Problem μ Max Min

Transcription

1 SIMG-733 Optics-9 Solutions to Midterm Eam Statistics and Comments: Man people are still not sketching before writing down equations. Some wrote down equations with no eplanation, e.g., the depth of field equation, which is the what rather than the (more important) wh. I advise that it tpicall does little good to memorize the equations ou should get a feeling about the graphical situation, which often gives ou the equation. The graphical solution likel will stick with ou much longer than the memorized equation(s). Amended Histogram (after updated grades) Average per problem (each out of possible ): Problem μ Ma 5 8 Min This indicates that the greatest difficult was with problem # (on depth of field), where, again, few people drew a good sketch of the situation. Problems 3 (complementar aperture),4 (aperture function to generate derivative), and 5 (diffraction grating) gave the net most difficult. Note that the maimum score of was attained b someone on all problems ecept # and #4.

2 . In the homework, ou demonstrated that the cascade of propagation of light from a -D function f [, ] over the distance z in the Fresnel region, followed b second propagation over the distance z in the Fresnel region ields the same result as a propagation of light from f [, ]over the distance z + z. Consider the -D case, i.e., consider propagation of light from the -D function f [] over the distances z and z. Is this propagation equivalent to that over the distance z + z?eplain. The doubl diffracted amplitude pattern in -D ma be evaluated via either a convolution over the distance z + z or over the same distance in two steps. If the first, the impulseresponseis: h [, ; z + z ] iλ (z + z ) ep +πi z + z ep +iπ + λ λ (z + z ) If the second, the impulse response is the convolution: Ã! Ã e +πi z λ e +iπ + λ z e +πi z λ e +iπ iλ z i e +πi z +z +iπ λ λ e z iλ z z! + +iπ λ e z e +iπ λ z +iπ λ e z You showed that these were equal. But if the problem is recast in -D where the impulse response is the -D analogue: h [; z ] ep +πi z ep +iπ iλ z λ λ z h [; z ] ep +πi z ep +iπ i λ so that we must check to see if: h [; z ] h [; z ] h [, ; z + z ] iλ (z + z ) ep +πi (z + z ) λ ep +iπ λ (z + z ) ep +πi z iλ z ep iλ z i λ ep +iπ λ z +πi (z + z ) λ ep ep +πi z i λ +iπ ep λ z +iπ ep +iπ ¾ F ½ep +iπ λ z ¾ F ½ep +iπ p h λ z ep +i π i ep iπλ z ξ 4 p h ep +i π i ep iπ ξ 4

3 ep +iπ F ep +iπ i λ z np h λ z ep +i π 4 ep iπλ z ξ pλ z ep h +i π i ep iπ ξ o 4 n F h λ z z ep +i π i ep iπλ (z + z ) ξ o (iλ z z ) F ep iπλ (z + z ) ξ ª (iλ z z ) r iλ z z ep z + z h p λ (z + z ) ep i π 4 +iπ λ (z + z ) 6 i ep +iπ λ (z + z ) +iπ iλ (z + z ) ep λ (z + z ) The Fourier transform in the second dimension provides the necessar scale factors to obtain the correct result. 3

4 . Images of objects at different distances from the optic are created b a lens with a certain focal length f 5mmand aperture diameter d 35mm. The lens is used in monochromatic light with wavelength λ.5 μm. The lens is used to image an object with depth, i.e., its distance from the lens ranges from (z ) far to (z ) near. Alternativel, ou could consider several objects in the field of view at distances in this interval. (a) Eplain the concept of depth of field for this lens, use diagrams. The f/number of this lens is: f/# f 5 mm d 35 mm 7.4 (the prescription was modeled on the Nikon 5mm f/.4 normal lens for 35mm cameras If the lens is used wide open, then the cone of light eiting the lens towards the image plane spans a full angle of: 35 mm 5 θ tan tan 5 mm 7 B using our rule of thumb, we estimate that the diameter of the Air disk pattern in visible light is approimatel the f/# measured in micrometers: D.4 μm D ~.4μm 35mm Δθ 5mm 4

