nds = n 1 d 1 sec θ 1 + n 2 d 2 sec θ 2 δopl =0

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Rafe Nelson
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1 1 Exercise The optical path length is given by OPL = Z C which for an optical ray, must be stationary nds = n 1 d 1 sec θ 1 + n d sec θ δopl =0 so the first derivative of the optical path length with respect path variations, such as that parameterized by θ 1 or θ, must be zero. Take the derivative with respect to θ 1 but note that θ 1 or θ are not independent. d dθ OPL = n 1 d 1 sec θ 1 tan θ 1 + n d sec θ tan θ =0 dθ 1 dθ 1 dθ n 1 d 1 sec θ 1 tan θ 1 = n d sec θ tan θ (1) dθ 1 This interdependence is described by the constraint d 1 tan θ 1 + d tan θ = d Taking the derivative of the constraint we find Divide (1) by () which simplifies to Snell s Law d 1 sec θ 1 + d sec θ dθ dθ 1 = 0 d 1 sec θ 1 = d sec θ dθ dθ 1 () n 1 tan θ 1 sec θ 1 = n tan θ sec θ n 1 sin θ 1 = n sin θ 1

2 Exercise a y a P Q z very substantial simplification results when one realizes that the triangle within the sphere is isosceles and thus the angle of refraction entering the sphere equals the angle of incidence leaving the sphere. pplying Snell s law to these two interfaces then leads to a very simple result for the exiting angle of refraction in terms of the entering angle of incidence. (Note that the figure is drawn for clarity and does not represent the actual numerical result.) The angle of incidence entering the sphere Snell s Law entering the sphere y =sinθ sin θ = n sin θ n and leaving n sin θ n =sinθ whereihaveusedthesameangleθ for entering incidence and exiting refraction. Since the two Snell s law applications are consistent it is clear that the two angles are indeed equal. straight line has angle π, as does the sum of the angles of a triangle. Thus find φ 1 π = θ +(π θn)+φ1 φ 1 = θ n θ

3 The exit point is then P = a(cos φ 1, sin φ 1 ) θ is the supplement (i.e. their sum is π) of the supplement of the sum of φ 1 and φ θ = φ 1 + φ φ = θ φ 1 φ = θ θ n tan φ = Q z = P y Q z P z P y + P z tan φ Plugging in the numbers 3 Exercise 1.3- f = Q z a = P y + P z a tan φ = a sin φ 1 + a cos φ tan φ 1 a sin φ1 = a +cosφ tan φ 1 1 y = 0.7,n=1.8,a=1 φ 1 = 0.035,φ =0.75 f = The marginally confined ray is incident on the end of the fiber with angle θ a. Use paraxial Snell s Law to find the initial angle θ 0 inside the fiber θ a n (0) θ 0 where so n (ρ) = n 0 1 α ρ ρ = p x + y θ 0 θ a n 0 3

4 What Saleh and Teich intend is that the marginally confined ray follows a sinusoidal trajectory with amplitude a. Seefigure (I think rays with an even larger amplitude will be totally internally refracted at the outer boundary and remain confined, but they will not be paraxial rays, so let us proceed to obtain the simpler result.) ssuming that the trajectory shown in the figure is a meridional ray and lies entirely within the y-z plane. Then initial conditions for the solution to the paraxial ray equations are: θ x0 = x 0 = y 0 =0and θ y0 θ a /n 0. The solution is then y (z) θ 0 sin (αz) αn 0 Setting the amplitude equal to a θ a a αn 0 we find the numerical aperture, N, which in the paraxial approximation is just the angle θ a N =sinθ a θ a n 0 aα For the step index fiber we have from equation (1.-15) q N = n 1 n for the fair comparison let n 1 = n (0) = n 0 n = n (a) =n 0 p 1 α ρ Then q N = n 0 n 0 (1 α a ) = n 0 aα which is the same as the GRIN fiber, but you knew it would be...didn t you? The GRIN fiber could be approximated be a series of small steps, and apparently, even one step is a fair approximation. 4 Exercise One round trip in the resonator is a propagation of distance d followed by a reflection from a spherical mirror of radius followed by another propagation of distance d and a reflection from a spherical mirror of radius R. The combined 4

5 matrix for the round trip is d d M = R d 1 d = R 1+ d R 1+ d R 1 ³ 1+ d = d + d 1+ d ³ ³ R + + 4d d R R + 1+ d R " # = Compute the b parameter d+ d +d +R +4d 4d +4d +dr + R R R 1+ d b = 1 ( + D) = 1 µ d + R1 + 4d +4d +dr + R R = d +d +dr + R R For stable (harmonic) trajectories to exist we must have b 1 d +d +dr + R R < 1 Tochecktheresult,lookatthesimplercaseof = R = R. Thenwehave d +4dR + R R < 1 d (d +R)+1 R < 1 d (d +R) R < 0 R < d/ Which makes sense since R = d/ would be the inside of a sphere and you know some stable trajectories for that case, e.g. rays that pass through the center of the sphere. 5 Exercise.-4 For a spherical wave the complex amplitude is U (r) = r e jkr 5

6 and the intensity (i.e. the power per unit area) is I = U (r) = r e jkr = r e jkr = For a surface normal to the direction of radiated power, the power is given by the surface integral of the intensity Z P = I(r)d Using a sphere of radius a for the integral (which indeed is normal to the radiated power direction for a spherical wave) Z Z P = I(a)d = I(a) d = I(a)4πa Thus Plugging in the numbers I(a) = P 4πa P = 100 W, a =1m I = 7.96 W/m How about the intensity of sun light at the earth? Seems like a fair amount of power! 6 Exercise.-6 P = W, a = m I = 1415 W/m paraxial wave can be defined as slowly varying modulation, (r), ofaplane wave parallel to the optic axis, e jkz U (r) = (r) e jkz We want to plug this form into the Helmholtz equation to see what equation (r) must obey. To do this we need the partial derivatives of U (r). First the x derivative (r) e jkz jkz = e x x (r) e jkz x = e jkz x r 6

7 The y derivative is similar. The z derivative requires the product rule (r) e jkz = ( jk) e jkz + z z e jkz (r) e jkz z = jke jkz + z z e jkz = k e jkz jk z e jkz + z e jkz jk z e jkz = k e jkz jk z e jkz + z e jkz Plugging it all into the Helmholtz equation we find U + k U =0 e jkz x + e jkz y k jk z e jkz + z e jkz + k e jkz =0 Combining terms and dropping the common factor, e jkz,wefind but (r) is slowly varying so that x + y jk z + z =0 z k z and we can drop the last term on the RHS and arrive at the paraxial Helmholtz equation T jk z =0 where we define the transverse laplacian operator T x + y 7

Lecture 19 Optical MEMS (1)

EEL6935 Advanced MEMS (Spring 5) Instructor: Dr. Huikai Xie Lecture 19 Optical MEMS (1) Agenda: Optics Review EEL6935 Advanced MEMS 5 H. Xie 3/8/5 1 Optics Review Nature of Light Reflection and Refraction