SIMG Optics for Imaging Solutions to Final Exam

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1 SIMG Optics for Imaging Solutions to Final Exam. An imaging system consists of two identical thin lenses each with focal length f = f = +300 mm and diameter d = d =50mm. The lenses are separated by t =50mm. Between the lenses and at equal distances from each is a single iris with diameter d iris =0mm. (a) Determine the effective focal length of the system. = + t f eff f f f f µ f eff = 300 mm + 50 mm 300 mm (300 mm) = f eff = +00 mm (b) Determine the locations of the focal points F and F 0 and principal points H, H 0 of the system, i.e., determine the distances FV, V 0 F 0, HV, andv 0 H 0 Trace a ray through the system from an object at infinity: + = = z z 0 f 300 mm z = = z 0 =300mm = z = 50 mm + µ = = z z 0 f 300 mm = z0 = 300 mm =+00mm 50 mm = V 0 F 0 =+00mm H 0 F 0 = f eff =+00mm=H 0 V 0 + V 0 F 0 = H 0 V 0 +00mm = H 0 V 0 =+00mm so the image-space principal point is 00 mm in front of the image-space vertex. Since the system is symmetric, the same distances apply in object space FV =+00mm VH =+00mm

2 (c) Sketch the system showing the locations of the vertices, focal points and principal points. (d) For an object at infinity, determine which element is the aperture stop of the system. From the drawing, you can see that the height of the marginal ray at the midpoint (75mmfromthefirst lens) between the lenses is 3/4 of the height at the first lens. Since the semidiameter of the first lens is 5 mm, the height of the ray at the iris is 8.75 mm, which is significantly larger than the semidiameter of the iris ( 0 mm), sothe iris is the stop. (e) Determine the locations of the entrance and exit pupils of the system for an object at infinity and locate them on the sketch in part (c) Again, since the system is symmetric, the pupils are symmetrically placed. We need to find the image distance created by the second lens for the object at the iris: z = 75mm f = 300mm µ z = 300 mm = 00 mm 75 mm So the pupils are virtual, i.e., the exit pupil is behind the image-space vertex and so is not accessible. The magnification of the pupils is: M T = z 00 mm = z 75 mm = 4 3 So the diameter of the entrance pupil is: d NP = mm = d NP = 80 3 mm = 6.7mm

3 (f) Determine the focal ratio (f/#) of the system. The focal ratio is the ratio of the focal length to the diameter of the entrance pupil: f/# = +00 mm = f/# =7.5 mm 80 3 (g) If the stop is circular with diameter d iris as stated, determine and sketch the the profile of the diffraction spot for wavelength λ 0 =0.5 μm D 0 =.44 λ0 f eff d NP = μm 7.5 = D 0 = 9.5 μm (h) Determine the locations of the object and image such that transverse magnification of the system is M T =+. trick question, these are at the locations of the principal planes, but these are virtual locations so you can do this by putting a virtual object at H to get an virtual image at H 0. (i) Determine the locations of the object and image such that transverse magnification of the system is M T =. Thesearethe equal-conjugate points: OH = f eff = 00mm=400mm = OV = OH + HV = OH VH =400mm 00 mm = 300 mm = V 0 O 0 = OV =300mmfor M T = So the object is 300 mm in front of the object-space vertex. Test this result by finding the image location measured from the image-space vertex: + = µ = z 0 z z 0 = f 300 mm = 300 mm z = t z 0 = =+ z 0 = f =+300mm 3

4 (j)determinethelocationsoftheobjectandimagerelativetotheverticesofthe system such that transverse magnification of the system is M T =. 4 Ifthetransversemagnification is, the object distance is four times larger than 4 the image distance: M T = 4 = z = z = OH =4z = + = + = z z z 4z z f eff = = 4z 4z 4z 00 mm = z = mm=50mm= H0 O 0 =50mm = V 0 O 0 = H 0 O 0 H 0 V 0 =50mm 00 mm = V 0 O 0 =50mm z = 4 z = OH =000mm OH = 000mm = OV + VH = OV +00mm = OV =900mm Check it by finding the images from each lens in turn and the associated transverse magnifications: z 0 = µ µ = f z 300 mm =+450mm 900 mm M = 450 mm 900 mm = z = t z 0 =50mm (450 mm) = 300 mm µ z 0 = µ = f z 300 mm =+50mm 300 mm 50 mm M = 300 mm =+ µ M T = M M = µ + = 4 (k) If the two lenses are made of the same crown glass with n D =.5 at λ D = nm, explainwhatyouexpecttobetrueaboutthefocallengthofthe system at the F line (λ F =486.3 nm) andatthecline(λ C =656.8 nm); I m not looking for numerical values, but rather for the trend. For blue light, we know that the refractive index will be larger, so the focal length with be shorter. The system focal length will still be: µ f eff = + t f f f f where t is fixed, but the shorter focal lengths for f and f mean that the system focal length is shorter for blue light and longer for red light (l) If the system is set up to image an object at, what can you say about the depth of field if the diameter of the iris is decreased? We know that the depth of field goes as the square of the focal ratio, so the depth of fieldshouldincrease. 4

