Telescopes and Optics II. Observational Astronomy 2017 Part 4 Prof. S.C. Trager

Transcription

1 Telescopes and Optics II Observational Astronomy 2017 Part 4 Prof. S.C. Trager

2 Fermat s principle

3 Optics using Fermat s principle Fermat s principle The path a (light) ray takes is such that the time of travel between two fixed points is stationary with respect to small changes in that path In other words, the travel time of a ray is infinitesimally close to that of a neighboring path This is a generalization of the principle of least time or principle of least action

4 P1 (y,z) P0 To be precise, given a surface lying between two points P0 and P1, consider two paths from P0 to P1 If the travel time from P0 to P1 is τ, then d /dy = d /dz =0 where y and z are the coordinates of the intersection of the path with the surface

5 P1 (y,z) P0 If we replace the words time of travel from Fermat s principle above with the words optical path length (OPL), cdt, we see that Fermat s principle is recovered The OPL is defined as d(opl) = cdt=(c/v)vdt= nds where dt is an infinitesimal travel time, v is the the speed of light in a medium of index of refraction n, and ds is the infinitesimal geometrical path length

6 P1 (y,z) P0 Then OPL = c Z dt = n Z ds Note that n can be a function p of position here, so consider n(y,z) and ds = dy 2 + dz 2 Letting y =dy/dz, Fermat s principle is δ(opl)=0 or q Z P1 n(y, z) (1 + y 02 )dz =0 P 0

7 P1 (y,z) P0 If we call F (y, y 0,z)=n(y, z) q 1+y 02, then where Z P1 P 0 F (y, y 0,z)dz = Z P1 P 0 F (y, y 0,z)dz =0, 0 y 0 d dz ( y)

8 P1 (y,z) P0 Substituting and integrating by parts, we have 1 P1 0 y 0 ydz =0 Z P1 P 0 where the second term vanishes because δy=0 at P0 and P1, and then Z P1 P P 0 d dz 0 ydz =0

9 P1 (y,z) P0 This must vanish for arbitrary δy, d 0 =0 q Substituting (after some algebra) q 1+y 0 n dz F (y, y 0,z)=n(y, z) y 0 p 1+y 02 1+y 02,! we have y 0 p 1+y 02 dn dz =0

10 P1 (y,z) P0 This must vanish for arbitrary δy, d 0 =0 q Substituting (after some algebra) q 1+y 0 n dz F (y, y 0,z)=n(y, z) y 0 p 1+y 02 1+y 02,! we have y 0 p 1+y 02 dn dz =0

11 P1 (y,z) Now, this looks ugly (and it is), but the presence of terms like P0 q (1 + y 02 ) suggests that trigonometric solutions would be useful: q tan = dy/dz = y 0, sin = dy/ds = y 0 / (1 + y 02 ) q cos = dz/ds =1/ (1 + y 02 ), d(sin )/dz = cos (d /dz) Here α is the angle made by the tangent of the path with the z-axis

12 P1 (y,z) P0 Then noting that dn we can @z n cos d dz =0

13 P1 (y,z) P0 Finally, note that we can write the curvature K of a path as K = d ds = d dz dz ds = cos d dz and therefore nk = n cos d dz @z (15)

14 nk = n cos d dz @z (15) This is the equation for the local curvature of a light ray subject to Fermat s principle in a medium in which n is a smoothly varying function of position (in the y,z-plane) As a special case, consider n=constant. Then K=0, and the path of a light ray in a homogeneous medium is a straight line as expected

15 z top of the atmosphere n= ground n= A useful example: atmospheric refraction Assume the atmosphere is a flat, layered medium with n=n(z) only, where the z-axis points to the center of the Earth, and hence the curvature of the atmosphere is negligible Then Eq. (15) becomes nk = n cos d dz

16 z α0 top of the atmosphere n= ground n= Because the change in n from the top of the atmosphere to the surface is very small, the path of a ray from a star will not be significantly deviated if α is not close to 90º Then we can integrate the expression on the last slide to find = tan 0 n where α0 is the zenith angle at the top of the atmosphere

17 z α0 top of the atmosphere n= ground n= For a ray passing down through the atmosphere, δn>0 and so δα<0 Thus the ray is bent towards the z-axis If α0=45º, then δα= rad 60 Note that n=n(λ), so different wavelengths get bent by different amounts Very important for wide-band spectroscopy!

18 s1 α1 A α2 θ(λ) s2 a1 L n(λ) L a2 t Another useful example is a dispersing prism with n=n(λ) in air (n=1) There is some wavelength whose rays follow paths parallel to the prism base For these rays, the diagram is symmetric about the prism s vertical bisector, so that s 1 = s 2 = s, 1 = 2 =, and a 1 = a 2 = a

19 s s1 α1 α A α2 α θ(λ) s2 s a1 a LL n(λ) LL a2 a t t Another useful example is a dispersing prism with n=n(λ) in air (n=1) There is some wavelength whose rays follow paths parallel to the prism base For these rays, the diagram is symmetric about the prism s vertical bisector, so that s 1 = s 2 = s, 1 = 2 =, and a 1 = a 2 = a

20 s s1 α1 α A A α2 α θ(λ) θ(λ) s2 s a1 a L L n(λ) n(λ) L L a2 a t t The OPL of the bottom ray (at the prism s base) is just nt The OPL of the top ray (at the prism s vertex) is 2L cos Fermat s principle says that these two OPLs must be the same, so nt =2L cos

21 s s1 α1 α A A α2 α θ(λ) θ(λ) s2 s a1 a L L n(λ) n(λ) L L a2 a t t We re interested in the change of θ with wavelength, so let s differentiate the OPL equation above: t dn = 2L sin d = 2L sin d d d d From the figure, we see that L sin = a and = A 2, so d /d = 1/2 Thus d d = t a dn d d d

22 s s1 α1 α A A α2 α θ(λ) θ(λ) s2 s a1 a L L n(λ) n(λ) L L a2 a t t For most optical glasses, we can write (approximately) where C 0 and C 1 are constants Therefore n( )=C 0 + C 1 / 2 d d = 2t a The negative sign means that θ decreases as λ increases, so that blue light is deviated more than red light The angular dispersion dθ/dλ is larger for shorter wavelengths C 1 3

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24 P y l y Δ C f B O z Let s return to the case of reflecting mirrors What shape must a mirror have to satisfy Fermat s principle? Let s consider three cases: A concave mirror with one conjugate at A concave mirror with both conjugates finite A convex mirror with both conjugates finite

25 P y l y Δ C f B O z Case 1: concave mirror, one conjugate at infinity For convenience, let f, l, and Δ be positive (note that this violates our sign convention!) Applying Fermat s principle to a ray on the optical axis and one at height y, we see that 2f = l +(f ) and so l = f +

26 P y l y Δ C f B O z We also know that (from Pythagoras) that l 2 = y 2 +(f ) 2 Eliminating l from these two equations we see that y 2 =4f = 4fz This is the equation of a parabola with a vertex at (0,0)

27 P y l y Δ C f B O z Using Eq. (7) from the last set of slides and applying the sign convention we have y 2 =2Rz where R is the radius of curvature and both R and z are negative In three-space, we replace y 2 with x 2 +y 2, and we find that a paraboloid satisfies Fermat s principle in this case

28 y P l y Δ s B R C B O Case 2: concave mirror, both conjugates finite s z In this case we find (using the same analysis) y 2 2z b2 a + z2 b2 a 2 =0 (16) where 2a = s + s 0 and b 2 = ss 0 This is the equation for an ellipse with center (0,a)

29 Case 3: convex mirror, both conjugates finite In this case we find (using the same analysis) y 2 +2z b2 a z 2 b2 a 2 =0 where 2a = s + s 0 and b 2 = ss 0 This is the equation for a hyperbola with center (0,0)

30 All of these forms can be written in a simple way: First, recall that Eq. (7) from the last set of slides says (in the paraxial approximation) Next, consider an ellipse with eccentricity e=c/a where c is the distance from one focus to the center of the ellipse: 1 s + 1 s 0 = 2 R or ss 0 s + s 0 = R 2 = b2 c 2 = a 2 b 2 2a

31 For our elliptical mirror above, then, we can write e 2 s s 0 2 = s + s 0 and 1 e 2 = 4ss0 (s + s 0 ) 2 = b2 a 2 We can then write Eq. (16) as Although we ve derived this equation for an ellipse with 1<e<1, it s actually the correct formula for any conic section y 2 2Rz +(1 e 2 )z 2 =0 (17)

32 We can write this equation for a surface of revolution as 2 2Rz +(1+K)z 2 =0 where K is Schwarzschild s conic constant and ρ 2 =x 2 +y 2. Then conic section e 2 K prolate ellipsoid <0 >0 sphere 0 0 oblate ellipsoid paraboloid 1 1 hyperboloid >1 < 1

33 Note that the above calculation suggests that for both classic Cassegrain and Gregorian telescopes, the primary should be a paraboloid, while the secondary should be an ellipsoid for a Gregorian a hyperboloid for a Cassegrain

34 Resolution to A θ Δ θ D θ B A z to B L Consider a perfect optical system one that satisfies Fermat s principle Light from two neighboring sources say, stars A and B separated by angle θ fill the aperture D

35 Resolution to A θ Δ θ D θ B A z to B L The wavelength theory of light says that two image points cannot be separated resolved if the difference in light travel time to them from opposite sides of an aperture is less than ~one period of the wave

36 Resolution to A θ Δ θ D θ B A z to B L In other words, the points can t be resolved if the OPL difference between the rays is less than ~one wavelength

37 Resolution to A θ Δ θ D θ B A z L to B The OPL difference in the figure is Δ, so we require Δ λ This then implies that min /D The exact result from diffraction theory is min =1.22 /D

38 Aberrations Departures from Gaussian optics are called aberrations They occur in imperfect optical systems Aberrations come in two main types: chromatic aberrations due to wavelength variation in the index of refraction monochromatic aberrations, which are independent of wavelength

39 Aberrations Monochromatic aberrations come in two types: those that deteriorate the image those that deform the image These are inherent to each optical element and can be corrected in multiple-element systems (hopefully)

40 Aberrations Since Gaussian optics are based on the paraxial approximation, where sin θ θ, this will clearly break down when considering rays that are either at a large distance y from the optical axis like the marginal rays or at a large angle θ to the optical axis

41 Aberrations So let s expand sin θ in a Taylor series: 3 sin = 3! + 5 5! Taking the first (paraxial) and second terms, we have third-order (Seidel) aberration theory Higher-order (Zernike) terms can of course still be present

42 Aberrations In third-order theory, there are five primary aberrations, which scale as y m n where m+n=3

43 For a telescope of radius R yf, we call the aberrations aberration spherical aberration scaling for a scaling for a type ray telescope deterioration (y/r) 3 F 3 coma deterioration θ(y/r) 2 θf 2 astigmatism deterioration θ 2 (y/r) θ 2 F 1 field curvature deformation θ 2 (y/r) θ 2 F 1 distortion deformation θ 3 θ 3

44 Note that the last four aberrations are dependent on θ these are off-axis aberrations

45 Spherical aberration Spherical aberration is independent of θ occurs even for on-axis sources! This aberration occurs when rays do not come to a focus at the same point on the optical axis a point source makes a blurred disk

46 This is the spherical aberration of the Hubble Space Telescope before (left) and after (right) correction image on left covers ~2 ; image on right covers 0.05 This was due to a 1.3mm washer inserted during polishing of the primary mirror, which caused the margins of the primary to be too flat by ~λ/2!

47 Coma Rays from an off-axis source converge at different points on the focal plane This occurs off-axis, and is asymmetric because rays from the opposite side of the optical axis land on the same side of the axis Coma from a parabolic mirror

48 This is a picture of a far off-axis part of a field with significant coma note the head-tail structure like a comet Coma also appears when optical elements are misaligned

49 Astigmatism Astigmatism is caused by rays in the horizontal plane and the vertical plane coming to different foci The best focus is then the circle of least confusion Astigmatism from a parabolic mirror

50 This is the image of a star in an astigmatic system

51 Field curvature Outside of the paraxial region the image plane is curved the mapping from object to image follows spherical surfaces You can correct this by adding lenses or mirrors to flatten the field If you can t remove the curvature, you need to have a curved detector (or curved slit in a spectrograph)

52 Distortion This is a radial change in plate scale with field angle Does not change the focus but deforms the images Can be calibrated out, but it s annoying!

53 Chromatic aberrations Caused by rays of light with different wavelengths coming to different foci

54 Note the large blue halos and smaller yellow and red halos this is chromatic aberration

55 You can correct chromatic aberration by creating achromatic doublets (or triplets) by combining positive and negative lenses Doublets bring light from two wavelengths to a common focus Triplets bring light from three wavelengths to a common focus generally sufficient for broad-band sources without causing undue aberrations

56 Aberration compensation Achromatic doublets and triplets demonstrate a general rule for compensating aberrations: one can generally correct n primary aberrations with n reasonably separated powered optical elements for example, we correct spherical aberration by using a single mirror shaped as a paraboloid instead of a sphere

57 Aberration compensation But with a single mirror, the other aberrations coma, astigmatism, field curvature, distortion remain We correct these by added more (and more) elements For example, by choosing the correct shapes (the conic constants Ki) for the primary and secondary, we can correct two aberrations

58 Aberration compensation Schwarzschild (1905) and Ritchey & Chrétien (1910) showed that two hyperboloidic mirrors with the focus of the hyperboloid of the primary shared with the focus of the hyperboloid of the secondary form an aplanatic telescope an aplanat is an optical system free of both spherical aberration and coma A Ritchey-Chrétien (RC) telescope is then limited by astigmatism, not coma like a classical Cassegrain HST is an RC telescope

59 Non-spherical surfaces are difficult to manufacture! But paraboloids can be made by spinning the liquid glass so many large telescopes are now constructed as classical Cassegrain or Gregorian telescopes

60 The WEAVE PFC PFC=prime focus corrector 6 lenses!