2.710 Optics Spring 09 Solutions to Problem Set #6 Due Wednesday, Apr. 15, 2009

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY.710 Optics Spring 09 Solutions to Problem Set #6 Due Wednesday, Apr. 15, 009 Problem 1: Grating with tilted plane wave illumination 1. a) In this problem, one dimensional geometry along the ais is considered. The Fresnel diffraction pattern, the field just behind the grating illuminated by the plane wave, is { ( )} { } m π g + (, z = 0)= g t ()g (, z = 0) = ep i sin π ep i θ. (1) λ Note that the transmission function can be epanded as { ( )} ( ) { } m m π g t () = ep i sin π = J q ep iq. () q= Using eq. (), we can rewrite eq. (1) as ( { ( ) } m ) π qλ g + (, z = 0)= J q ep i θ +. (3) λ q= { ( ) } qλ Since ep i π θ + λ represents a tilted plane wave whose propagation angle is θ + qλ/, 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 J q (m/). The propagation direction of the zero order 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 far zone, the Fraunhofer diffraction pattern is { } g(,z)= g + (, z = 0) ep i π ( ) d. (4) λz 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. () into (4), we obtain the field distribution of the Fraunhofer diffraction as [ ] ( ) { ( )} { } g( m,z)= Jq ep i π θ + qλ ep i π ( ) d λ λz q= ( m ) [ { ( q θ ) } { } ] = J q ep iπ + ep iπ d λ λz q= ( ) ( [ ]) m q θ = J q δ +. (5) λz λ q= 1

2 (a) on ais plane wave (b) tilted plane wave Figure 1: The whole diffraction patterns rotate by θ as the incident plane wave rotates The intensity of the Fraunhofer diffraction pattern is ( [ ]) I(,z)= g(,z) = ( m ) J q θ q δ +. (6) λz λ q= In the far region, we should observe a infinite number of diffraction orders. The intensity of the diffraction order is proportional to J q (m/) and the offset between two neighboring diffraction orders is (λz)/. The zeroth order is located at = zθ. 1.c) In both cases (Fresnel and Fraunhofer diffraction), the diffraction patterns of the grating probed by a on-ais and tilted plane waves are identical ecept the angular shift by the incident angle θ, as shown in Fig. 1.

3 Problem : Grating spherical wave illumination.a) Using the same approach as in Prob. 1, we obtain z [ ( )] iπ 0 { } 1 e λ + y g + (, y, z = 0)= g t (, y)g (, y, z = 0)= 1+ m cos π ep iπ. iλz 0 λz 0 (7).b) Since both the cosine term and eponential terms in eq. (8) vary with, we use following relation to understand eq. (8); ( ) m ( { } { }) 1+ m cos π =1+ ep iπ + ep iπ. (8) Hence, eq. (8) can be rewritten as superposition of three spherical waves; [ z iπ 0 { e 1 + y } λ g + (, y, z = 0)= ep iπ + iλz 0 λz 0 { ] m + y } { m + y } ep iπ + iπ + ep iπ iπ 4 λz 0 4 λz 0 [ z iπ 0 { 1 + y } { m ( + λz 0 /) + y } { } e λ λz 0 = ep iπ + ep iπ ep iπ + iλz 0 λz 0 4 λz 0 { ] m ( λz 0 /) + y } { } λz 0 ep iπ ep iπ. (9) 4 λz 0.c) Figure (a) conceptually shows the diffraction pattern epressed in eq. (10). The first eponential term represents the zero order diffraction, which is identical to the incident spherical wave originated at ( =0,y =0,z = z 0 ) ecept amplitude attenuation. The second and third eponential terms indicate two spherical waves originated at (±λz 0 /, 0, z 0 ) with additional phase factor of e iπλz 0/, which is independent on and y..d) If the illumination is a spherical wave emitted at ( 0, 0, z 0 ) as shown in Fig. (b), then we epect that the origins of the three spherical waves will be shifted by 0 ; i.e., the three origins are ( 0, 0, z 0 ), ( 0 λz 0 /, 0, z 0 ), and ( 0 + λz 0 /, 0, z 0 )if the paraial approimation holds. More rigorously, the Fresnel diffraction pattern is computed as [ ( )] { } iπ z 0 1 e λ ( 0 ) + y g + (, z = 0)= 1+ m cos π ep iπ, (10) iλz 0 λz 0 3

4 (a) on ais spherical wave (b) off ais spherical wave Figure : The diffraction patterns rotate in the same fashion as the incident spherical wave rotates and using the same epansion we eventually obtain [ { } e iπ z 0 λ 1 ( 0 ) + y g + (, y, z = 0)= ep iπ + iλz 0 λz 0 { ] m ( 0 ) + y } { m ( 0 ) + y } ep iπ + iπ + ep iπ iπ 4 λz 0 4 λz 0 [ { } e iπ z 0 λ 1 ( 0 ) + y = ep iπ + iλz 0 λz 0 { m ( 0 + λz 0 /) + y } { ( )} 0 λz 0 ep iπ ep iπ λz 0 { ] m ( 0 λz 0 /) + y } { ( )} 0 λz 0 ep iπ ep iπ +. (11) 4 λz 0 As epected, the diffraction pattern is consisted of three spherical waves originated at ( 0, 0, z 0 ), ( 0 λz 0 /, 0, z 0 ), and ( 0 + λz 0 /, 0, z 0 ), respectively. 4

