Optics Quiz #2 April 30, 2014

Transcription

1 .71 - Optics Quiz # April 3, 14 Problem 1. Billet s Split Lens Setup I the ield L(x) is placed against a lens with ocal length and pupil unction P(x), the ield (X) on the X-axis placed a distance behind the lens is given by Eq. (5-14) in Goodman (modiied to be 1D and I have reinserted the constant phase actor due to propagation over in the z direction): k exp[ k ]exp X k ( X ) (x)px exp L xx dx Eq. 1 Part (a) Lens L (ocal length ) is being illuminated by a spherical wave originating at location z = -. Thereore, the ield L(x) incident on L is (using the paraxial approximation): exp[ k(z )] x L (x) E(x, z ) E exp k (z ) (z ) z exp[ k ] x L (x) E exp k The lens is partially obstructed by a block o width (). Thereore, the ield pattern (X) at the CCD detector can be viewed as ollows. Let (X) be the ield pattern that would result i the lens L was o ininite extent (with no obstructing block). Meanwhile, let (X) be the ield pattern that would result i the lens L had inite extent (again, with no obstructing block). (X) is essentially the ield pattern blocked by the opaque block. Thus, by linearity, (X) will be (X) subtracted rom (X): (X ) ( X ) ( X ) Eq. e can use Eq. 1 to calculate (X) and (X) individually. L(x) will be the same or each calculation. However, the pupil unction P(x) will dier as ollows: P (x) 1 Lens with ininite ( ) extent x P (x) rect Lens with inite extent Page 1 o 17

2 .71 - Optics Quiz # April 3, 14 To calculate (X), we plug in L(x) and P (x) into Eq. 1: k k exp exp X exp[ k ] x ( X ) E exp k exp k k exp k exp X x ( X ) E F exp k X / k exp k exp X 1 ( X ) E F exp x k exp k exp X ( X ) E exp X k expk exp X k ( X ) E exp X exp k ( X ) E X X / X X X / xx dx As expected or a spherical wave originating rom the ront ocal point o a lens, (X) is a plane wave. To calculate (X), we plug in L(x) and P(x) into Eq. 1: k exp k exp X exp[ k ] x x k ( X ) xx dx E exp k rect exp k exp k exp X x x k ( X ) E exp k rect exp xx dx I the irst exponential inside the integral is suiciently constant (i.e., = 1) over the interval o the rect unction (i.e., analogous to the Fraunhoer condition > /λ), then it can be neglected: k exp k exp X x k ( X ) E rect exp xx dx Page o 17

3 .71 - Optics Quiz # April 3, 14 k exp k exp X x ( X ) E F rect k exp k exp X ( X ) E sinc X k exp k exp X sin X ( X ) E X X X / k exp k exp X exp X exp X ( X ) E X k k k exp k exp X exp X exp X ( X ) E X exp k ( X ) E k k exp (X X ) exp ( X X ) X k k exp X X exp X X exp k k 4 4 ( X ) E exp 4 X k k exp X exp X exp k k (X ) E exp 4 X To ind the ield pattern (X) at the CCD detector, we plug (X) and (X) into Eq. : k k k exp exp X exp X exp k 1 4 (X ) E X Page 3 o 17

4 .71 - Optics Quiz # April 3, 14 Parts (b) and (c) In the expression or (X), all three terms have the orm o a spherical wave. Thus, (X) essentially represents the coherent superposition o three spherical waves as ollows: 1 st term: a plane wave, equivalent to a diverging spherical wave originating at (x, z) = (, - ), i.e., on-axis at ininity (to the let) nd term: a diverging spherical wave originating at (x, z) = (-/, ), i.e., the bottom edge o the opaque block, with a phase actor and a 1/X attenuation actor 3 rd term: a diverging spherical wave originating at (x, z) = (+/, ), i.e., the top edge o the opaque block, with a phase actor and a 1/X attenuation actor For each pair o waves, bright and dark ringes will occur when the phase dierence (δ) between the pair is a multiple m and (m+.5) o π, respectively: 1 st and nd waves (Plane & Spherical): bright ringes occur at Xm with increasing separation with m k m 1 1 X X m m st and 3 rd waves (Plane & Spherical): bright ringes occur at Xm with increasing separation with m k m X X m m 4 4 nd and 3 rd waves (Spherical & Spherical): bright ringes occur at Xm with constant separation with m (as in Young s double slit experiment with slit separation ) k m m 3 3 X X X m The total irradiance I on the CCD detector will be the aggregate o these ringe patterns, weighted by the irradiances (I1, I, and I3) o each wave. Note that I and I3 decrease with increasing X, due to the 1/X attenuation actor. Page 4 o 17

