Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

42,741 views

Published on

No Downloads

Total views

42,741

On SlideShare

0

From Embeds

0

Number of Embeds

1,162

Shares

0

Downloads

0

Comments

0

Likes

44

No embeds

No notes for slide

- 1. Prisms and its propertiesGauri S. Shrestha, M.Optom, FIACLE
- 2. apex undeviatedRefracting Surface β deviated incident base
- 3. Definitions Prisms: Two plane surfaces inclined at an angle with respect to each other Refracting Surfaces: The two interfaces of refraction in a prism. Inclined at an angle equal to the apex angle. Reflecting Surface: In some prism setups, the internal ray hits the second refracting surface such that total internal reflection occurs (Reflecting Prism)
- 4. Definitions Apical Angle (α): The angle between the two refracting surfaces in a standard refracting prism. Also referred to as the “Refracting Angle”. Apex: The tip of the prism where the two refracting surfaces meet. The apical angle is the apex of the prism Base: The bottom of the prism or the side opposite the apex or apical angle. The orientation of an ophthalmic prism is described relative to the base
- 5. Deviation Angle The angle from the original light ray path direction to the direction of the same light ray after passing through the prism We will use the nomenclature of (β)
- 6. β1=θ1-θ1’ apex undeviated β1 α β2=θ2’- θ2 β2 path normal βt θ1 θ 2’ θ 1’ θ2incidentray prism base ray
- 7. n1 sin θ 1 = n2 sin 1’ θ α = θ 1’ + θ 2n2 sinθ 2 = n3 sinθ 2’ βt = β1 + β2 βt = θ 1 +θ 2’ - α
- 8. βT = β1 + β2βT = (θ1 – θ1’) + (θ2’ – θ2)β T + θ1 ’ + θ2 = θ2 ’ – θ2β T + α = θ1 + θ2 ’β T = θ1 + θ2 ’ - α
- 9. Prism Sign Convention THEORETICAL - + + -+ - - +If n2>n1, the deviation is always towards the base of the prism. By signconvention, a base down prism has a positive apex angle and a negativedeviation angle. A base up prism has a negative apex angle and a positivedeviation angle.
- 10. A flint glass (n=1.617) prism in air has a 50 degree apex angle. What is the deviation for a ray with an incident angle of 70 degrees on the base side of normal? Solution:1 1.00 sin (70) = 1.617 sin θ1’ θ1’ = 35.53 degrees Note: This prism bends this ray 43.83 degrees toward2 α = θ1’ + θ2, 50 = 35.53 + θ2 , the base. By reversibility, a θ2 = 14.47 degrees ray with an incident angle of 23.83 degrees will have a 703 1.617 sin (14.47) = 1.00 sin θ2’ θ2’ = 23.83 degrees degree final angle of refraction and 43.83 degree β = θ1 + θ2’ - α deviation angle4 β = 70 + 23.83 – 50 β = 43.83 degrees
- 11. Minimum Deviation • internal ray perpendicular to bisector of apex angle • incident and emergent angles equal in magnitude (θ1 = θ2’) θ1 θ2’ βmin
- 12. (βt)
- 13. (α + min) βn2 sin = 2n1 αNote that βmin is sindependent on only the 2apical angle and theindexes of refraction
- 14. What is the minimum deviation through a flint glass prism (n= 1.617) with an apical angle of 50 degreesn2/n1 = sin [(α + βmin)/2] sin (α/2)1.617/1.00 x sin 25 = sin [(50 + βmin)/2]βmin = 36.2 degrees
- 15. Maximum Deviation• set incident angle or emergent angle equal to 90° on bsn• compute prismatic deviation by four step process
- 16. What is the deviation angle for a ray that has grazing incidence on the base sideof the normal? (θ1=90 degrees) (MAXIMUM DEVIATION!) Assume the sameflint glass prism (n=1.617) with an apical angle of 50 degrees Solution: 1.00 sin (90) = 1.617 sin θ1’ Note: Since grazing1 θ1’ = 38.20 degrees inicidence gives the largest incident angle possible, the α = θ1’ + θ2 59.31 degree deviation is the2 50 = 38.20 + θ2 θ2 = 11.80 degrees MAXIMUM DEVIATION that this prism gives. By reversibility, the 59.31 degree maximum deviation will also3 1.617 sin (11.80) = 1.00 sin θ2’ occur for an incident angle of θ2’ = 19.31 degrees 19.31 degrees4 β = θ1 + θ2’ - α β = 90 + 19.31 - 50 β = 59.31 degrees