5 3. (a) The Fourier series coefficients of a periodic function t() are given by: L 1 a t( )d 0 = L L L ( ) nπ a t( ) cos d n = L L L/ ( ) L nπ b n = t( ) sin d L L L/ where L is the period of t(). The function t() can then be written as an infinite sum: ( ) ( ) nπ nπ t() = a 0 + a n cos + b n sin L/ L/ n=1 n=1 For the given function, L 1 4 = d 1 a 0 = L L 4 L ( ) ( ) L ( ) 4 πn L πn 4 = cos πn a d n = sin L = sin L L L πn L πn 4 L ( ) [ ( ) 4 ] L L 4 πn 4 L πn b d n = sin = cos L = 0 L L L πn L L 4 4 }{{} 0 ( ) 1 sin πn a 0 =, b n = 0, a n = πn where n = 1,, 3,... Note that when n is even, a n = 0, when n = 1 + 4m, a n = +1, and when n = 3 + 4m, a n = 1, where m is a positive integer. (b) A single bocar is given by ( ) t 1 () = rect L ( ) L L T 1 (u) = F(t 1 ()) = sinc u 5

6 (c) An infinite array of bocars of width L L with a spacing of between them can be epressed as a convolution of a comb() function and a rect() function: ( ) ( ) t () = rect comb L L A truncated centered portion containing N bocars is then given by ( ) [ ( ) ( ) ] ( ) t() = t () rect = rect comb rect NL L L NL The Fourier transform of t() becomes [ ( ) ] L L T (u) = T (u) (NL)sinc(NLu) = sinc u comb(lu) (NL)sinc(NLu) (d) Single bo car: T (u) = L sinc( L u) Infinite array: 6

7 Finite array (N): 4. Tilted ellipse: The Fraunhofer diffraction pattern is similar to the Fourier transform of the functions (with a scaling factor u = /λl) A single bo car creates a sinc() diffraction pattern. Having an infinitely long array would generate a set of δ() functions, i.e. single dots whose amplitude is modulated by a sinc() envelope profile identical to that generated by one bocar. The spacing of the δ() s is the reciprocal of the period of the array. A finite array of bocars 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 bocar array. Limiting the size of the array is equivalent to imposing a window onto an infinite array. This window spreads the δ() functions into sinc() functions. The spread of each of these sinc() s is inversely proportional to the width of the window. ( ) + y Circle: f (, y ) = circ r a b Ellipse: 1 = ; y 1 = y r r = 1 ; y = y 1 a b Tilted: = 1 cos θ y 1 sin θ y = 1 sin θ + y 1 cos θ ( ) ( ) ( r 1 ) + ( r y 1 ) ( 1 ) ( y1 ) f 1 ( 1, y ) = circ a b = circ + r a b 7

8 F 1 (u 1, v 1 ) = F(f 1 ) = ab jinc(π (au 1 ) + (bv 1 ) ) A rotation by θ in the space domain is equivalent to a rotation by θ in the frequency domain; hence, u 1 = u cos θ + v sin θ, v 1 = u sin θ + v cos θ The Fourier transform of an ellipse tilted by an angle θ is F (u, v) = ab jinc(π a (u cos θ + v sin θ) + b ( u sin θ + v cos θ) ) (a) Sketch of Fourier transform (b) The Fraunhofer diffraction pattern is given by ( ) F (u, v) = ab jinc π a ( cos θ + y sin θ) + b ( sin θ + y cos θ) λl λl λl λl (, y ) λl λl 8

9 Problem 5: Blazed grating In this problem, the given phase profile of the grating is shown in Fig. 3. So it is important to properly construct the comple transparency of the grating. Figure 3: phase profile of the blazed grating 3.a) Since the grating is a periodic function without any truncation, Fourier series can be used and the Fourier coefficient (its square for intensity) represents the diffraction efficiency. { } 1 π { q } a q = ep i ep iπ d 0 { } = 1 ep i π (1 q) d = 1 ep {i π(1 q)} 1 π 0 i (1 q) = e iπ(1 q) sin(π(1 q) = e iπq sinc (q 1), (1) π(1 q) where q denotes the diffraction order. Hence, the efficiency of the diffraction order q is proportional to I q a q = sinc (q 1). (13) Note that only the first order is survived and other orders are canceled out. The blazed grating produces the same number of diffraction orders as non blazed gratings; but the blazed grating concentrates more light into a specific order due to the phase profile. (You can imagine that the one period of the grating behaves as a prism to bend light!) The same conclusion can be obtained with Fourier transform. We first choose one period of the grating whose comple transparency (not phase profile!) as ( ) iπ t () = rect e, (14) and it is shown in Fig. 4(a). To make the grating periodic, it convolves with a comb function (impulse train) whose period is shown in Fig. 4(b). Note the comb function is shifted by / to correctly represent the given grating. Then, the comple transparency of the grating can be written as [ ( ) ] ( ) t() = rect e iπ comb. (15) 9