5 .71 - Optics Quiz # April 3, 14 Parts (d) I the illumination were moved o axis by a distance o x, then the ield L(x, y) incident on the lens L would become: exp[ k ] (x x ) L (x) E exp k According to the Shit Theorem, (X) (rom Part a) would become: exp k k ( X ) E exp Xx (X) (rom Part a) would become (plugging P(x) and the new L(x) into Eq. 1): k exp k exp X exp[ k ] (x x ) x k ( X ) E exp k rect exp xx dx k exp k exp X x xx x x k ( X ) E exp k exp k exp k rect exp xx dx k exp k exp ( X x ) x x k (X ) E exp k rect exp x( X x ) dx Once again, i the irst exponential inside the integral is suiciently constant (i.e., = 1) over the interval o the rect unction (i.e., analogous to the Fraunhoer condition > /λ), then it can be neglected as ollows: k exp k exp ( X x ) x ( X ) E F rect X ( X x ) / k exp k exp ( X x ) ( X ) E sinc ( X x ) k exp k exp ( X x ) sin ( X x ) ( X ) E ( X x ) Page 5 o 17

6 .71 - Optics Quiz # April 3, 14 k exp k exp (X x ) exp (X x ) exp (X x) ( X ) E (Xx ) k k k expk exp (X x ) exp ( X x ) exp ( X x) ( X ) E Xx k k exp (X X x ) exp (X X x exp k k ( X ) E exp x Xx exp k k ( X ) E exp x 4 k k k k exp x exp X X exp x exp X X 4 4 X x exp k k ( X ) E exp x 4 k k k k exp x exp X exp x exp X Xx To ind the ield pattern (X) at the CCD detector, we plug (X) and (X) into Eq. : exp k ( X ) E x k k k k exp k X exp x exp X exp x exp X k exp x 4 X x ) The ield (X) is still a coherent superposition o three spherical waves: 1 st term: a plane wave propagating at angle x/, equivalent to a diverging spherical wave originating o-axis at ininity (to the let) nd term: a diverging spherical wave originating at (x, z) = (-/, ), i.e., the bottom edge o the opaque block, with a phase actor and a 1/(X+x) attenuation actor 3 rd term: a diverging spherical wave originating at (x, z) = (+/, ), i.e., the top edge o the opaque block, with a phase actor and a 1/(X+x) attenuation actor Page 6 o 17

7 .71 - Optics Quiz # April 3, 14 Relative to Part (a), the (X) here has changed as ollows: 1 st term: this term is still a plane wave, but is now propagating at an angle o -x/ nd term: this term is still a diverging spherical wave originating at (x, z) = (-/, ), but now has an additional phase actor o exp[kx/] and a 1/(X+x) attenuation actor (instead o ust 1/X) 3 rd term: this term is still a diverging spherical wave originating at (x, z) = (+/, ), but now has an additional phase actor o exp[-kx/] and a 1/(X+x) attenuation actor (instead o ust 1/X) For each pair o waves, bright and dark ringes will occur when the phase dierence (δ) between the pair is a multiple m and (m+.5) o π, respectively: 1 st and nd waves: bright ringes occur at Xm with increasing separation with m k 1 1 m x X x Xx 4 X m m x k 4 m X x 4 1 st and 3 rd waves: bright ringes occur at Xm with increasing separation with m k m x X x Xx X m m x 4 4 nd and 3 rd waves: bright ringes occur at Xm with constant separation with m k m m 3 3 X X x Xm x Thus, relative to Part (a), the bright (and dark) ringes on the CCD detector will be shited by a distance o -x, i.e. in the opposite direction o the source shit. The attenuation on I and I3 will also be shited by -x, due to the 1/(X+x) attenuation actor (instead o ust 1/X in Part a). Page 7 o 17

