Tutorial Solutions 3 Point Spread Functions This set of questions look in detail at the point spread function of lenses and its implications. The details of problems 6 and 7 are beyond the scope of this course but the results are very important since they put produce real practical numbers. The computer simulation programs in problem 8 are strongly recommended as they, hopefully, give insight into what a PSF looks like. dx 3.1 Bessel Functions Given that the expansion of Bessel functions of integer order is, ∞ k n ·2k µ ´ = µ ´ 1 x 2 µ= Jn ´x ∑ · · µ k!Γ´n k 1 k =0 µ ´ µ write out the first three terms of the expansion for J0 ´x and J1 x . Use this expansion to determine the value of µ J1 ´x when x = 0 x µ ´ µ ´ µ= = ! Plot the functions J0 ´x ,J1 x and J1 x x for x 10 10. (Maple would be a good idea). Solution Part a: Firstly note the identity, that for n an integer Γ · µ= ´n 1 n! so for we can write the expansion as: ∞ k n ·2k µ ´ = µ ´ 1 x 2 µ= Jn ´x ∑ · µ k! ´n k ! k =0 So the first three terms are x2 x4 µ = · ::: J ´x 1 0 4 64 x x3 x5 µ = · J ´x 1 2 16 192 So from these expansions, we get that 2 4 µ J1 ´x 1 x x · x = 2 16 192 Department of Physics and Astronomy Revised: August 2000 so that when x = 0, µ J1 ´0 1 0 = 2 Part b: the plots are: µ ! Plot of J0 ´x from 10 10: 1 besj0(x) 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -10 -8 -6 -4 -2 0 2 4 6 8 10 µ ! Plot of J1 ´x from 10 10: 0.6 besj1(x) 0.4 0.2 0 -0.2 -0.4 -0.6 -10 -8 -6 -4 -2 0 2 4 6 8 10 µ= ! Plot of J1 ´x x from 10 10: Department of Physics and Astronomy Revised: August 2000 0.5 besj1(x)/x 0.4 0.3 0.2 0.1 0 -0.1 -10 -8 -6 -4 -2 0 2 4 6 8 10 3.2 PSF of Simple Lens Given that the Intensity PSF of a lens is given by the scaled Power Spectrum of the pupil function derive an expression for the PSF of a circular lens at geometric focus. This is the full solution to the round lens outlined in lectures. Solution ; µ δ ∞ For a lens with pupil function p´x y then if the object is a -function at , the amplitude in the back focal plane is: ZZ κ ´ ; µ= ´ ; µ ´ · µ u x y Bˆ p s t exp ı xs yt dsdt 2 0 f where κ 2 2 ´ · µ Bˆ = B exp ı x y 0 0 2 f the PSF is the intensity in the back focal plane, given by 2 ; µ=j ´ ; j g´x y u2 x y So for a simple round lens of radius a,wehave 2 2 2 ; µ = · p´x y 1forx y a = 0else If we shift to Polar Coordinates, with ρ θ = ρ θ s = cos & t sin we get that Z Z a 2π κ ; µ= ´ ρ θ · µρ θ ρ ρ θ u2 ´x y Bˆ0 exp ı x cos y sin d d 0 0 f Department of Physics and Astronomy Revised: August 2000 ; µ ´ ; µ We have that p´s t is circularly symmetric, so u2 x y must also be circularly symmetric. Note: if we rotate a circular lens we do not do anything to the PSF! As a consequent, we can ; µ = calculate u2 ´x y along one radial line. Select the line y 0, so we get, Z Z a 2π κ ; µ= ρ ρ θ u2 ´x 0 Bˆ0 exp ı xρcosθ d d 0 0 f We now use the standard result that Z 2π θµ θ = π ´ µ exp ´ırcos d 2 J0 r 0 so if we let κ r = xρ f we get that Z a κ ρ ; µ= π ρ ρ u2 ´x 0 2 Bˆ0 J0 x d 0 f now noting the second standard result that d µ= ´ ´ µµ rJ ´r rJ r 0 dr 1 so by integration we have Z r µ= ´ µ rJ1 ´r J0 t tdt 0 If we substitute, κ α = xρ f we can write κ Z xa 2 f f ; µ= π ´αµ α α u2 ´x 0 2 Bˆ0 J0 d 0 κx This is now in the right form to be integrated, to get κax J1 2 f ; µ= π ´ ˆ u2 x 0 2 B0a κax f ; µ= The PSF is usually normalised so that u2 ´0 0 1 and the phase term associated with Bˆ0 ignored, so that κax 2J1 f ; µ= u2 ´x 0 κax f µ= = = Note the factor of 2, since J1 ´0 0 1 