DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING NORTHEASTERN UNIVERSITY ECEU646/ECEG105 OPTICS FOR ENGINEERS FALL 2008.

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1 DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING NORTHEASTERN UNIVERSITY ECEU646/ECEG105 OPTICS FOR ENGINEERS FALL 2008 Solutions This homework set includes problems for both ECEU646 and ECEG105. Problems marked (G)are for ECEG105 graduate students only. Problems marked (U)are for ECEU646 undergraduate students only. The other problems are for students in both courses. See the syllabus for assignment dates. Homework Set 1: Geometric Optics Problem 1 Dogleg. A beam of laser light is incident on a beamsplitter of thickness, t, and angle, θ, as shown in Figure 1. The index of refraction of the medium on each side is the same; n, and that of the beamsplitter is n. When I remove the beamsplitter, the light falls on a detector. When I replace the beamsplitter, the light follows a dogleg path, and I must move the detector in a direction transverse to the beam. How far? a. Derive the general equation. = t sin θ t tanθ cos θ = t sin θ t sin θ cos θ cos θ ( = t sin θ 1 n 1 sin 2 ) θ n 1 sin = t sin θ 1 2 θ n 1 sin 2 θ n 1 sin 2 θ(n/n ) 2 ) = t sin θ 1 n2 n 2 sin 2 θ (n ) 2 n 2 sin 2 θ The geometry is shown in Figure 2 b. Check your result for n = n and for n /n. Calculate and plot vs. angle, for t = 5mm, n = 1, with n = 1.5, For n = n = 0, as is obviously expected. n t sin θ because the ray will follow the normal. See the MATLAB code and plot in Figure 3 c. and again with n = 4. Problem 2 Some Lenses.

2 Chuck DiMarzio, Northeastern University document:11270hw 2 Figure 1: Layout for Problem 1 Figure 2: Solution for Problem 1, Part a

3 Chuck DiMarzio, Northeastern University document:11270hw 3 Figure 3: Solution for Problem 1, Parts b and c Figure 4: Sample Lenses for Problem 2

4 Chuck DiMarzio, Northeastern University document:11270hw 4 a. Consider the thin lens in panel A of Figure 4. Let the focal length be f = 10cm. Plot the image distance as a function of object distance from 50 cm down to 5 cm. b. Plot the magnification. c. Consider the thin lens in panel B of Figure 4. Let the focal length be f = 10cm. Plot the image distance as a function of object distance from 50 cm down to 5 cm. d. Plot the magnification. e. In panel C of Figure 4, the lenses are both thin, and have focal lengths of f 1 = 20cm, and f 2 = 10cm, and the separation is d = 5 cm. For an object 40 cm in front of the first lens, where is the final image? f. What is the magnification? The following is a screen dump from MATLAB for all parts of this problem. Figure 5 shows the resulting plots. echo on % file m10353b.m % for homework in Optics for Engineers % by Chuck DiMarzio % Northeastern University % July 2003 % problem on lenses figure; f=10; s=[5:1:50]; sprime=1./(1/f-1./s); Warning: Divide by zero. > In C:\document\working\10353\m10353b.m at line 13 subplot(2,2,1);plot(s,sprime); title( Part a ); xlabel( s, Object Distance, cm ); ylabel( s, Image Distance, cm ); % m=-sprime./s; subplot(2,2,2);plot(s,m); title( Part b ); xlabel( s, object distance ); ylabel( m, Magnification ); % f=-10; sprime=1./(1/f-1./s); subplot(2,2,3);plot(s,sprime); title( Part c ); xlabel( s, Object Distance, cm ); ylabel( s, Image Distance, cm ); % m=-sprime./s;

5 Chuck DiMarzio, Northeastern University document:11270hw 5 title( Part d ); subplot(2,2,4);plot(s,m); xlabel( s, object distance ); ylabel( m, Magnification ); print -dtiff eps % disp( Part e ); Part e s1=40; f1=20; d=5; f2=10; s1prime=1/(1/f1-1/s1) s1prime = 40 s2=d-s1prime s2 = -35 s2prime=1/(1/f2-1/s2) s2prime = disp( the image is this far to the right of the second lens ) the image is this far to the right of the second lens disp( ); % disp( Part f ); Part f m1=-s1prime/s1 m1 = -1 m2=-s2prime/s2 m2 = m=m1*m2 m = disp( m is the total magnification ) m is the total magnification Problem 3 Focussing a Laser Beam. The configuration in panel D of Figure 4 can be used to focus a laser beam to couple it into an optical fiber. We want the sine of the half-angle subtended by the last lens (shaded region) to equal the numerical aperture of the fiber, NA = 0.4. Let the second lens have a