5 (b) Eplain the effect of stopping down the lens (reducing the diameter of the aperture while maintaining the image distance). D ~ 3μm 35mm 7mm Δθ Depth of Focus 5mm Stopping down the lens both increases the diameter of the Air disk and decreases the cone angle of the light from the lens to the image. In this wa, the depth of focus is increased b two factors of the f/#. (c) Describe the effect of stopping down the lens on the qualit of images obtained of an object at the distance z that eactl satisfies the imaging equation for a fied focal length f and fied image distance z ; in other words, what are the relative qualities of the images before and after stopping down the lens. Since the lens is stopped down to create a smaller aperture diameter, the diameter of the Air disk is increased, so the entire image is blurrier, but the range of distances that appear equall blurrier is now larger 5

6 3. Determine the forms of the Fraunhofer diffraction patterns for an aperture function consisting of transparent circular holes is similar to that for the complementar aperture, i.e., replace the transparent holes with opaque circular spots and opaque background with a transparent background. I ll just do the general case; we need to evaluate the patterns resulting from two two aperture functions f [, ] where f and f [, ] f [, ]. The Fraunhofer diffraction patterns are proportional to the Fourier transforms of the aperture functions: F {f [, ]} ξ,η F, F {f [, ]} ξ,η F {[, ] f [, ]} λ z ξ,η λ z δ, F, The observed irradiance in each case is proportional to the squared magnitude: I [, ], F I [, ], F δ, F, δ, δ, F, + F, + F, δ, δ, F, + F, + F, δ, ( Re {F [, ]}) + F, So the two irradiance patterns are identical ecept at the origin of coordinates! In the specific problem, ou can write the aperture function as, sa: f [, ] COMB, r CY L d F, COMB p λ z λ z, ³ π 4 SOMB + ³, F COMB p λ z, ³ π 4 SOMB + ³ 6

7 F, f [, ] [, ], F δ, δ, δ, πd 4 λ z, COMB COMB, ³ COMB + COMB SOMB ³ Ã πd! 4 + COMB, ³ λ z λ z λ z δ,, ³, ³ δ r CY L p π 4 SOMB + ³ p π 4 SOMB + ³ p π 4 SOMB + ³, p π 4 SOMB + ³ 7

8 4. Construct an aperture function p [, ] that, if acting on coherent (monochromatic) light will produce an output amplitude g [, ] of the form: g [, ] f [, ] Describe how a practical such aperture ma be constructed from glass, etc., and also describe an limitations in the success of the result. We know that the aperture function must be proportional to the transfer function of the desired impulse response in coherent light. Since the impulse response is a derivative, the transfer function must be: H [ξ,η] F [δ [] δ []] +πiξ [η] p [, ] +πi +πi [] So we need the pupil function to induce an amplitude function that is proportional to +πi π ep +i π on the incoming light. It is helpful to plot the function to be generated, which I ll do as a -D function of : -ais profile of aperture function to evaluate the -D derivative; the function induces an increase in the magnitude proportional to the distance from the origin and a change in phase of π radians in the domain from negative to positive coordinates. Note that the magnitude cannot eceed unit (maimum transmittance) and the aperture width must be finite. So we must var the magnitude from at the origin to unit at the edge of the aperture 8

9 and have a phse change of π radians as ou pass from negative to positive coordinates. We create the variation in magnitude b apodizing the aperture (varing its transmittance) b using a photographic transparenc. Note that the maimum magnitude is limited to unit (since this is a passive sstem that cannot inject power) and that the aperture size must be constrained to satisf the Fraunhofer condition. The phase change of π radians requires an additional phase dela on one side of the origin, which ma be implemented via a thin piece of glass (which has practical issues). Perhaps contrar to our intuition, this is a practical device. 9

