MIT2_71S09_practice2_sol

MIT2_71S09_practice2 - 2.71/2.710 Optics Practice Exam 2 Solutions Spring `09 1 What is the Fraunhofer diffraction pattern of a 1-D slit of size a

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Unformatted text preview: 2.71/2.710 Optics Practice Exam 2 - Solutions Spring `09 1. What is the Fraunhofer diffraction pattern of a 1-D slit of size a? Slit description (1D): Fourier transform of slit: Diffracted far field: Fraunhofer diffraction pattern (intensity): a F(u) = asinc(au) F ax 2 2 2 |g(x )| = a sinc z g(x ) = e z f (x) = rect x i x 2 +y 2 x z 2. What is the Fraunhofer diffraction pattern of this sinusoidal amplitude grating, where is the grating period? 1 Solution: f (x) = x 1 1 + cos 2 2 = DC term (0th order) + diffracted orders x 1 1 i2 x 1 = + e + e-i2 2 4 4 plane wave, u=0 plane wave, 1 u= plane wave, 1 u=- 1 1 1 1 F(u) = (u) + u - + u+ 2 4 4 2 +y 2 1 x 1 x i x z - g(x ) = e + 2 z 4 z 1 1 1 x 1 + + 4 z Note: without being too rigorous mathematically, we treat the intensity corresponding to the -function field as a "very bright and sharp" spot. 3. How does the result of problem 2 change if the illumination is a plane wave incident at angle 0 with respect to the optical axis? (0 << 1) sin 0 Solution: Let u0 = , so the plane wave is ei2u0 x (at z = 0). 2 F e i2u0 x x 1 1 + cos 2 2 1 1 1 1 1 = (u-u0 )+ u - u0 - + u - u0 + 2 4 4 4. What is the Fraunhofer pattern of this truncated sinusoidal amplitude grating? Assume that a >> . x x 1 f (x) = 1 + cos 2 rect 2 a Solution: x 1 f1 (x) = 1 + cos 2 2 x f2 (x) = rect a 1 1 1 1 1 F1 (u) = (u) + u - + u+ 2 4 4 F2 (u) = asinc(au) According to the convolution theorem, F{f1 (x) f2 (x)} = F1 (u) F2 (u). Recall that: (u - u0 ) A(u) = (u - u0 )A(u - u)du = A(u - u0 ) - 1 1 1 1 1 asinc(au) F1 (u) F2 (u) = (u) + u - + u+ 2 4 4 1 1 1 1 1 = asinc(au) + asinc a u - + asinc a u + 2 4 4 Note: When you take ||2 , cross-terms can be ignored. Why? a2 ax a2 x 1 a2 x 1 2 2 2 2 |y(x )| sinc + sinc a - + sinc a + z 4 z 16 16 z Fraunhofer pattern (intensity) 3 Fraunhofer pattern of truncated grating 5. What is the Fraunhofer diffraction pattern of two identical slits (width a) separated by a distance d >> a? f (x) = rect x- a d 2 + rect x+ a d 2 Use the scaling and shift theorems, and linearity: F(u) = asinc(au)e-i2u 2 + asinc(au)ei2u 2 = 2asinc(au) cos(ud) 4 d d |g(x )| = 4a sinc 2 2 2 ax z cos x d z `modulated' sinc pattern 6. In the 4F system shown below, the sinusoidal transparency t(x) is illuminated by a monochromatic plane wave on-axis, at wavelength = 1m. Describe quantitatively the fields at the Fourier plane (x ) and the output plane (x ). 1 x t(x) = 1 + cos 2 2 10m Solution: The field immediately past the transparency is produced by the on-axis plane wave multiplied by t(x), the transmission function of the transparency. gin (x) = 1 t(x) x The field at the Fourier plane is Gin where Gin (u) is the Fourier transform of f1 gin (x ), i.e. 1 1 1 1 1 Gin (u) = (u) + u - + u+ 2 2 10m 2 10m 5 The field at the Fourier plane will consist of three peaks corresponding to the three -functions of the Fourier transform. The locations of these peaks are found as follows: +1st order: 0th order: -1st order: 1 x 1 f1 1m 10cm = x = = = 1cm 10m f1 10m 10m 10m x = 0 x = = -1cm u= The field at the output plane is: f1 10cm 1 x gin - x = gin - x = gin (-5x) = 1 + cos 2 f2 2 2m 2cm So the field has been laterally demagnified by the imaging system. Notice that lateral demagnification implies angular magnification according to the following diagram: 6 7. Repeat the calculations of problem 6, except this time with illumination of a tilted plane wave incident at angle = 0.25 rad with respect to the optical axis. Solution: This time gin (x) = ei sin x t(x), where sin = 0.25 (paraxial approxi mation). Therefore, 1 x i2 0.25x 1 + cos 2 gin (x) = e 1m 10m 2 25x 1 i2 01.m 1 i2 10x 1 -i2 10x m + m = e 1+ e e 2 2 2 x 1 1 x 1 1 x 1 1 1 = ei2 4m + ei2( 10 + 4 ) m + ei2(- 10 + 4 ) m 2 4 4 1 1 1 1 Gin (u) = u - + (u - 0.35m-1 ) + (u - 0.15m-1 ) 2 4m 4 4 7 2 Note 1. We can get this result faster by use of the shift theorem of Fourier transforms: x 1 -i2 4m gin (x) = e t(x) Gin (u) = T u - 4m So it is the result of Problem 6 shifted to the right by 0.25m-1 . Note 2. Physical explanation: The diffracted order angles are: -1 0.25 - 0.1 = 0.15 rad 0 0.25 + 0 = 0.25 rad +1 0.25 + 0.1 = 0.35 rad Compare this with the diagram in the answer to Problem 6! 