Unformatted text preview: 2.710 Optics Final Exam Solutions Spring `09 1. Consider the following system. (a) If we position an onaxis point source at the center of the object plane (front focal plane of L1), a collimated ray bundle will emerge to the right of L1 and its diameter is set by S1; therefore, S1 is the aperture stop (A.S.). Similarly, S2 limits the lateral extent of an imaged object (consider an offaxis point source) and thus, it's our field stop (F.S.). (b) The entrance pupil is the image of the A.S. by the preceding optical components. To find its location we use the imaging condition, 1 1 1 + = So Si f1 2f1 So = 3 Si =
2 So f1 2f1 /3 = So  f1 f1 ( 2  1) 3 Si = 2f1 (virtual) So the entrance pupil is located at 2f1 to the right of L1. To find its radius, we compute the lateral magnification, ML =  For the exit pupil (Ex.P.), f1 So = + f2 , 3 ML = Si = ( f31 + f2 )f2
f1 3 Si = 3 rEnP = 3a1 So +  f2 f2 = 3 = f2 + 3 2 f2 (to the right of L2) f1  3ff12 ( f31 + f2 ) ( f31 + f2 ) f2 (inverted) f1 rExP = 3 f2 a1 f1 The exit window is the same as S2. The entrance window is the image of S2 through the preceding optical ele ments (i.e. combination of L1 and L2). It is f1 to the left of L1. 1 (c) Solution: The numerical aperture is: tan sin NA a1 f1 The field of view (FOV) is: FOV = 2 = 2Xs 2 f1 a2 2 a2 = = 3f1 3 f 1 f2 3 f2 (d) The location of S1 limits the FOV because of the requirement for the C.R. to go through the center of the aperture stop (A.S.). It can be seen that the least restrictive A.S. location is at the Fourier plane (f1 to the right of L1 f2 to the left of L2). 2. Given the following GRIN medium, 2  r2 n(r) = 1 0<r<1 , r1 r= x2 + z 2 (a) The Hamiltonian equations are (r < 1):
dx ds dz ds dPx ds dPz ds = = = = H Px H Pz  H x  H z =  1 2Px 2 2 = = = 2 Px +Pz 1 2Pz 2 2 2 Px +Pz 1 2x 2 n  1 2z 2 n P =  Px =  2xx z2 2 n =  Pz n = =
x n z n P =  2xz2 z2 = = x 2x2 z 2 z 2x2 z 2 where,
2 H = n(q)  [Px + Pz2 ]1/2 = 0 2 = [2  x2  z 2 ]1/2  [Px + Pz2 ]1/2 = 0 2 n = Px + Pz2 2 2 (b) Recall that we just found n2 = Px + Pz2 . dPx ds 2 + dPz ds 2 x2 + z 2 x2 + z 2 = 2 n n n2 Px + Pz2 r2 2  n2 2 = 2 = 2 = 2 1 2 2 Px + P z Px + P z Px + Pz2 = x 2 + z 2 = n = 0, the Screen Hamiltonian is not conserved. This may be verified by (c) Since z direct substitution: 2 2 h =  n 2  Px =  2  x 2  z 2  Px and we see that, h z = =0 2  z2  P 2 z 2x x 3. Consider the Michelson interferometer shown below. (a) We begin by writing the analytic expression (in phasor form) of a spherical wave with origin at (xp , zp ), using the paraxial approximation: E0 ei (zzp ) i (xxpp))2 ES = e (zz i(z  zp ) For simplicity we take xp = zp = 0. At the observation plane, the interference pattern is given by: I = ES1 + ES2 2 = ES1 2 + ES2 2 + ES1 ES2 + ES1 ES2
2 3 where E0 ei 1 i x2 ES1 = e 1 4 i1 1 = z0 + 2z1 + zc 2 E0 ei 1 i x2 ES2 = e 2 4 i2 2 = z0 + 2z2 + zc 2 ES1 2 = So, E0 2 16(1 )2 ES2 2 = E0 2 16(2 )2 E0 2 E0 2 E0 2 I= + + [ei + ei ] 16(1 )2 16(2 )2 162 1 2 E0 2 1 1 2 = + + cos , = 2  1 162 2 2 1 2 1 2 2 x2 2 x2 2 = + 2 ; 1 = + 1 22 21 2 2 1 1 = x  + 2z , z = z2  z1 22 21 E0 2 1 1 2 2 2 I= + + cos x  162 2 2 1 2 21 2 1 2 = 2z = 2  1 Imax Imin E0 2 = 162 E0 2 = 162 1 1 2 + 2+ 2 1 2 1 2 1 1 2 + 2 2 1 2 1 2 (b) If the flat mirror M2 is replaced by a convex spherical mirror of radius 2(z0 + z2 ), the spherical wave gets collimated and the interference pattern becomes: 2 2 E0 i E0 2 E0 2 ES1 + = E0 I= e + + 2 sin )2 2 4 16( 16(1 ) 8 1 Plane Wave Chirp Function 4 = z0 + z2 2 = 2 = 1  4. Consider the 4f system shown below, (a) The pupil mask can be implemented by placing two pinholes (small apertures), one centered with respect to the optical axis and the second one at 1 cm offaxis. The 2nd pinhole is phase delayed by a piece of glass of thickness t, where 2 == t(1.5  1) t = (b) The input transparency is gin = gt = gillumination 1 q= sinc(q)ei2 qx At the Fourier plane, q Gin = sinc(q) u  u= x q= f q 1 f Gin (x ) = sinc x  q , 2 q= 2 5 f 0.5 104 10 = =1 5 104 After the pupil mask, only the 0th and +1 orders pass. The +1 order gets phase delayed by ei = 1. 1 1 1 Gout (x ) = (x )  sinc (x  1) 2 2 2 1 1 = (x )  (x  1) 2 1 1 i2u gout (u ) =  e x 2 u = f 1 1 i 2x 2 1 1 1 2
x Iout =  e f = + 2  cos 2 4 f The contrast is v = 0.906 =
1 1 4 1 2 + = 4 . 4 + 2 5. (a) To compute the OTF, we first need the ATF: u1 = x 1cm = = 0.2m1 = 200mm1 f 0.5m 10cm
est. x 0.2cm u = = 0.025m1 = 25mm2 f 0.5m 10cm H(u) = H(u) H(u) (autocorrelation) u u  u1 u u u  u1  u  rect rect  rect du = rect u u u u 6 = rect u u rect u  u u du  rect u u  rect u  u1  u u du  rect u  u1 u rect u  u u du + rect u  u1 u rect u  u1  u u du So the resultant OTF and MTF are (after normalization): (b) Only the DC and the 1st harmonics at u = 200mm1 (period = 5m), i.e. 1 1 1 2x I(x ) =  2 cos 2 2 5m 2x 2 = DC term  H(200mm ) 1st harmonic 2 cos 5m (c) Solution: 7  F sinc2 (ux), so the iPSF is: = sinc2 x 40m x 1  cos 2 5m 8 MIT OpenCourseWare http://ocw.mit.edu Spring 2009 2.71 / 2.710 Optics For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
View
Full
Document
This note was uploaded on 02/24/2012 for the course MECHANICAL 2.710 taught by Professor Sebaekoh during the Spring '09 term at MIT.
 Spring '09
 SeBaekOh

Click to edit the document details