MIT2_71S09_usol6

# MIT2_71S09_usol6 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.71 Optics Solutions to Problem Set #6 Spring '09 Due Wednesday, Apr. 15, 2009 Problem 1: Grating with tilted plane wave illumination 1. a) In this problem, onedimensional geometry along the xaxis is considered. The Fresnel diffraction pattern, the field just behind the grating illuminated by the plane wave, is m x 2 exp i x . (1) g+ (x, z = 0) = gt (x)g- (x, z = 0) = exp i sin 2 2 Note that the transmission function can be expanded as m m x 2 Jq = exp iq x . gt (x) = exp i sin 2 2 2 q=- Using eq. (2), we can rewrite eq. (1) as g+ (x, z = 0) = q=- (2) Jq m 2 2 q exp i + x . (3) Since exp i 2 + q x represents a tilted plane wave whose propagation angle is + q/, eq. (3) implies that the transmitted field just behind the grating is consisted of a infinite number of plane waves, where q denotes diffraction order and the amplitude of the diffraction order q is Jq (m/2). The propagation direction of the zeroorder is identical as one of the incident tilted plane wave. 1.b) The field behind the grating is identical to eq. (1). When the observation plane is in the farzone, the Fraunhofer diffraction pattern is 2 (4) g(x , z) = g+ (x, z = 0) exp -i (x x) dx. z Note that we neglected the scaling factor and phase term because the scaling factor change overall magnitude of diffraction pattern and the phase term does not contribute to intensity. Substituting eq. (2) into (4), we obtain the field distribution of the Fraun hofer diffraction as m 2 q 2 Jq exp i + exp -i (xx ) dx g(x , z) = 2 z q=- m q x Jq exp i2 + x exp -i2 x dx = 2 z q=- m x q = Jq - + . (5) 2 z q=- 1 (a) onaxis plane wave (b) tilted plane wave Figure 1: The whole diffraction patterns rotate by as the incident plane wave rotates The intensity of the Fraunhofer diffraction pattern is I(x , z) = |g(x , z)| = 2 q=- 2 Jq m x q - + . 2 z (6) In the farregion, we should observe a infinite number of diffraction orders. The in 2 tensity of the diffraction order is proportional to Jq (m/2) and the offset between two neighboring diffraction orders is (z)/. The zeroth order is located at x = z. 1.c) In both cases (Fresnel and Fraunhofer diffraction), the diffraction patterns of the grating probed by a on-axis and tilted plane waves are identical except the angular shift by the incident angle , as shown in Fig. 1. 2 Problem 2: Grating spherical wave illumination 2.a) Using the same approach as in Prob. 1, we obtain 0 x ei2 z 1 x2 + y 2 g+ (x, y, z = 0) = gt (x, y)g- (x, y, z = 0) = 1 + m cos 2 exp i . 2 iz0 z0 (7) 2.b) Since both the cosine term and exponential terms in eq. (8) vary with x, we use following relation to understand eq. (8); x m x x =1+ exp i2 + exp -i2 . (8) 1 + m cos 2 2 Hence, eq. (8) can be rewritten as superposition of three spherical waves; z i2 0 e 1 x2 + y 2 g+ (x, y, z = 0) = exp i + iz0 2 z0 x x x2 + y 2 m x2 + y 2 m + exp i - i2 exp i + i2 4 z0 4 z0 z0 i2 e 1 x2 + y 2 (x + z0 /)2 + y 2 m z0 = exp i + exp i exp -i 2 + iz0 2 z0 4 z0 2 2 m (x - z0 /) + y z0 exp i exp -i 2 . (9) 4 z0 2.c) Figure 2(a) conceptually shows the diffraction pattern expressed in eq. (10). The first exponential term represents the zeroorder diffraction, which is identical to the incident spherical wave originated at (x = 0, y = 0, z = -z0 ) except amplitude attenuation. The second and third exponential terms indicate two spherical waves 2 originated at (z0 /, 0, -z0 ) with additional phase factor of e-iz0 / , which is in dependent on x and y. 2.d) If the illumination is a spherical wave emitted at (x0 , 0, -z0 ) as shown in Fig. 2(b), then we expect that the origins of the three spherical waves will be shifted by x0 ; i.e., the three origins are (x0 , 0, -z0 ), (x0 - z0 /, 0, -z0 ), and (x0 + z0 /, 0, -z0 ) if the paraxial approximation