8 Field at the output plane: gout (x ) = gin (-5x ) = e x -i21.25 m x 1 + cos 2 2m Notice the intensity |gout (x )|2 is the same as in problem 6. The extra phase factor indicates that the overall output field is propagating "downwards". 8. Repeat problem 7 with a truncated grating of size 1 mm. Solution: Now the input field is x x 1 x i2 4m gin = e 1 + cos 2 rect 2 10m 1mm truncates the grating to total size of 1 mm We need to compute Gin (u). We will do it in two steps: x 1 x (a) Compute the Fourier transform of 2 1 + cos 2 10m rect 1mm by applying the convolution theorem: x 1 x f= 1 + cos 2 rect 2 10m 1mm 1 1 1 1 1 F = (u) + u - + u+ sinc(1mm u) (1mm) 10m 4 10m 2 4 neglect from now on 1 1 1 1 1 = sinc(1mm u) + sinc 1mm u - + sinc 1mm u + 2 4 10m 4 10m Let's plot the first term of this expression before continuing. (b) Apply the shift theorem to take into account the ei2 4m factor (see also problem 7): Gin (u) = 1 sinc 1mm u - 2 0.25 m x 1 + 4 sinc 1mm u - 9 0.35 m 1 + 4 sinc 1mm u - 0.15 m Field at output plane: gout (x ) = gin (-5x ) = e x -i21.25 m x x 1 + cos 2 rect 2m 0.2mm It is still a truncated grating, shrunk by a factor of 5 compared to the original grating. 9. In the optical system of problem 6 (infinitely large grating, on-axis plane wave illumi nation) we place a small piece of glass at the Fourier plane as follows: What is the output field? What is the output intensity? Solution: The piece of glass delays the +1st order field by a phase equal to: = 2 (n - 1)d 0.5 501m = 2 = 501 = 1m 10 (phase is mod 2) So, immediately after the Fourier plane, the field is: 1 2 (u) + 1 ei u - 2 1 10m 1 + 2 u + 1 10m = 1 2 1 (u) - 2 u - 1 10m 1 + 2 u + 1 10m We simplify the phase delay, ei = ei = -1. Next, switch to coordinates x = f1 u: 1 1 1 (x ) - (x - 1cm) + (x + 1cm) gF (x ) 2 2 2 Now consider the second half of the 4F system: Since L2 is acting as a Fourier-transforming lens, x x x 1 1 i2 2m 1 -i2 2m gout (x ) = GF = 1- e + e f2 2 2 2 1 x = 1 - i sin 2 field 2 2m 1 x 2 2 |gout (x )| = 1 + sin 2 intensity 2 2m 10. Consider the 4F optical system shown in Figure B, where lenses L1, L2 are identical with focal length f . A thin transparency with arbitrary transmission function t(x) is placed at the input plane of the system, and illuminated with a monochromatic, coherent plane wave at wavelength , incident on-axis. At the Fourier plane of the system we place the amplitude filter shown in Figure C. The filter is opaque everywhere except over two thin stripes of width a, located symmetrically around the y axis. The distance between the stripe centers is x0 > a. 11 (a) Which range of spatial frequencies must t(x) contain for the system to transmit any light to its image plane? Solution: The system admits frequencies +a 2 u f x0 a x0 a + 2 -2 2 - u - 2 f f - f 12 x0 2 a 2 x0 2 (b) Write an expression for the field at the image plane as the convolution of t(x) with the coherent impulse response of this system. Solution: x0 x0 u - 2f u + 2f H(u) = rect + rect a a f f a ax a h(x) = sinc e + sinc f f f 2a ax x0 x = sinc cos f f f The output is: 2a g(x ) = f x0 x -i2 2f ax f ei2 2f x0 x t(x - x)sinc - ax f cos x0 x f dx = t(x) h(x) x 13 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. ...
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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.

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