α and β are the incidence and refraction angles respectively, and hence the Snell s law is: n sin α = n sin β n α = n β (paraxial approximation) Then we need to find out what α and β are. The upper left right angle gives: γ+α θ = π/, and the lower right right angle gives: γ + β + ( θ ) = π/, where γ = π/ y/r. 3
Therefore, we have: α = θ + y R β = θ + y R. Substitute these into the Snell s law above: n (θ + y R ) = n (θ + y R ) θ = n n n R y + n n θ Compare the above equation with equation, we have: C = (n n ) n R, D = n n.. The imaging condition is easily seen from a thin lens: Figure 5 Suppose a ray is propagating from left to right, the overall ray transfer matrix is then: [ ] z M = 0 [ ] 0 z 0 f z z + z z z f f = z. f f If the imaging condition: /z + /z = /f is satisfied, M becomes: z 0 f M = z, (3) f f 4
from which we can see that all rays (arbitrary θ ) generating from the same ponit (fixed y ) will converge to a single point on the other side of the optical system (fixed y ). Thus, image can form. Then, this problem reduces to show that the ray transfer matrix of the thick lens is equivalent to equation 3, given the redefined z, z, f and the corresponding imaging condition. Figure 6 The overall matrix for the thick lens is: [ ] [ ] d 0 [ ] [ ] d M = 0 n 0 [ ] R n 0 n d 0 [ ] nr n A B =, C D where A = n R [ n nr d]d n nr d B = d + d + d n n nr dd n nr dd n R [ n nr d]d d C = n R [ n nr d] D = n R [ n nr d]d n nr d Note that by definition: n nr d = h f n R [ n nr d] = f d i + h = z i 5
By playing with the algebra, we have: [ ] z A B f = C D f z + z z z f z, f (the most unapparent part may be B, where you need to see that d/n = h h ) which is an imaging system if /z + /z = /f is satisfied. 3. We need to show: A ka,, (4) z which has to be satisfied for both amplitude and phase of A, given From equation 5: θ 0 = W 0 z 0. (5) W 0 z 0 λz0 /π z 0 kz 0 The Gaussian complex evenlope A is: A(ρ, z) = A e jφ = A 0 w 0 k z 0 ρ w e w e j(k ρ R ζ) 6
For the amplitude part, A z = A w w z = ( w + ρ w ) A w 3 z (if we only look at ρ of our interest: set ρ = w) < w A w z = w w z 0 A + (z/z0 ) = z A z + z0 At this point, we can further show that the above expression has maximum value at z = z 0, by setting A / z = 0: Therefore, by substitue z = z 0 For the phase part, z 0 A z = 0 z 0 z z 0 + z = 0 z = z 0 A z < z 0 A k A φ z = R kρ R (z) z + (z/z 0 ) z 0 = kρ R (z) ( z 0 z ) + (z/z 0 ) z 0 (set ρ = w) = kw0( + z /z0) z ( + z0/z ) ( (let x = z 0/z ) + z /z 0 = kw0( + x ) x z0( + x) ( x) + /x = k w 0 z 0 x x + z 0 x x + 7 z 0 )
It is easy to see that functions (x )/(x + ) and x/(x + ) have magnitude of about unity. Since w 0 z 0, z 0 k we can see that: φ z k Therefore, equation 4 is indeed satisfied. In fact, for a Gaussian beam, the part that least satisfies the paraxial approximation is that with minimum radius of curvature, where z = ±z 0. Therefore, it is sufficient to show the approximation only at z = ±z 0. 4. The radius of curvature is given by: R(z) = z[ + ( z 0 z ) ] = z + z 0 z Suppose the given two points have z of z and z + d: R = z + z 0 z (6) R = z + d + z 0 z + d. (7) We then have two unkonwns z and z 0 and two independent equations. Unknowns can be solved. Equation 6 gives: z 0 = R z z Substitute this into equation 7 and arrange: Then correspondingly, z = d(r d) R R d z 0 = d(r + d)(r d)(r R d) (R R d) 5. (a) To be a resonant mode, the Gaussian should stay invariant after a round trip. In terms of the ABCD law: q = Aq + B Cq + d Rearrange the equation and we have: Cq (A D)q B = 0 8
By the quadratic formula: q = (A D) ± (A D) + 4BC C = (A D) ± (A D) + 4AD 4 C = A D C ± ( A + D C ) C (det M = ) Since for a Gaussian, q = z iz 0, where z 0 is positive, the allowed q can only be: q = A D C i C (A + D C ) (b) This part is easier done if we use the /q version of the ABCD law: Rearrange the equation and we have: q = C + D/q A + B/q B( q ) + (A D) q C = 0 By the quadratic formula: (A D) ± (A D) + 4BC = q B = A D B + i B (A + D B ) For a Gaussian: q(z) = R(z) + iλ πw (z). By comparing two expression of /q, it is obvious that: R(z i ) = B A D w(z i ) = = π λ ( A+D B B ) λ B π ( A+D ) 9
6. This is a practical case of Problem 5, and we only need to find out the specific M matrix and apply results derived in Problem 5. Let choose the round trip matrix M as shown blew: Figure 7 where the ray starts with going rightward, then reflects, go leftward and reflects again. (You can refer to Steck Chapter.7 for this) Therefore, the M matrix is: M = 0 [ ] d 0 [ ] d 0 0 R R + d d( + d ) = R R + + 4d + 4d + d + 4d R R R R R R R R From Problem 5, we then know that: R(z ) = B A D = 4d( + d/r ) 4d( + d/r )/R = R The rest would be to find the round trip M matrix M corresponding to the point z. Note that we have been working with a Gaussian that is initially going rightward, and we need to stick with it to be consistent. Hence: [ ] d M = 0 [ ] d 0 0 0 R R + d + 4d + 4d d( + d ) = R R R R R + + 4d + d, R R R R R 0
from which it can be easily shown that: R(z ) = R Now, since you have known radii of curvature of two points and their separation, by results of Problem 4, the Gaussian can be identified. 7. (a) The transmission spectrum is shown below: Figure 8: Transmission spectrum This structure cannot work as a cavity because the transmission of the modes cannot approach unity. (b) The symmetric structure works. Figure 9: Transmission spectrum
Figure 0: Mode profile of /λ = 4698 /cm Figure : Mode profile of /λ = 5655 /cm Figure : Mode profile of /λ = 6608 /cm