10 (a) phase of the one period of the grating (b) desired comb function Figure 4: The grating can be constructed by the convolution of a triangular phase profile and comb function. To find the diffraction efficiencies, we compute Fraunhofer diffraction pattern, which is identical to taking Fourier transform of the comple transparency. {[ ( ) ] ( )} iπ F rect e comb = { ( ) } { ( )} iπ F rect e F comb = {[ ] [ ( )]} { } 1 iπ u iπ u sinc (u) e δ u comb (u) e = { ( [ ]) } 1 iπ [u 1 ] iπ sinc u e comb (u) e u = { ( [ ])} e iπ[u 1 ] sinc u 1 comb (u). (16) Thus, the intensity of the diffraction pattern is proportional to ( [ ]) 1 I(u) sinc u comb (u), (17) where the sinc and comb functions are shown in Fig. 4(a) and (b). Fig. 4(c) shows the Fraunhofer diffraction pattern replacing with = uλz at an observation plane. Again, the diffraction efficiency of the diffraction order q, which is defined by the comb function, is found to be I q sinc (q 1). (18) Note that the lateral shift of the grating does not make any significant change in the intensity of the Fraunhofer diffraction pattern; i.e., if the grating is shifted by / laterally, then the comple transparency would be [ ( ) ] ( ) iπ t() =rect e comb. (19) 10

11 (a) (b) (c) Figure 5: The Fraunhofer diffraction pattern Using the same procedure, we obtain the same result as {[ ( ( [ ]) iπ ) ] ( iπ F e rect e )} comb = e iπ sinc u 1 comb (u) (0) and I q sinc (q 1). (1) 3.b) If the phase contrast is reduced from π to φ 0, then the comple transparency of the new grating is written as [ ( ) ] ( ) t() = rect e iφ 0 comb. () Using the same procedure, we obtain { ( ) } { ( )} F {t()} = F rect e iφ 0 F comb = {[ ] [ ( )]} { } iπ u φ 0 iπ u sinc (u) e δ u comb (u) e = π { ( [ ]) } sinc u φ 0 e iπ [u φ 0 ] π comb (u) e iπ u = π { ( [ ])} e iπ[u φ 4π 0 ] sinc u φ 0 comb (u). (3) π 11

12 Hence, the efficiency of the diffraction order q is proportional to ( ) I q sinc q φ 0. (4) π Note that now we have an infinite number of diffraction orders, and the efficiencies can be tuned by changing the phase modulation. The Fourier series approach should epect the same result. 3.c) Since a prism ehibits refraction, all incoming light bends based on the Snell s law. The prism does not produce additional diffraction orders. Also the dispersion of the prism is normal dispersion, in which long wavelength bends less than short wavelength, as shown in Fig. 6(a). The blazed grating, even though its shape of the profile is similar to the prism, it ehibits diffraction and produces multiple diffraction orders. The zero order diffraction keeps propagating along the same direction as the incident wave but no dispersion happens, Other higher orders, whose efficiency is dependent on the modulation of the phase (in this problem, φ 0 ), are p in the same way of the incident light ecept the amplitude attenuation. This phase grating produces an infinite numbers of diffraction, and the efficiency of the diffraction order depends on the modulation of the phase (in here, φ 0 ). The dispersion of the grating is anormal, in which longer wavelength diffracts more than short wavelength. Another way to understand is that since u = /(λz). Thus, although spatial frequencies are same, they appear at different location due to = λzu, where longer wavelength length deviates more. (a) Prism (b) Blazed grating Figure 6: Comparison of a prism and blazed grating 1

13 6. The grating can be epressed as a convolution of a rect() function and a comb() function, where the comb() is an infinite sum of equally spaced δ() functions. The truncated grating is the product of a D rect() function (the window) and the infinite grating. Upright grating: + n=+ δ(y n) rect(y/d) From problem #, we know that this convolution results in: sin(n π d ) δ(u n π n=, v) }{{ n } sinc envelope A rotation by 60 in space results in a rotation by 60 in the frequency domain: u = u cos 60 + v sin 60, v = u sin 60 + v cos 60 ( ) π d sin(n ) 1 3 π 3 1 δ n u + v n, u + v = F 1(u, v) n= The Fourier transform of the truncated grating is the convolution of F 1 with the Fourier transform of the rectangular window: F (u, v) = F 1 (u, v) ab sinc(au)sinc(bv) 13

14 The Fraunhofer diffraction pattern is Sketches of Fourier Transforms iπ e l λ [F (u, v)] iλl (, y ) λl λl (a) Infinite grating (b) Rotated infinite grating (c) Rectangular aperture (d) Truncated grating (b c) 14

15 MIT OpenCourseWare /.710 Optics Spring 009 For information about citing these materials or our Terms of Use, visit: