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Optics for Residents

Learn about astigmatic lenses, the second-order aberration with toroidal surfaces, resultant image formation, Maddox Rod usage, cross diagram transformation, and the Circle of Least Confusion in optics.

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Optics for Residents

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  1. Optics for Residents Astigmatic Lenses Amy Nau, OD

  2. Astigmatic Lenses • Spherical lenses form a point image for each object point • Stigmatic = point-like • Toroidal are not-point-like • Astigmatic! • This is a second order aberration

  3. Have two radii of curvature Toroidal Surfaces r1 r3 r2

  4. Vertical plane r1=r2 2 1 Horizontal plane Toroidal Surfaces The surface is created by two radii of unequal length, Each in a plane at right angle to the other P=n1-n2/r For any object vergence, a toric Surface creates two separate images.

  5. r1 The astigmatic image- plano cyl r1 r1 r2 Looks like a stack of thin plus lenses, each Of the same refracting power

  6. X’ r1 X’ X’ The astigmatic lens- plano cyl Cylinder axis Say each lens has P=5D. Then if we put an Object 50cm in front of them, EACH lens forms A point image 33cm away. The final composite image will be a series of Points oriented in a straight vertical line x HORIZONTAL LENS FORMS VERTICAL LINE IMAGE

  7. Astigmatic Lenses-plano cyl X’ Horizontal line image X’ X’ X’ x VERTICAL LENS FORMS HORIZONTAL LINE IMAGE

  8. Vertical plane 2 1 Horizontal plane Astigmatic Lens- plano cyl r1 is shorter than r2, so the power of r1 will be greater than r2. F=n1-n2/r Therefore, the VERGENCE of the two Powers will be different Recall do+di=1/f and F=1/f So, if F1>F2, then di2 is farther from The lens than di1

  9. Astigmatic Lens- plano cyl All toroidal surfaces have two major meridians- the one with the max power and the one with minimal power 90 degrees away. Each will form a line image, so what happens in the plano (no power) meridian? Each horizontal slice Has parallel faces w/o Curvature and thus no Refractive power. Same Alignment as axis. Vertical lens

  10. Problem type Find the powers of a plano cyl lens using radius information. Determine image position using radius information.

  11. Maddox Rod This is used to test EOM imbalances. Real horizontal line image X point source THE MADDOX ROD IS A STACK OF THESE LENSES ALL TOGETHER. Position of virtual, vertical line image, same position as X

  12. X Maddox Rod • This vertical line image is virtual • Cannot be focused on a screen • CAN be seen when looked at through the lens towards X. Since the eye is very close to the lens, the horizontal line is not seen. The eye then sees the virtual, VERTICAL image line that appears to be Located at the object point (where the light is).

  13. Maddox Rod Eye sees VIRTUAL horizontal image Eye sees VIRTUAL vertical image Remember the eye behind the red lens (OD) deviates in the direction OPPOSITE to that of the virtual red line.

  14. Problem Type • Know generally how Maddox rod works. • Know what occurs clinically!

  15. The Cross Diagram Take the example some power, P x 180 This has a maximum power is located in the vertical meridian. {An equivalent expression would be P@90- this is how K’s are expressed.} +1.00 = +2.00 x 180 +1.00 +2.00 +3.00 = +1.00 pl +1.00

  16. +3.00 +1.00 Transpositon +3.00 = -2.00 x 090 pl +3.00 = +3.00 -2.00 Combined cyl

  17. Transposition +3.00 = -2.00 x 090 is same as +1.00= +2.00 X180 +2.00 +1.00 +3.00 = pl +1.00 +1.00

  18. Problem Types Be able to convert an Rx into a cross diagram Be able to convert a cross diagram into an Rx Know how to transform between plus and minus cyl Draw cross diagrams in plus and minus form

  19. X’v X’h The Circle of Least Confusion +3.00 x 180 +5.00 x 090 x 20cm 33cm

  20. X’v X’h The Circle of Least Confusion Interval of Sturm - distance between the two line images x CLC

  21. CLC • Vergence at the clc is the average emergent vergence leaving the lens. • Lc=L1+L2/2 • The location of the clc is the reciprocal of Lc

  22. The Circle of Least Confusion 2 10 8 +8.00+2.00 x180 8 pl 8 Object is placed 1 m in front of this lens. L’=F+L L’=10-1=9D and l’=11.11cm L’’=F+L L’’=8-1=7D and l’’=14.28cm Location of clc = reciprocal of average verg. L’c=L’+L’’/2 = (9+7)/2 = 8D; and l’c=1/L’c so, l’c=1/8 =12.5cm