- 17. (βt)
- 18. Limitations of Refraction ThroughPrism Light will not be refracted through a prism if the internal angle at the second refracting surface is greater than the critical angle What incident angle at the first surface of the prism will yield the critical angle at the second refracting surface? Solve for critical angle and work backwards to find the limiting incident angle Incident angles less than this will not be refracted through the prism and will undergo TIR
- 19. Reflecting Prisms Created by an angle of incidence which after refracting through the first surface of the prism is incident on the second surface at an angle creating total internal reflection at the second surface Can be a positive or negative incident angle, depending on the apical angle and the index of refraction
- 20. Reflecting Prisms
- 21. For the same prism (n = 1.617, apical angle = 50 degrees), what is the deviationangle for an incident angle of 10 degrees on the base side of the normal? Solution: 1.00 sin (10) = 1.617 sin θ1’1 θ1’ = 6.17 degrees α = θ1’ + θ22 50 = 6.17 + θ2 θ2 = 43.83 degrees3 1.617 sin (43.84) = 1.00 sin θ2’ sinθ2’ = 1.12 θ2’ = Error (The sin function cannot be greater than zero)4 Total Internal Reflection
- 22. (βt)
- 23. Prism Deviation and Apex Angles The largest deviation possible through a prism would be the condition where: θ1= θ’2 = 90° α = θ’1 + θ2 `=θc + θc θ’1 = θ2 = θc α = 2θc If the apex angle is greater than two times the critical angle, no incident rays will be refracted through the prismThis demonstrates that an apex angle greater than twotimes the critical angle creates TIR at the second surfaceno matter the incident angle
- 24. What is the maximum apex angle that a prism (n=1.50) can have so that theprism can still refract incident light through without TIR?sin θc = n’/nsin θc = 1.00/1.50θc = 41.81 degreesα(max) = 2(41.81) = 83.62 degrees
- 25. Determining Index of RefractionUsing Thick Prisms Spectrometer can be used to narrow the wavelength of light to minimize chromatic aberrations. Dmin can be determined
- 26. (α + min) βn2 sin = 2n1 α Note that in air, the index of sin the prism can be determined by knowing the apex angle and the minimum angle of 2 deviation
- 27. A 48 degree prism has a minimum deviation of 27degrees. What is the index of refraction of the materialwhich makes up this prism?n2 = sin [ (α + βmin) / 2 ] sin (α / 2)n2 = sin [ ( 48 + 27) / 2 ] sin (48 / 2)n2 = 1.50
- 28. As the apex angles get smaller and smaller, the deviation angles are constant around the paraxial region 90 80°Deviation Angle (degrees) 80 70 70° 60 50 50° 40 30° 15° 30 4°(thin) 20 10 -90 -60 -30 0 +30 +60 +90 Incident Angle (degrees)
- 29. Thin Prisms Have a clinically constant deviation angle in the paraxial area (± 20 °) Don’t have to worry about where a patient is looking through the prism Defined as prisms with an apical angle of less than 8°-15° Defined as prisms with a prismatic power of approximately 15 to 25 or less Δ Δ
- 30. Deviation by Thin Prisms npβ = (n )α -1 s
- 31. What is the deviation angle of an 8 degree apical angle prism(n=1.49) in water? In air?QUESTION: Do you predict the deviation to be greater in Water or Air?Solution: (in water)β = [(np/ns) – 1] αβ = [(1.49/1.33) -1] 8β = 0.96°(in air)β = [(np/ns) – 1] αβ = [(1.49/1.00) -1] 8β = 3.92°
- 32. Prism Diopter Unit
- 33. Displacement (cm) β d lDistance (m)
- 34. d in cm∆ = l in m = 100tanβ