10 5. Light with wavelength λ propagates from a point source located on ais at a ver large distance to the left of the plane z, i.e., at some coordinate z. The wavefronts from the source observed at the origin are appropriatedl modeled as plane waves; assume that the amplitude of the sinusoidal oscillations is. The aperture at the plane z is a square with side dimension of 5 mm. Within the square, the aperture function consists of alternating transmitting and opaque regions with equal widths of μm. The light transmitted through the aperture then traverses a large distance z to the observation plane in the Fraunhofer region. (a) Write down an equation for and sketch t [, ]; label relevant quantities. + X n t [, ] RECT [ n] RECT [] δ [ n] RECT [ n] + X n RECT [] RECT [] RECT [] (RECT [] δ [ n]) + X n + X n + X n δ [ n] RECT [] COMB h h ³ n i δ h i δ n i t [, ; z ] μm RECT μm COMB μm ³ h RECT 5 mm i h RECT This is a square-wave grating with about 5 bars per mm along the -direction. The drawing shows an approimation (though with fewer than number of bars) 5 mm i This ma look familiar it is a mathematical model of a diffraction grating. The (hidden) goal of the rest of the problem is to show that the grating disperses white light.

11 (b) Derive the epression for the amplitude pattern observed at the distance z from the aperture. Include relevant numerical factors in the epression. You need not sketch this, but it ma help to do so. g [, ; z ] ep +πi z F {t [, ]} i λ ξ,η ep +πi z T, i λ λ z K T, t [, ] μm RECT μm COMB μm ³ h RECT 5 mm i h RECT T [ξ,η] ( μm SINC [ μm ξ] μm COMB [ μm ξ]) 4 μm (5 mm) SINC [5 mm ξ] SINC [5 mm η] g [, ] T, λ z 5μm SINC μm COMB μm λ z (5 mm) SINC 5 mm SINC 5 mm which is a wide SINC function along the -direction modulated b a COMB function whose separation is half as wide convolved with a ver narrow -D SINC function; the scale factor for the narrow SINC is smaller than that of the wide SINC function b the factor of: μm mm so there are 5 narrow SINCs within the width parameter of the wide SINC; therefore we approimate the narrow SINC function as a Dirac delta function: g [, ] T, λ z 5μm SINC μm COMB μm λ z + X SINC μm δ ³ k k μm SINC ³ λ z X + δ k μm μm μm k 5 mm i

12 which is a SINC function multiplied b a series of Dirac delta functions. Note the wavelength dependence in the translation parameter in the Dirac delta function the translation gets larger as λ increases. In words, the output is a modulated COMB function whose amplitude decreases with increasing radial distance in the direction. The even orders of the COMB vanish under the action of the SINC. The elements of the COMB represent the diffracted light at those positions. (c) Evaluate AND SKETCH the irradiance pattern at the same location. The irradiance is the squared magnitude: I [, ] g [, ] SINC ³ μm + X k δ k μm (a) approation of transmittance function t []; (b) approimation of irradiance pattern for a single incident wavelength. The displacement of the first order is proportional to the wavelength, which is the working mechanism for diffraction gratings. (d) Describe in words AND SKETCH the irradiance pattern that would be observed if the point source radiates the same amplitude at different wavelengths λ spanning the entire visible spectrum. If multiple wavelengths, ou get different Dirac delta function (or, more accuratel, approimations thereto) for each wavelength that are displaced from the origin b distances proportional to λ; each appears in the color of its wavelength, so ou get the spectrum of the white light.