holds. More rigorously, the Fresnel diffraction pattern is computed as 0 x ei2 z 1 (x - x0 )2 + y 2 1 + m cos 2 g+ (x, z = 0) = exp i , (10) 2 iz0 z0 3 (a) onaxis spherical wave (b) offaxis spherical wave Figure 2: The diffraction patterns rotate in the same fashion as the incident spherical wave rotates and using the same expansion we eventually obtain z0 ei2 1 (x - x0 )2 + y 2 exp i + g+ (x, y, z = 0) = iz0 2 z0 m (x - x0 )2 + y 2 x m (x - x0 )2 + y 2 x + exp i - i2 exp i + i2 4 z0 4 z0 z0 ei2 1 (x - x0 )2 + y 2 = exp i + iz0 2 z0 (x - x0 + z0 /)2 + y 2 2x0 z0 m exp i exp -i - + 2 + 4 z0 2 2 (x - x0 - z0 /) + y 2x0 z0 m exp i exp -i + 2 . (11) 4 z0 As expected, the diffraction pattern is consisted of three spherical waves originated at (x0 , 0, -z0 ), (x0 - z0 /, 0, -z0 ), and (x0 + z0 /, 0, -z0 ), respectively. 4 3. (a) The Fourier series coefficients of a periodic function t(x) are given by: L 2 1 t(x )dx a0 = L -L 2 L 2 2 nx an = t(x ) cos dx L L -2 L/2 L 2 nx bn = t(x ) sin dx L -L L/2 where L is the period of t(x). The function t(x) can then be written as an infinite sum: nx nx t(x) = a0 + an cos + bn sin L/2 L/2 n=1 n=1 For the given function, L 1 4 1 a0 = dx = L -L 2 4 L L 2 4 2nx 2 L 2nx 4 = 2 sin n an = cos dx = sin L L -L L L 2n L n 2 -4 4 L L 2 4 2nx 2 L 2nx 4 bn = sin - cos = 0 dx = L -L L L 2n L L -4 4 0 sin n 1 2 a0 = , bn = 0, an = where n = 1, 2, 3, . . . n 2 2 Note that when n is even, an = 0, when n = 1 + 4m, an = +1, and when n = 3 + 4m, an = -1, where m is a positive integer. (b) A single boxcar is given by t1 (x) = rect 2x L L u 2 L T1 (u) = F(t1 (x)) = sinc 2 5 (c) An infinite array of boxcars of width L with a spacing of L between them can be 2 2 expressed as a convolution of a comb() function and a rect() function: x 2x t2 (x) = rect comb L L A truncated centered portion containing N boxcars is then given by x x x 2x t(x) = t2 (x) rect = rect comb rect NL L L NL The Fourier transform of t(x) becomes 2 L L T (u) = T2 (u) (NL)sinc(NLu) = sinc u comb(Lu) (NL)sinc(NLu) 2 2 (d) Single box car: T (u) = L sinc( L u) 2 2 Infinite array: 6 Finite array (N): The Fraunhofer diffraction pattern is similar to the Fourier transform of the functions (with a scaling factor u = x/L) A single box car creates a sinc() diffraction pattern. Having an infinitely long array would generate a set of () functions, i.e. single dots whose amplitude is modulated by a sinc() envelope profile identical to that generated by one boxcar. The spacing of the ()'s is the reciprocal of the period of the array. A finite array of boxcars generates a set of sinc() functions whose peaks are modulated by another sinc() function and whose spacing is the reciprocal of the period of the boxcar array. Limiting the size of the array is equivalent to imposing a window onto an infinite array. This window spreads the () functions into sinc() functions. The spread of each of these sinc()'s is inversely proportional to the width of the `window.' 4. Tilted ellipse: x2 + y 2 Circle: f2 (x2 , y2 ) = circ r a b Ellipse: x1 = x2 ; y1 = y2 r r r r x 2 = x 1 ; y2 = y 1 a b Tilted: x = x1 cos - y1 sin y = x1 sin + y1 cos r x 2 y 2 ( a x1 )2 + ( r y1 )2 1 1 b f1 (x1 , y2 ) = circ = circ + r a b 7 F1 (u1 , v1 ) = F(f1 ) = |ab|jinc(2 (au1 )2 + (bv1 )2 ) A rotation by in the space domain is equivalent to a rotation by in the frequency domain; hence, u1 = u cos + v sin , v1 = -u sin + v cos The Fourier transform of an ellipse tilted by an angle is F (u, v) = |ab|jinc(2 a2 (u cos + v sin )2 + b2 (-u sin + v cos )2 ) (a) Sketch of Fourier transform (b) The Fraunhofer diffraction pattern is given by x x y y 2 ab jinc 2 a ( cos + sin )2 + b2 (- sin + cos )2 F (u, v) x y = | | ( , ) 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. ...
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