5 . An object with a square-wave amplitude transmittance with a period of 00 cycles is mm imaged by a lens with a circular pupil with =0mmand f =00mm.Thedistance from object to lens is z =00mm.Thewavelengthλ 0 =μm (a) Sketch a profile of the measured irradiance at output plane (the detector is sensitive at λ 0 =μm). Solution: The image distance is obviously 00 mm (equal conjugates) so the transverse magnification is M T =. The transfer function for coherent illumination will be a cylinder function with scaled frequency: µ µ r p (r) =CY L = H (ρ) =CY L λ 0z ρ = CY L ³ ρ The edge of the pupil is located at: λ 0 z r = and so the frequency at the edge of the transfer function satisfies the condition: ³ ρ = λ 0 z = ρ = ' 00 cycles λ 0 z mm ' 00 cycles μm 00 mm mm = 00 mm μm 00 mm = 40 mm > The square wave pattern may be written as: µ µ x f [x, y] = RECT 5 μm 0 μm COMB So its frequency spectrum is: x 0 μm [y] F [ξ,η] = (5μm SINC [5 μm ξ] COMB [0 μm ξ]) δ [η] + X µ δ ξ k SINC [5 μm ξ] 0 μm k= + X µ = δ ξ k k SINC 5 μm 0 μm 0 μm k= + X µ = δ ξ k k SINC 0 μm k= This is an array of Dirac delta functions modulated by a SINC function, so the input image may be written as the sum of cosine terms: Ã f [x, y] = " #! x 3 + SINC cos π + SINC cos π x 0 μm + 0 μm 3 = + " # π π cos x 0 μm 3π cos π x 0 μm + 3 5

6 The frequency of the first non-zero order of the input spectrum is: ξ = 0 μm =00cycles mm This frequency is (just barely) outside the finite support of the transfer function, so you see NO modulation, just a uniform constant field: g [x] = [x] g[x] ^ x [micrometers] 6

7 (b) Repeat for wavelength λ =0.50 μm. Solution: 00 cycles mm 0.5 μm 00 mm = 0 mm = At this shorter wavelength, the constant and the fundamental sinusoidal frequency terms are passed to the output, which consists of a biased cosine function: g [x, y] = + SINC x cos π 0 μm = + π π cos x 0 μm g[x] ^ x [micrometers] Approximate irradiance profile if using λ 0 =0.5μm, since the fundamental frequency of the square wave is passed after attenuation by. The units on the x-axis are μm 7

8 (c) If the same lens diameter in (a) is used for incoherent illumination at λ 0 =0.5μm, sketch the expected irradiance Solution: In the incoherent case, the transfer function is the appropriately scaled autocorrelation of the pupil, which in this case is a circular triangle. The cutoff frequency is twice as large, but the nonzero components are further attenuated. The output image consists of the unattenuated DC term and a sinusoid with amplitude approximately equal to g [x] = + π cos x π 0 μm g[x] ^ x [micrometers] Approximate irradiance profile for incoherent light at λ 0 =0.5μm, since the fundamental frequency of the square wave is passed after attenuation by. The units on the x-axis are μm 8

3. Consider a pinhole camera where the pupil function (circular pinhole of diameter ) is small compared to the object distance z,whichmaybeapproximatedasinfinite.

9 3. Consider a pinhole camera where the pupil function (circular pinhole of diameter ) is small compared to the object distance z,whichmaybeapproximatedasinfinite. Assume that the object is illuminated by quasimonochromatic light with mean wavelength λ μ. (a) Assuming that the pinhole is small, but sufficiently large so that the ray optics model is valid, find an expression for the spatial part of the optical transfer function of the camera. Determine the spatial frequency of the first zero of the OTF (which we may call the cutoff frequency ). If you see it, this problem is rather easy. Under the ray optics description, then the incoherent impulse response is proportional to the projection of the pinhole onto the observation plane. For an object at infinity, this is: µ r h [x, y; λ μ,z ] CY L Note that the ray-optics impulse response depends on neither wavelength λ μ nor propagation distance z. The OTF is just the appropriately scaled (and, arguably, normalized) -D Fourier transform: µ ¾ r F ½CY L = π d 0 4 SOMB (ρ) =H [ξ, η; λ μ,z ] The normalized OTF is: H [ξ,η; λ μ,z ] H [0, 0; λ μ,z ] = SOMB (ρ) =SOMB µ ρ d 0 The cutoff frequency (first zero) is located at the frequencies that satisfy ρ cutoff =. = ρcutoff =. which is not a function of wavelength in the ray optics model! Note that the cutoff frequency decreases with increasing diameter in the ray-optics model. 9