13 6. An object f [, ] is illuminated b a monochromatic plane wave with wavelength λ. Immediatel after the object is a lens with focal length f whose diameter is sufficientl large so that all light transmitted through the object passes through the lens; in other words, the lens is in contact with the object. The light propagates a distance z f, where it encounters an identical lens with focal length f. The observation plane immediatel follows (is in contact with) the lens. (a) Sketch the sstem. f (b) Determine the amplitude and irradiance patterns at the observation plane in terms of the parameters of the sstem and f [, ]. The first lens imposes a quadratic-phase factor on the light from the object transparenc f [, ] g [, ] f [, ] p [, ]ep iπ + +πi i ep fz ep +iπ + p [, ]ep iπ + but the lens is assumed large enough so that p [, ] [, ] g [, ] +πi i ep fz f [, ] ep iπ + ep +iπ + ep iπ + 3

14 For simplicit, consider in -D: g [] f [] ep iπ ep +iπ ep iπ Z + " # ep iπ f [α]ep iπ α ( α) ep +iπ dα Z + ep iπ f [α]ep iπ α ep +iπ + α α dα Z + ep iπ ep +iπ f [α] ep iπ α ep +iπ α ep iπ α dα Z + f [α] () ep iπ α dα Z + f [α] ep πi α dα F [ξ] ξ F This is an M-C-M chirp Fourier transform, so we can see that the -D output mabewrittenintheform: g [, ] +πi i ep fz F, F, 4

15 7. Consider two monochromatic point sources emit radiation with wavelength λ and the same phase. The sources are separated b distance along the -ais. Light emitted b from one of the sources immediatel passes through a pane of glass with refractive inde n,thickness, and plane-parallel sides. The thickness is such that to the light emerging from the glass is eactl out of phase relative to the light from the other source. Light from both sources then propagates a distance z to the observation plane. (a) Determine the thickness of the glass and comment on the practicalit of using glass to induce the phase dela. The optical thickness of the glass is n. The term out of phase means that the phase difference is π radians (NOT π as several respondents did). This means that the phase of the light eiting the glass must be π radians: φ [z,n ] π π n λ n λ λ n The optical thickness of the glass must be half a wavelength: n λ λ n If n.5 (as we usuall assume for glass), then the thickness is: λ.5 λ μm μm which is too thin to be practical, but the principle is sound. (b) Determine the irradiance pattern observed if the distance z is in the Fresnel diffraction region. The source function is a pair of Dirac delta functions with a phase shift applied to one: f [, ] δ, d + δ, + d ep [+iπ] δ, d δ, + d 5

16 The amplitude in the Fresnel diffraction region is obtained b propagating: g [, ; z ] f [, ; z,λ ] ep +πi z ep +iπ + i λ K f [, ; z,λ ] ep +iπ + where K ep +πi z i λ K δ, d δ, + d ep +iπ + K ep +iπ Ã " # " ep +iπ + #! ep +iπ K ep +iπ Ã " ep +iπ + # " d ep +iπ + #! + d " K ep +iπ + d # ep +iπ ep iπ d ep +iπ d " i K ep +iπ + d # ep +iπ sin π d The irradiance is proportional to the squared magnitude: I [, ; z.λ ] " i K ep +iπ + ep +iπ 4 λ sin π d z Recall : sin [θ] ( cos [θ]) I [, ; z.λ ] 4 λ cos π ³ z I [, ; z.λ ] λ z cos π ³ d # sin π which is a biased cosine pattern with a minimum at the origin and a period: D λ z 6

17 (c) Determine the irradiance pattern observed if the distance z is in the Fraunhofer diffraction region. In the Fraunhofer region, the observed pattern is proportional to the squared magnitude of the -D Fourier transform of the source function: F F,, I [, ; z,λ ] ep +πi z i λ F, f [, ] δ, d δ, + d F [ξ,η] [ξ] ep πi η d ( ) i [ξ] sin [π ξ] i sin π 4 sin π ³ D I [, ; z,λ ] λ z cos π cos π [ξ] ep i ³ λ z +πi η d sin π (d) (OPTIONAL BONUS) Describe how ou can create a practical pane of glass that induces an appropriate phase dela in the light from one source. The ke point to recognize here is that we get the same result in monochromatic light if the phase induced b the glass is an odd multiple of π radians, so we can make the glass much thicker. ³ 7