6 Chuck DiMarzio, Northeastern University document:11270hw 6 Figure 5: Plots for Problem 2

7 Chuck DiMarzio, Northeastern University document:11270hw 7 focal length of 16 mm, and assume that the initial laser beam diameter is D 1 = 1.4mm. We want the magnification of the second lens to be 1/4. a. What is the focal length of the first lens? We know that s 2 = m 2 = 1 s 2 4 The focus between the lenses is located to the left of the second lens by an amount, s 2 given by = 1 s 2 s 2 f s 2 s 2 /4 = 5 = 1 s 2 16 s 2 = 5f 2 = 80 s 2 = 20 where all distances are in millimeters. The beam diameter at the second lens is related to the tangent of the angle in the shaded area, whose sine is the numerical aperture. NA d 2 = 2 1 NA 2 s 2 = 17.5mm. Thus, by similar triangles, noting that s 1, d 1 f 1 = d 2 s 2 f 1 = d 1 s 2 = = 6.4mm. d b. What is the spacing between the two lenses, assuming thin lenses? f 1 + s 2 = = 86.4mm. c. Now, assume that the lenses are both 5 mm thick, the first is convex plano and the second is bi convex. What is the spacing between the vertices of the lenses now? What is the distance to the image point from the second lens vertex? Where should the fiber be placed relative to the second lens? You may use an approximation we discussed in class, and the fact that the lenses are glass. The principal planes of the first lens are at the left (convex) vertex and one third of the thickness (5/3 mm) to the right of that same vertex. This the second principal plane of the first lens is 10/3 mm to the right of the plane surface. The principal planse of the biconvex lens are one third of the thickness in from each vertex. Thus for the distance between principal planes to be z 12 = 86.4mm, the distance between vertices must be Likewise, the distance to the fiber is w 12 = = 81.4mm. w 2 = s 2 5/3 = 18.33mm.

8 Chuck DiMarzio, Northeastern University document:11270hw 8 d. Use matrices and find the focal length of this pair of lenses and the location of the principal planes. Be sure to specify distances and directions from lens vertices. The matrix for the first lens, from front principal plane to back principal plane is f 1 Note that we could have worked from vertex to vertex assuming an index of refraction and computing the surface curvatures, but there is no real need to do this, and in the case of complicated lenses, which are often proprietary designs, you won t know these parameters. The complete matrix is ( ) z f 2 f 1 Note that we use z 12 rather than w 12, because the lens matrices are from principal plane to principal plane. Now, using MATLAB % file m10353c.m - revised 2 Nov 03 % for homework in Optics for Engineers % by Chuck DiMarzio % Northeastern University % The focussing problem, part d: Principal Planes d1=1.4; na=0.4; f2=16; m2=-1/4; s2=f2*(1+1/(-m2)) s2 = 80 s2prime=-s2*m2 s2prime = 20 tang=0.4/sqrt(1-0.4^2) tang = d2=2*s2prime*tang d2 = f1=s2*d1/d2 f1 = disp( above solves a ); above solves a

9 Chuck DiMarzio, Northeastern University document:11270hw 9 z12=f1+s2 z12 = w12=z12-5/3-10/3 w12 = disp( above solves b,c ); above solves b,c lens1=[1,0;-1/f1,1] % minus sign added 2 Nov 03 lens1 = lens2=[1,0;-1/f2,1] % minus sign added 2 Nov 03 lens2 = sep=[1,z12;0,1] sep = m=lens2*sep*lens1 m = f=-1/m(2,1) f = d=(1-m(2,2))/m(2,1) d = dprime=(1-m(1,1))/m(2,1) dprime = disp( Above solve d ); Above solve d Now, we need to interpret the results. The front principal plane of the combined lens is a distance D in front of the front principal plane of the first lens. Because the first lens is convex plano, the result is that the principal plane is D = 8.7mm to the left of the first vertex of the first lens. The back principal plane is D to the right of the back principal plane of the second lens, D 5/3mm = 19.9 mm to the right of the second vertex of the

10 Chuck DiMarzio, Northeastern University document:11270hw 10 second lens.