8 .71 - Optics Quiz # April 3, 14 Appendix to Problem 1 My answers above were contingent on the assumption that > /λ, which allowed or easy calculation o ( X), i.e., the ield pattern block by the opaque block. To be more clear, I will now show what would happen i I did not make that assumption. Part (a) ithout assuming > /λ, L( X) does not change but ( X) must be recalculated. Plugging L P ( X) and the original ( X) into Eq. 1, the irst exponential inside the integral cannot be neglected now: k exp k exp X x x k ( X ) E exp k rect exp k exp k exp X x x ( X ) E F exp k rect X / k exp k exp X x x ( X ) E F exp k F rect k exp k exp X ( X ) E exp X sinc( x ) k exp k exp X k exp sinc ( X ) E X X x xx dx x X / X / Note that we are let with a convolution o a converging spherical wave with a sinc, rather than a sinc by itsel (as in my Part (a) previously). To ind the new ield pattern (X) at the CCD detector, we plug (X) and (X) into Eq. : k X exp k exp 1 k exp sinc ( X ) E E X X x Page 8 o 17

9 .71 - Optics Quiz # April 3, 14 Part (b) and (c) ithout assuming > /λ, the expression or (X) can no longer be viewed as a superposition o three spherical waves. The irst time is still a plane wave, i.e., a plane wave originating at ininity. However, the second term is now the product o a diverging spherical wave and the convolution o a converging spherical wave with a sinc. It is no longer ruitul to expand the sinc into complex exponentials in hope o creating two spherical waves (as in my Part (a) previously). Instead, (X) is the superposition o a plane wave (as beore) and the Fresnel diraction pattern o a rectangular aperture with plane wave incidence (not two spherical waves). Part (d Despite the changes to Parts a, b, and c, the answer here does not change. I the illumination were moved o axis by a distance o x, then ( X) would still shit a distance o x. (X) can be written as: k expk exp (X x ) k ( X ) E exp ( X x ) k exp k exp ( X x ) ( X ) E exp ( X x ) sinc X (X) can be written as: k By observation, both o these terms are simply shited versions (plus some phase change) o the x = case. Thus, the combined (X) will also be shited. Page 9 o 17

10 .71 - Optics Quiz # April 3, 14 Problem. 4F Imaging System Setup In a 4F imaging system, the primary ields o interest are: input(x, y): the ield pattern at the x-y input plane pupil(x, Y): the ield pattern incident on the X-Y pupil plane pupil(x, Y): the ield pattern immediately behind the X-Y pupil plane (x, y ): the ield pattern at the x -y plane The ield pattern input(x, y) is simply the input image (to be speciied in Parts a, b, and c). I ield pattern input(x, y) is placed a distance d = in ront o L1 with ocal length and pupil unction P(X, Y) = 1 (i.e., neglecting the inite extent o L1), the ield pattern pupil(x, Y) on the X-Y pupil plane placed a distance behind L1 is given by Eq. (5-19) in Goodman (Note: I have reinserted the constant phase actor due to propagation over in the z direction): k e k pupil ( X,Y ) input (x, y)exp (xx xy ) dxdy k e pupil ( X,Y ) F input (x, y) x X / y Y / k e X Y pupil ( X,Y ) ˆ input, Thus, pupil(x, Y) is simply the Fourier transorm o the input ield. Meanwhile, the pupil plane has a mask with amplitude transmission unction t(x, Y). I pupil(x, Y) is incident on the pupil plane, then the ield pattern pupil(x, Y) immediately behind the pupil plane is the product o pupil(x, Y) and t(x, Y): pupil (X,Y ) pupil ( X,Y )t( X,Y ) I ield pattern pupil(x, Y) is placed a distance d = in ront o L with ocal length and pupil unction P(X, Y) = 1 (i.e., neglecting the inite extent o L), the ield pattern (x, y ) on the x -y outuput plane placed a distance behind L is given Eq. (5-19) in Goodman (Note: I have reinserted the constant phase actor due to propagation over in the z direction): e k (x', y') pupil ( X,Y )exp (x' X y'y ) dxdy k Page 1 o 17

11 .71 - Optics Quiz # April 3, 14 k e k (x', y' ) pupil ( X,Y )t( X,Y ) exp (x' X y'y ) dxdy e X Y k (x', y' ) ˆ input, t( X,Y ) exp (x' X y'y ) dxdy e X Y (x', y' ) F ˆ input, t( X,Y ) X x'/ y'/ Y Eq. 1 e X Y (x', y' ) F ˆ input, Ft(X,Y ) X x'/ y'/ Y e input X Y X Y X Y y'/ (x', y' ), tˆ(, ) x'/ e x' y' input (x', y' ) x',y' tˆ, Eq. e (x', y' ) input x', y' PSF x', y' Thus, the ield (x, y ) can be calculated in two equivalent ways (whichever is more convenient or the given input ield): Eq. 1: (x, y ) is the Fourier transorm o the product o the Fourier transorm ˆ input ( X /,Y / ) o the input image and the amplitude transmission unction t(x, Y) o the mask. Eq. : (x, y ) is the convolution o the inverted input image input(-x, -y ) and the Fourier transorm tˆ(x' /, y' / ) o the mask s amplitude transmission unction ((i.e., the point spread unction PSF(x, y )). In either case, the intensity pattern can then be calculated as ollows: I ( x', y' ) ( x', y' ) Eq. 3 Page 11 o 17

