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Prisms and its properties


Gauri S. Shrestha, M.Optom, FIACLE
apex
                     undeviated
Refracting Surface



                                  β



                             deviated
  incident
                     base
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)
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
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)
β1=θ1-θ1’
                     apex
                                     undeviated
     β1       α         β2=θ2’- θ2
                               β2    path
 normal                                     βt
    θ1                               θ 2’
              θ 1’      θ2
incident
ray
                                        prism
                  base                    ray
n1 sin θ 1 = n2 sin 1’
                  θ

       α = θ 1’ + θ 2

n2 sinθ 2      = n3 sinθ 2’
            βt = β1 + β2
            βt = θ 1 +θ 2’ - α
βT = β1 + β2
βT = (θ1 – θ1’) + (θ2’ – θ2)
β T + θ1 ’ + θ2 = θ2 ’ – θ2
β T + α = θ1 + θ2 ’
β T = θ1 + θ2 ’ - α
Prism Sign Convention
            THEORETICAL
                      -                   +
                   + -+               -        -
                                          +

If n2>n1, the deviation is always towards the base of the prism. By sign
convention, a base down prism has a positive apex angle and a negative
deviation angle. A base up prism has a negative apex angle and a positive
deviation angle.
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 toward
2
       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 70
3
      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 angle

4     β = 70 + 23.83 – 50
      β = 43.83 degrees
Minimum Deviation

 • internal ray perpendicular to bisector
   of apex angle
 • incident and emergent angles equal in
   magnitude (θ1 = θ2’)

            θ1
                                    θ2’



                                      βmin
(βt)
(α + min)
                           β
n2            sin
   =                        2
n1                          α
Note that βmin is       sin
dependent on only the       2
apical angle and the
indexes of refraction
What is the minimum deviation through a flint glass prism (n
= 1.617) with an apical angle of 50 degrees

n2/n1 =      sin [(α + βmin)/2]
              sin (α/2)


1.617/1.00 x sin 25 = sin [(50 + βmin)/2]
βmin = 36.2 degrees
Maximum Deviation
• set incident angle or emergent angle
  equal to 90° on bsn
• compute prismatic deviation by four
  step process
What is the deviation angle for a ray that has grazing incidence on the base side
of the normal? (θ1=90 degrees) (MAXIMUM DEVIATION!) Assume the same
flint glass prism (n=1.617) with an apical angle of 50 degrees

     Solution:
     1.00 sin (90) = 1.617 sin θ1’       Note: Since grazing
1    θ1’ = 38.20 degrees
                                         inicidence gives the largest
                                         incident angle possible, the
    α = θ1’ + θ2                         59.31 degree deviation is the

2   50 = 38.20 + θ2
    θ2 = 11.80 degrees
                                         MAXIMUM DEVIATION that
                                         this prism gives. By
                                         reversibility, the 59.31 degree
                                         maximum deviation will also
3
    1.617 sin (11.80) = 1.00 sin θ2’
                                         occur for an incident angle of
    θ2’ = 19.31 degrees
                                         19.31 degrees

4   β = θ1 + θ2’ - α
    β = 90 + 19.31 - 50
    β = 59.31 degrees
(βt)
Limitations of Refraction Through
Prism
 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
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
Reflecting Prisms
For the same prism (n = 1.617, apical angle = 50 degrees), what is the deviation
angle 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’ + θ2

2   50 = 6.17 + θ2
    θ2 = 43.83 degrees



3
    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
(βt)
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 prism



This demonstrates that an apex angle greater than two
times the critical angle creates TIR at the second surface
no matter the incident angle
What is the maximum apex angle that a prism (n=1.50) can have so that the
prism can still refract incident light through without TIR?