2 from previous question. The system is rotational symmetry, so in two dimensions we have that κar 2J1 f ; µ= u2 ´x y κar f Department of Physics and Astronomy Revised: August 2000 2 2 2 · where r = x y . The intensity PSF is then ¬ ¬ 2 ¬ ¬ κar ¬ ¬ 2J1 f ¬ ¬ ; µ= g´x y κ ¬ ¬ ar ¬ ¬ f which is the expressions given in lectures and in Physics 3 Optics course. (See next question for plots of this function). dx 3.3 Annular Aperture Many telescopes, and mirror objectives have a central stop, giving an annular aperture. Extend the above derivation to give an expression for the PSF of annular lens. Hint: The method is almost identical to the full aperture case except for the limits of integration. For the special case of the central stop being half the diameter of the aperture find the location of the first zero, and compare this with full aperture case. Plot the cross-section of the intensity PSF of both apertures, and comment on the result. Hint: You will been a numerical solution to obtain the location of the first zero, and use Maple to produce the plots. Solution The annular aperture or outer radius a and inner b is described as 2 2 2 2 ; µ = · p´x y 1forb x y a = 0else being of shape a b The amplitude in the back focal plane is given by the scaled Fourier Transform of the pupil function. The main part of the integration is identical to the previous question for the open circular aperture except that the radial integration is between b ! a. The amplitude in tha back focal plane is now Z a κ ; µ= π ρ ρ u2 ´x 0 2 Bˆ0 J0 xρ d b f This can separated and written as, Z Z a κ 0 κ ρ ρ ; µ= π ρ ρ · ρ ρ u2 ´x 0 2 Bˆ0 J0 x d J0 x d 0 f b f Department of Physics and Astronomy Revised: August 2000 µ ´ µ We now note that J1 ´x is even so that xJ1 x is odd. Using this we can rearrange the integral to give Z Z a κ b κ ; µ= π ρ ρ ρ ρ u2 ´x 0 2 Bˆ0 J0 xρ d J0 xρ d 0 f 0 f Both terms here have the same form, that being identical to the form for the normal circular aperture. This can then be integrated to give, ¿ ¾ κax κbx J1 J1 2 f 2 f 4 5 ; µ= π ´ ˆ u2 x 0 2 B0 a κax b κbx f f ; µ= so again, we “spin” about the optical axis, normalise to get u2 ´0 0 1 and ignore the phase term associated with Bˆ0 to get ¾ κar κbr ¿ J1 J1 2 2 f 2 f 4 5 ; µ= u2 ´x y 2 2 a κar b κbr µ ´a b f f and then the intensity PSF is given by 2 ; µ=j ´ ; µj g´x y u2 x y Part b: Look at locations of first zeros. For the full apetrure b = 0, we get the usual case ¬ ¬ 2 ¬ ¬ ¬ ¬ κ ar 2 ¬ ¬ ¬ J1 ¬ αµ f J1 ´ ¬ ¬ ¬ ¬ ; µ= = g´x y κ ¬ ¬ ¬ ar ¬ α ¬ ¬ f κ = α = : π where α = ar f The first zero occurs at 0 1 22 , so, as covered in lectures, the first zero is at λ 0:61 f r = 0 a = Look at the special case of b = a 2. By substitution we get that 2 2 αµ ´α= µ 4 2 J1 ´ a J1 2 ; µ= g´x y 2 2 2 a = µ α α= ´a a 4 4 2 which can we rewritten as: 64 1 1 2 ; µ= ´αµ ´α= µ g´x y J J 2 9 α 1 2 1 The first zero is thus when 1 αµ= ´α= µ J ´ J 2 1 2 1 This has a numerical solution of α : < : π 0 3 144 1 22 so the first zero occurs at λ 0:5 f r 0 a which is narrower then the full aperture case, so the PSF of the annular aperture in narrower, so the resolution is increased.Thisisnot what you would expect! The radial, normalsied, plot is Department of Physics and Astronomy Revised: August 2000 1 Annular(x) 0.9 Full(x) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 The full aperture is the wider plot with the lower secondard maximas, while the annular aperture has a narrower central peak but higher secondary maximas.