12 .71 - Optics Quiz # April 3, 14 Part (a) In this case, the ield (x, y ) is most easily ound using Eq. 1. I the input plane is illuminated by a coherent plane wave, the input ield input(x, y) is a constant and its Fourier transorm ˆ (, ) is a δ unction: input X Y input (x, y) 1 ˆ input (, Y )F1 (, y ) Meanwhile, the pupil plane s mask is a rectangular grid o squares, speciically a a x b (Y-width by X-height) rectangle o (Λ d) x (Λ d) squares repeating every Λ in each direction. Thereore, t(x, Y) is the convolution o the square with a comb unction bounded by a rectangle: X Y X Y X rect Y t(x,y )rect rect comb comb rect d d b a Note the grid has a transparent square at the center, such that t(, ) = 1. Plugging ˆ input ( X ) and t(x, Y) into Eq. 1 X x e X Y ˆ ( x ', y ' ) F, t( X,Y ) input X x'/ Y y' / e X Y ( x ', y ' ) F, t( X,Y ) X x'/ e X Y ( x ', y ' ), t(,) F Y y' / X x'/ Y y' / e X Y ( x ', y ' ) F, since t(,) 1 X x'/ (x', y' ) e e ( x ', y' ) e (x', y' ) X x'/ Y y' / Y y' / Page 1 o 17

13 .71 - Optics Quiz # April 3, 14 By Eq. 3, the intensity pattern I(x, y ) is: I (x', y' ) ( x', y' ) 1 Putting the results together: e (x', y' ) I ( x', y' ) 1 Thus, the ield (x, y ) is a plane wave (with some phase delay) and the intensity pattern I(x, y ) is a constant. This result is easily explained. The input ield is a plane wave, i.e., with only a DC component. Because the mask is transparent at the point (X, Y) = (, ) (via the center square), the DC component o the input is passed by the mask. Thereore, the entire input is preserved and a plane wave appears again at the. Below, I have plotted the intensity pattern (generated in Matlab), which is simply a constant background. I(x, y ) Page 13 o 17

14 .71 - Optics Quiz # April 3, 14 Part (b) In this case, the ield (x, y ) is most easily ound using Eq.. I a coherent point source is placed at the origin o the input plane, the input ield input(x, y) is a δ unction. input ( x, y) ( x, y) The Fourier transorm tˆ( X ) o the mask t(x, Y) can be calculated as ollows: tˆ( X ) Ft(X,Y ) X Y X Y X Y tˆ( X ) F rect rect comb comb rect rect d d b a X Y X Y X Y tˆ( X ) F rect rect F comb comb F rect rect d d b a tˆ( X ) ( d) sinc( d) X sinc( d) Y comb X comb Y absincb X sinca Y tˆ( X ) ab ( d) sinc( d) X sinc( d) Y ( X m)( Y n) sincb X sinca Y m n m n tˆ( X ) ab ( d) sinc( d) X sinc( d) Y X Y sincb X sinca Y m n m n tˆ( X ) ab ( d) sinc( d) X sinc( d ) Y sinc b X sinc a Y m n m n tˆ( X ) ab ( d) sinc( d) X sinc( d ) Y sinc b X sinc a Y m n m n tˆ( X ) sinc( d ) X sinc( d ) Y sinc b X sinc a Y m n where the constant actor abλ (Λ - d) has been dropped or simplicity in the last step above. Thus, tˆ( X ) is a grid o skinny D sinc unctions modulated by a wider D sinc unction envelope. Plugging input(x, y) and tˆ( X ) into Eq. : e x' y' input (x', y') x', y' tˆ, e x' y' (x', y') x', y' tˆ, Page 14 o 17