sin θc = n’/n
sin θc = 1.00/1.50
θc = 41.81 degrees

α(max) = 2(41.81)
         = 83.62 degrees
Determining Index of Refraction
Using Thick Prisms
 Spectrometer can be used to narrow the
  wavelength of light to minimize chromatic
  aberrations.
 Dmin can be determined
(α + min)
                                     β
n2                sin
   =                                  2
n1                                    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
A 48 degree prism has a minimum deviation of 27
degrees. What is the index of refraction of the material
which makes up this prism?
n2 = sin [ (α + βmin) / 2 ]
         sin (α / 2)


n2 = sin [ ( 48 + 27) / 2 ]
       sin (48 / 2)


n2 = 1.50
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)
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
Deviation by Thin Prisms



          np
β =    (n        )α
                -1
            s
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] α
β = [(1.49/1.33) -1] 8
β = 0.96°

(in air)
β = [(np/ns) – 1] α
β = [(1.49/1.00) -1] 8
β = 3.92°
Prism Diopter Unit
Displacement (cm)




               β
                     d



      l
Distance (m)
d in cm
∆   =
         l in m

    = 100tanβ
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
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 prism
diopters.

Solution: (in water)
β = [(np/ns) – 1] α
β = [(1.49/1.33) -1] 8
β = 0.96°                Δ = 100 tan β = 1.68Δ

(in air)
β = [(np/ns) – 1] α
β = [(1.49/1.00) -1] 8
                         Δ = 100 tan β = 6.85Δ
β = 3.92°
A glass ophthalmic prism (n=1.52) displaces the image of an object 0.037m at
2.2m. 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 m
Δ = 1.68Δ                                                β
                                                                  d
Δ = 100 tan β
1.68 = 100 tan β
β = 0.96° or 57.6’
                                                l
                                          Distance (m)
Prism Base Directions re
Patients
BU
         Patient     BU


      OD             OS
BO         BI   BI        BO
     BD               BD
Patient
B@135       B@135
         B@45           B@45

        OD             OS

B@225   B@315 B@225     B@315
Patient
            90                     90

      135           45      135          45



180         OD          0   180
                                   OS         0




      225         315        225
                                         315
            270                    270
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
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
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%
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!
Prism Image Displacement
β = Rotational
                                             Stimulus
h

                         β
                                       EYE

                 l
The image displacement, the eye’s
rotational stimulus, and the actual
rotation of the eye can be expressed
in either degrees or prism diopters.    BD
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)
β>βe
h                Effective Displacement


        β        βe
                                     EYE

    l                 l’


            BD
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:
= h in cm     = 100 tan βe
β   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
Prism Effectivity
              variable object
 0.9

0.85

 0.8

0.75

 0.7
       25     30       35       40      45   50
                   object distance (cm)

                   10 cm             6 cm
                    Prism Distance
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
Prism Effectivity
                    variable prism distance
0.95


 0.9

0.85


 0.8

0.75


 0.7
       5   6    7     8     9    10     11   12   13   14   15

                    eye-prism distance (cm)

                    40 cm                    50 cm
                      Object Distance
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.
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.
What is the effectivity of a 15 prism diopter prism located 25mm in front of the
center of rotation of the eye, when the wearer reads at a distance of 40 cm from
the prism?


Solution:
βeΔ = 15 / 1 + (2.5/40)
βeΔ = 14.12Δ
Loss of approximately 1 prism diopter of effectivity
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
A 35 year old white male presents to your office with a complaint of double
vision at near. He wears glasses for his myopia and has been told that he
wears 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 180
You find that he needs an addition prism diopter prism at near to maintain clear,
single, and comfortable vision.


Visual need = 6Δ at near


What do his current glasses give him at distance and near (30cm and
vertex distance of 2cm)?
Distance = 5Δ, Near = βeΔ = 5 / 1 + (2.0/30)
βeΔ = 4.7Δ
A 35 year old white male presents to your office with a complaint of double
vision at near. He wears glasses for his myopia and has been told that he
wears prisms in his glasses. What are you going to do?