- 35. Prism Diopter 1.0Δ Unit used in clinical Displacement (cm) practice of prism use A one prism diopter prism will displace an β d image one centimeter at 1 cm a distance of one meter l Distance (m) 1 meter
- 36. What is the deviation angle of an 8 degree apical angle prism(n=1.49) in water? In air? Give the answer in terms of prismdiopters.Solution: (in water)β = [(np/ns) – 1] αβ = [(1.49/1.33) -1] 8β = 0.96° Δ = 100 tan β = 1.68Δ(in air)β = [(np/ns) – 1] αβ = [(1.49/1.00) -1] 8 Δ = 100 tan β = 6.85Δβ = 3.92°
- 37. A glass ophthalmic prism (n=1.52) displaces the image of an object 0.037m at2.2m. What is the deviation of the prism in prism diopters and degrees?Solution: Displacement (cm)Δ = d (cm) / l (m)Δ = 3.7 cm / 2.2 mΔ = 1.68Δ β dΔ = 100 tan β1.68 = 100 tan ββ = 0.96° or 57.6’ l Distance (m)
- 38. Prism Base Directions rePatients
- 39. BU Patient BU OD OSBO BI BI BO BD BD
- 40. PatientB@135 B@135 B@45 B@45 OD OSB@225 B@315 B@225 B@315
- 41. Patient 90 90 135 45 135 45180 OD 0 180 OS 0 225 315 225 315 270 270
- 42. QUIZ #2 QUESTION #1 A prism (n=1.60) deviates a light ray surrounded by air ____________ it deviates the light ray when the same prism is surrounded by water. A. more than B. less than C. the same as D. I don’t know. I was asleep in class today
- 43. QUIZ #2 QUESTION #2 What is the approximate incident angle(s) that yield the maximum deviation for the prism shown? A. 6° B. 32° C. 90° D. A and B E. A and C
- 44. QUIZ #2 Question #3 What is the amount of light that is reflected from a water/air plane interface? A. 2% B. 3% C. 4% D. 7% E. 14%
- 45. QUIZ #2 Question #4 At plane refractive surfaces, a virtual object produces which type of image? A. Virtual Image B. Real Image C. Virtual or Real image depending on the vergence effect of the plane refractive surface D. A virtual object CANNOT be incident on a plane refractive surface!
- 46. Prism Image Displacement
- 47. β = Rotational Stimulush β EYE lThe image displacement, the eye’srotational stimulus, and the actualrotation of the eye can be expressedin either degrees or prism diopters. BD
- 48. QUESTION Does the prism have the same effect on the eye when the prism is moved away from the eye? Answer: NO If viewing a near object Answer: YES If viewing a distant object (at infinity)
- 49. β>βeh Effective Displacement β βe EYE l l’ BD
- 50. Prism Effectivity Effective Displacement = Eye Rotation Stimulus As a prism is moved away from the eye the ‘effect’ that it has on the eye decreases (assuming the image is closer than infinity) The rotation stimulus decreases For a given prism power Δ (βΔ), the eye rotation stimulus βΔe – Effective Displacement in Prism Diopters Can be defined by the following equation:
- 51. = h in cm = 100 tan βeβ Δ e l + l’ in m β>βe l ∆ = l + l’ Effective Displacement β βe EYE l l’ ∆ = 1 + l’ As l increases, β = βe BD l l increases, β = βe AsAs ldecreases, βe approaches zero
- 52. Prism Effectivity variable object 0.90.85 0.80.75 0.7 25 30 35 40 45 50 object distance (cm) 10 cm 6 cm Prism Distance
- 53. Prism Effectivity REMEMBER: If an object is at infinity, then there is no effective change in prism power at the eye no matter the distance the prism is away from the eye
- 54. Prism Effectivity variable prism distance0.95 0.90.85 0.80.75 0.7 5 6 7 8 9 10 11 12 13 14 15 eye-prism distance (cm) 40 cm 50 cm Object Distance
- 55. Prism Effectivity REMEMBER: If a prism is at the surface of the eye (really the center of curvature), there is no effective change in prism power at the eye no matter the distance of the object.
- 56. Effectiveness of Prism Measured by relative eye rotation Visual Prism Demand (need) is the amount of prism “effect” required by the eye to get desired result (single, comfortable binocular vision) Therefore, if a patient if viewing a near object and a prism is some distance in front of the eye (spectacles), the eye is possibly not having its PRISM DEMAND met.