10 (b) Now assume that the pinhole is so small that Fraunhofer diffraction governs the shape of the psf; calculate the spatial frequency of the first zero of the OTF. If Fraunhofer diffraction applies, then the incoherent impulse response is proportional to the squared magnitude of the aperture function: h [x, y; λ μ,z ] F ½ µ r CY L ¾ ρ r λμz = π d 0 4 SOMB ( r ρ) ρ = π d 0 4 SOMB ³ r λμz so the impulse response in the Fraunhofer case IS a function of the wavelength λ μ and of the propagation distance z. The OTF is normalized -D Fourier transform: H [ξ,η; λ μ,z ] π d 0 4 F SOMB ³ r λμ z µ = π d 0 4 CTRI λ μz λμz ρ = π d 0 4 CTRI ³ ρ λ μ z The CTRI function has compact support with unit radius, so the cutoff frequency satisfies the condition: λ μ z ρ cutoff = = ρ cutoff = λ μ z which is a function of wavelength and which INCREASES with increasing diameter, as we expect in the wave model. (c) OPTIONAL BONUS: Use the results of parts (a) and (b) to estimate an optimum size of the pinhole in terms of the various parameters, i.e., the diameter that produces the largest cutoff frequency (of the first zero). For a pinhole with diameter that is sufficiently large for geometrical optics to be valid, the cutoff frequency increases with decreasing diameter. Eventually the pinhole diameter will be small enough for the Fraunhofer approximation to be valid. The approximate diameter where the two expressions are both valid is where the cutoff frequencies match:. (d 0 ) optimum = ( ) optimum λ μ z = ( ) optimum =.λ μ z = ( ) optimum = p.λμ z =.04 p λμ z 0

11 4. The relationship of the the focal length and the object and image distances is: z + z = f (a) Derive the minimum separation between a real object and its real image Solution: The separation between the real object and the real image is µ s z + z = z + f z The minimum separation must satisfy the relation: = f + z ds =0 dz f µ d z dz f z µ = f µ 0+ z z = µ = z f z = = fz z f = = z =f fz substitute into imaging equation: µ ds = dz z =f = s min =f +f =4f The minimum separation between object and image is s min =4f

12 (b) An object is located at a distance z = f from a lens with focal length f =+ f. Locate the image and determine the transverse magnification. Draw the system including the locations of object and image. f + = µ = z = z f f = f f The image is virtual and the transverse magnification is: M T = z z = Ã f f! =+virtual, upright and magnified

13 5. A lens with f =+500mmis sawn into two pieces through a plane cutting through the optical axis (i.e., the cut is along a diameter). A point source of monochromatic light with λ 0 =500nmis placed on the optical axis at a distance z = 000 mm from the lens. The half lenses are gradually moved apart; each creates an image of the point source that are mutually coherent. The light is observed on a screen placed at a distance z =000mmfromthelens(intheFraunhoferdiffraction region). (a) Determine the period of interference fringes observed on the screen if the separation between the two lens halves is increased to 0.5mm. (b) Describe in words and/or equations what will happen if the lens and observation screen are moved farther from the point source. (a) SOLUTION: The lens creates two images of the point source. We know from the imaging equation that: + = = z = z f 000 mm 500 mm = z z f z f 000 mm 500 mm =000mm which you probably knew already since this is the distance for equal conjugates Since the transverse magnification is M T =, wecanfind the separation of the images of the point source by geometric considerations: 000 mm =0.5mm 000 mm =mm This is the distance between the sources in a Young s two-aperture experiment. 3

14 ThedistancetotheobservationscreenisL 0 =000mm, so the period of the irradiance fringe pattern is D = L 0λ 0 = 000 mm 500 nm mm = 0.5mm=D (b) SOLUTION: as the lens is moved away from the source, the magnification changes so the point sources appear closer together, which means that the fringe period will increase. 4

Introduction to aberrations OPTI518 Lecture 5

Introduction to aberrations OPTI518 Lecture 5 Second-order terms 1 Second-order terms W H W W H W H W, cos 2 2 000 200 111 020 Piston Change of image location Change of magnification 2 Reference for OPD