15 .71 - Optics Quiz # April 3, 14 e x' y' (x', y') tˆ, e ( d ) ( d ) x' m y' n (x', y') sinc x' sinc y' sincb sinca m n e ( d ) ( d ) b b (x', y') sinc x' sinc y' sinc x' m sinc y' n m n Plugging in Λ = 1 μm, d = μm, a = 5mm, b = 3 mm, λ =.5 μm, and = 1 cm: e 8μm 8μm (x', y') sinc x' sinc y'.5μm 1cm.5μm 1cm 3mm.5μm 1cm 5mm.5μm 1cm sinc x' msinc y' n m n.5μm 1cm 1μm.5μm 1cm 1μm e 8μm 8μm (x', y') sinc x' sinc y'.5μm 1mm.5μm 1mm 3μm.5μm 1mm 5μm.5μm 1mm sinc x' msinc y' n m n.5μm 1mm 1μm.5μm 1mm 1μm e -1-1 (x', y') sinc.16mm x'sinc.16mm y' -1-1 sinc6mm x'5mm msinc1mm y'5mm n m n Thus, the ield (x, y ) is a grid o skinny D sinc unctions separated by 5 mm in each direction and modulated by a wider D sinc unction envelope. Each skinny sinc unction has a main lobe with a width o Δx = / 6 mm -1 =.33 mm in the (vertical) x -direction and Δy = / 1mm -1 =. mm in the (horizontal) y -direction. The wider sinc unction envelope has a main lobe with a width o Δx = Δy = /.16 mm -1 = 1.5 mm. The intensity pattern I(x, y ) can be calculated by Eq. 3. Since each sinc unction is reasonably ar rom all neighboring sinc unctions relative to their widths (5 mm >.33 mm), let s assume or simplicity that there is negligible overlap among the sinc unctions in the intensity calculation: I -1-1 (x', y' ) sinc.16mm x'sinc.16mm y' -1-1 sinc 6mm x'5mm msinc 1mm y'5mm n m n Page 15 o 17

16 .71 - Optics Quiz # April 3, 14 e (x', y' ) sinc.16x'sinc.16y' sinc6x'5msinc 1y'5n m n I (x', y' ) sinc.16x'sinc.16y' sinc 6x'5msinc 1y'5n m n where x and y are in units o millimeters (mm). Because the input ield was a point source, the ield (x, y ) and intensity I(x, y ) here are equivalent to the coherent and incoherent PSF(x, y ), respectively, o this 4F imaging system. Below, I have plotted the intensity pattern (generated in Matlab) on a log1 scale, which makes it easy to see the grid o sinc unctions repeating every 5 mm in each direction. For more clarity, the bottom igures show the horizontal (@ x = ) and vertical (@ y = ) cross sections o the intensity pattern. In the zoomed out igure, one can easily see the amplitude o the skinny sinc unctions decreasing outward rom the center, due to the wider sinc envelope. In the zoomed in igure, the vertical cross section is broader than the horizontal cross section because the mask is wider than it is tall, i.e., a > b. log 1 [I(x, y )] Cross Sections Cross Sections (Zoomed into Central Sinc) Page 16 o 17

17 .71 - Optics Quiz # April 3, 14 Part (c) In this case, the ield (x, y ) is most easily ound using Eq.. I the transparent stamp o Tim the beaver is illuminated by a coherent plane wave, the input ield input(x, y) is a equivalent to the (amplitude transmission o the) stamp itsel: input ( x, y) Tim ( x, y) Plugging input(x, y) and tˆ( X ) (calculated in Part b) into Eq. : e (x', y') Tim x', y' sinc.16x'sinc.16 y' sinc6x'5msinc1y'5n m n I (x', y') (x', y') where x and y are in units o millimeters (mm). The ield (x, y ) is the convolution o the Tim stamp Tim(x, y ) and the coherent PSF. The result is a grid o Tim stamps repeating every 5 mm in each direction and modulated by a sinc unction envelope. Each Tim stamp is blurry due to convolution with a low-pass ilter, i.e., the skinny sinc unction repeated in each direction in the PSF. Below, I have plotted the intensity pattern (generated in Matlab). The let igure is plotted on a linear scale. On this scale, only the central Tim is clearly visible, due to the sinc envelope making the neighboring Tims considerably darker. For more clarity, the right igure is plotted on a log1 scale. On this scale, the grid o Tims, repeating every 5 mm in each direction, is clearly visible. I(x, y ) log 1 [I(x, y ]) Page 17 o 17

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