- 57. What is the effectivity of a 15 prism diopter prism located 25mm in front of thecenter of rotation of the eye, when the wearer reads at a distance of 40 cm fromthe prism?Solution:βeΔ = 15 / 1 + (2.5/40)βeΔ = 14.12ΔLoss of approximately 1 prism diopter of effectivity
- 58. Prism Effectivity Since prisms are usually prescribed in powers of no more than a few prism diopters, the loss in effectivity for near vision, is usually clinically insignificant. Exceptions: High prism amounts Prisms used in phoropters (Risley) are routinely used to measure deviations of high prism amounts Risley prisms have a longer vertex distance than glasses since it sits in front of the lens
- 59. A 35 year old white male presents to your office with a complaint of doublevision at near. He wears glasses for his myopia and has been told that hewears prisms in his glasses. What are you going to do?Current glasses prescription: OD: -5.00-1.25 x 180 5Δ BD OS: -4.50-1.00 x 180You find that he needs an addition prism diopter prism at near to maintain clear,single, and comfortable vision.Visual need = 6Δ at nearWhat do his current glasses give him at distance and near (30cm andvertex distance of 2cm)?Distance = 5Δ, Near = βeΔ = 5 / 1 + (2.0/30)βeΔ = 4.7Δ
- 60. A 35 year old white male presents to your office with a complaint of doublevision at near. He wears glasses for his myopia and has been told that hewears prisms in his glasses. What are you going to do?What prism power would yield 6Δ at the working distance and vertexdistance provided? βΔe = Δ / 1 + l’/l 6 = Δ / 1 + 2/30 Δ = 6.4Δ BD OD What if 6.4Δ is too much prism for distance??? Now what? SLAB-OFF PRISM or REVERSE SLAB-OFF PRISM: A way to incorporate different amounts of vertical prism between distance and near
- 61. Prism Resolution
- 62. y ∆ φ x
- 63. y ∆∆y φ x ∆x
- 64. P ∆ ,φ → R ∆ x ,∆ y ∆ x = ∆ cos φ ∆ y = ∆ sin φ
- 65. R ∆ x ,∆ y → P ∆ ,φ ∆ = ∆ x 2 +∆ y ½ 2 ∆y tan φ = ∆x
- 66. Prism Addition
- 67. Method 1 complete the parallelogram total vector is arrow from origin to intersection point
- 68. Method 2 move end of one vector to tip of other vector total vector is arrow from origin to tip of moved vector
- 69. ∆ y2∆ ytot ∆ y1 ∆ x1 ∆ x2 ∆ xtot
- 70. Find the single prism equivalent to a 2Δ BI combined with a 5Δ BU in front of theleft eye. 5.4Δ@112 5 Patient 90 90 135 45 135 45180 OD 0 180 OS 0 225 315 225 315 270 270 2 21.8 5 Solution: Using Formula – tan Φ = Δy/Δx Φ Δ2 = Δx2 + Δy2 Φ = tan-1 5/2 2 = 4 + 25 Φ = 68.2° Δ = 5.39Δ
- 71. Adding Prisms Applied to an Eye Adding prisms in opposite base directions is subtractive Adding prisms in same base directions is additive
- 72. Clinical Application Prism Prescribing and Analyzing Lensometry Ophthalmic Optics and Opticianry
- 73. Risley Prism A pair of counter-rotating prisms Base direction stays constant but the magnitude of the prism changes Can be rotated to set the base direction Found on all phoropters Used extensively in optometry
- 74. y ∆1 φ1 ∆ tot x φ2 = - φ1 ∆2 ∆2 = ∆1
- 75. ∆ tot = 2∆ 1 cos φ 1
- 76. Each component of a Risley Prism is a 10Δ prism. The Risley is set so that bothprisms are BD. What is the maximum prism power? What is the power whenthey are counter-rotated by 45° from the maximum. Solution: Maximum = 2 x 10 = 20Δ Δ = 2 Δ1 cos Φ1 Δ = 2 (10) cos 45 Δ = 14.1Δ BD
- 77. 0 BO BI 10Δ Risely Prism over OD y ∆1 φ120 ∆ x 20 φ2 = - φ1 ∆2 = ∆1
- 78. 20 10Δ Risely Prism over OD BU y 0∆2 = ∆1 ∆1 φ φ2 = - φ1 1 BD ∆ 20 x

No public clipboards found for this slide

Be the first to comment