  23. Spherical equivalent • Take ½ the cyl and add to sphere • -4.00+1.00x180 becomes • -3.50D • Good for patients who can’t tolerate cyl in spectacles • Good for contact lenses

  24. Problem types Understand the terminology Know how to calculate the length of the conoid of Sturm Know how to find the spherical equivalent in an Rx Know how to locate the CLC

  25. Images of Extended Objects Horiz meridian focus Vertical meridian focus E Circle focus

  26. Images of Extended Objects If the astigmatism is such that the two cyl axes are not Oriented vertically and horizontally but at some other position, Each image line is still parallel to the corresponding axis Axis 45 Axis 135

  27. Differential Motion of Image Lines It is possible to move only ONE part of a toric image P1 X 090 combined with P2 X 180 You can place a new cyl lens in front of the above, P3 x 180 PL p2 p3 p1 PL + PL The result is (p2 + p3) x 180. P1 is unaffected! If p3 is +, pulls the horizontal line image closer to the lens itself If p3 is - ,the vergence is decreased and the image is pushed away THIS IS HOW THE JCC LENS WORKS…………………

  28. You can move both the horizontal and vertical images simultaneously Differential Motion of Line Images p3 pl p2 + p3 pl p1 (p1+p3)x180 combined with (p2+p3) x 090 Changes the vergence in both meridians equally If it is of the proper strength, the concoid of Sturm Can be collapsed, thus eliminating the astigmatism

  29. Differential Motion of Line Images P1 P1 @ 090 P2 @ 180 P2 -p3 @ 180 +3D sph

  30. Differential Motion of Line Images Note that the image moves equally, but only in the dioptric sense—the linear distance moved by the more distant line must be greater that that moved by the closer line. This is due to proximity to the lens (vertex) A 1cm linear distance between 10 and 11 cm corresponds To a dioptric change of 0.9D A 1 cm linear distance between 25 and 26 cm corresponds To a dioptric change of .15D = .9D =.15D 10 11 25 26

  31. Cyl orientation (convention) The 0-180 axis (horizontal) 0 begins at the patients LEFT ear and rotates counterclockwise when you are facing the patient. This is true for both eyes. 120 deg 30 deg 180 0 Left reference ear

  32. Meridional Powers of Cyl Lenses What about the powers between the major meridians??? The power gradation from max to min is NOT a straight line change; the power gain moving from the axis meridian (min) to the maximum increases by the sin 2 of the angle away from the axis. 60 F(sin2f) In this 3D cyl, the power in the meridian 60 to the axis is 3(sin260) = 3(.866)2 = 3(.75) = +2.25D

  33. Meridional power of cyl lenses • Factoid • For any spherocylinder lens, the power in the meridian 45 degrees to the axis (that is, halfway between the max and min meridional powers) is always the spherical equivalent of that lens.

  34. Problem type • Be able to calculate an off axis power • What is the power at 45 degrees? • -1.00-2.00x180? A. -2.00D

  35. Jackson Cross Cylinder A toric lens that is composed of a + cyl and – cyl of equal powers ground on to one lens, with their axes at right angles To each other. The strength of the cyl is always two times, and of opposite sign to the power of the sphere +1.00 = -2.00 x 180 or -0.25 + 0.50 x 090 ALL CROSS CYL LENSES HAVE AN EQUIVALENT POWER OF ZERO THUS, THE CLC WILL NOT BE MOVED! - + Plus axis at 180 + + - Plus axis at 90 - - + The meridians marked are the axes!

  36. JCC for power refinement + axis at 180 - Axis at 090 Both focal lines will move away from each other And the astigmatism will increase (CLC increases in size) + axis at 090 - Axis at 180 Both focal lines will move towards each other And the astigmatism will decrease (CLC decreases In size) Note the position of the clc is stationary

  37. JCC for power refinement Clinically, the CLC is placed as Close to the retina as possible Using the sphere powers (usually the spherical equivalent). Then The patient can determine if the size of the blur circle increases or decreases As soon as the patient can no longer tell the difference, then the Interval of sturm is collapsed, and there is “no more” astigmatism.

  38. Problem type • Be able to recognize a JCC in Rx form • Be able to write JCC in Rx format • Understand how it works in general terms.

  39. Nature of torics Maddox rod optics Cross diagrams Transposition CLC Images formed by torics Manipulation of image position Meridional (off axis) powers Optics of the JCC Learning Goals

  40. Free Optics Textbook online • http://www.lightandmatter.com/bk5a.pdf

  41. Is it moving and shimmering? Buy a poster!Look at this illusion for a while and it willappear to be shimmering and moving.Also: Follow the outermost groove and watch itchange from a groove to a hump as you go around the wheel.

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