What prism power would yield 6Δ at the working distance and vertex
distance 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
Prism Resolution
y   ∆

    φ

        x
y        ∆


∆y
          φ
                  x
         ∆x
P ∆ ,φ   → R ∆ x ,∆ y

    ∆ x = ∆ cos φ

    ∆ y = ∆ sin φ
R ∆ x ,∆ y   → P ∆ ,φ

         ∆ = ∆   x
                  2
                      +∆ y ½
                        2



            ∆y
    tan φ =
            ∆x
Prism Addition
Method 1
 complete the parallelogram
     total vector is arrow from origin to intersection
      point
Method 2
 move end of one vector to tip of other vector
     total vector is arrow from origin to tip of moved
      vector
∆ y2
∆ ytot
     ∆ y1

            ∆ x1        ∆ x2
               ∆ xtot
Find the single prism equivalent to a 2Δ BI combined with a 5Δ BU in front of the
left eye.
                              5.4Δ@112
                                                   5
                    Patient
            90                        90

      135            45      135           45



180         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
Adding Prisms Applied to an Eye
 Adding prisms in opposite base directions is
  subtractive
 Adding prisms in same base directions is
  additive
Clinical Application
 Prism Prescribing and
  Analyzing
 Lensometry
 Ophthalmic Optics and
  Opticianry
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
y
        ∆1

             φ1 ∆
                  tot     x

              φ2 = - φ1
        ∆2

    ∆2 = ∆1
∆ tot = 2∆ 1 cos φ 1
Each component of a Risley Prism is a 10Δ prism. The Risley is set so that both
prisms are BD. What is the maximum prism power? What is the power when
they 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
0

     BO                          BI
                                     10Δ Risely Prism over OD
          y
                     ∆1

                       φ1
20
                                 ∆           x
                                                 20

                          φ2 = - φ1

                  ∆2 = ∆1
20
                                    10Δ Risely Prism over OD




                                          BU



                                y
                                          0
∆2 = ∆1




                               ∆1
                           φ
          φ2 = - φ1

                           1




                                      BD
                       ∆




                      20
                       x

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Prism and its properties

  • 1. Prisms and its properties Gauri S. Shrestha, M.Optom, FIACLE
  • 2. apex undeviated Refracting 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’ θ2 incident ray prism base ray
  • 7. n1 sin θ 1 = n2 sin 1’ θ α = θ 1’ + θ 2 n2 sinθ 2 = n3 sinθ 2’ βt = β1 + β2 βt = θ 1 +θ 2’ - α
  • 8. βT = β1 + β2 βT = (θ1 – θ1’) + (θ2’ – θ2) β T + θ1 ’ + θ2 = θ2 ’ – θ2 β 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 sign convention, a base down prism has a positive apex angle and a negative deviation angle. A base up prism has a negative apex angle and a positive deviation 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 toward 2 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 70 3 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 angle 4 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) β n2 sin = 2 n1 α Note that βmin is sin dependent on only the 2 apical angle and the indexes of refraction
  • 14. What is the minimum deviation through a flint glass prism (n = 1.617) with an apical angle of 50 degrees n2/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 side of the normal? (θ1=90 degrees) (MAXIMUM DEVIATION!) Assume the same flint glass prism (n=1.617) with an apical angle of 50 degrees Solution: 1.00 sin (90) = 1.617 sin θ1’ Note: Since grazing 1 θ1’ = 38.20 degrees inicidence gives the largest incident angle possible, the α = θ1’ + θ2 59.31 degree deviation is the 2 50 = 38.20 + θ2 θ2 = 11.80 degrees MAXIMUM DEVIATION that this prism gives. By reversibility, the 59.31 degree maximum deviation will also 3 1.617 sin (11.80) = 1.00 sin θ2’ occur for an incident angle of θ2’ = 19.31 degrees 19.31 degrees 4 β = θ1 + θ2’ - α β = 90 + 19.31 - 50 β = 59.31 degrees
  • 17. (βt)
  • 18. Limitations of Refraction Through Prism  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
  • 21. For the same prism (n = 1.617, apical angle = 50 degrees), what is the deviation angle 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’ + θ2 2 50 = 6.17 + θ2 θ2 = 43.83 degrees 3 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 prism This demonstrates that an apex angle greater than two times the critical angle creates TIR at the second surface no matter the incident angle
  • 24. What is the maximum apex angle that a prism (n=1.50) can have so that the prism can still refract incident light through without TIR? sin θc = n’/n sin θc = 1.00/1.50 θc = 41.81 degrees α(max) = 2(41.81) = 83.62 degrees
  • 25. Determining Index of Refraction Using Thick Prisms  Spectrometer can be used to narrow the wavelength of light to minimize chromatic aberrations.  Dmin can be determined
  • 26. (α + min) β n2 sin = 2 n1 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 27 degrees. What is the index of refraction of the material which 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 np β = (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] α β = [(1.49/1.33) -1] 8 β = 0.96° (in air) β = [(np/ns) – 1] α β = [(1.49/1.00) -1] 8 β = 3.92°
  • 33. Displacement (cm) β d l Distance (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 prism diopters. Solution: (in water) β = [(np/ns) – 1] α β = [(1.49/1.33) -1] 8 β = 0.96° Δ = 100 tan β = 1.68Δ (in air) β = [(np/ns) – 1] α β = [(1.49/1.00) -1] 8 Δ = 100 tan β = 6.85Δ β = 3.92°
  • 37. A glass ophthalmic prism (n=1.52) displaces the image of an object 0.037m at 2.2m. 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 m Δ = 1.68Δ β d Δ = 100 tan β 1.68 = 100 tan β β = 0.96° or 57.6’ l Distance (m)
  • 38. Prism Base Directions re Patients
  • 39. BU Patient BU OD OS BO BI BI BO BD BD
  • 40. Patient B@135 B@135 B@45 B@45 OD OS B@225 B@315 B@225 B@315
  • 41. Patient 90 90 135 45 135 45 180 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!
  • 47. β = Rotational Stimulus h β EYE l The image displacement, the eye’s rotational stimulus, and the actual rotation of the eye can be expressed in 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>βe h 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 βe β 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 0.9 0.85 0.8 0.75 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 distance 0.95 0.9 0.85 0.8 0.75 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 the center of rotation of the eye, when the wearer reads at a distance of 40 cm from the prism? Solution: βeΔ = 15 / 1 + (2.5/40) βeΔ = 14.12Δ Loss 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 double vision at near. He wears glasses for his myopia and has been told that he wears 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 180 You find that he needs an addition prism diopter prism at near to maintain clear, single, and comfortable vision. Visual need = 6Δ at near What do his current glasses give him at distance and near (30cm and vertex 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 double vision at near. He wears glasses for his myopia and has been told that he wears prisms in his glasses. What are you going to do? What prism power would yield 6Δ at the working distance and vertex distance 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
  • 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
  • 67. Method 1  complete the parallelogram  total vector is arrow from origin to intersection point
  • 68.
  • 69. Method 2  move end of one vector to tip of other vector  total vector is arrow from origin to tip of moved vector
  • 70. ∆ y2 ∆ ytot ∆ y1 ∆ x1 ∆ x2 ∆ xtot
  • 71. Find the single prism equivalent to a 2Δ BI combined with a 5Δ BU in front of the left eye. 5.4Δ@112 5 Patient 90 90 135 45 135 45 180 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
  • 72. Adding Prisms Applied to an Eye  Adding prisms in opposite base directions is subtractive  Adding prisms in same base directions is additive
  • 73. Clinical Application  Prism Prescribing and Analyzing  Lensometry  Ophthalmic Optics and Opticianry
  • 74. 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
  • 75.
  • 76. y ∆1 φ1 ∆ tot x φ2 = - φ1 ∆2 ∆2 = ∆1
  • 77. ∆ tot = 2∆ 1 cos φ 1
  • 78. Each component of a Risley Prism is a 10Δ prism. The Risley is set so that both prisms are BD. What is the maximum prism power? What is the power when they 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
  • 79. 0 BO BI 10Δ Risely Prism over OD y ∆1 φ1 20 ∆ x 20 φ2 = - φ1 ∆2 = ∆1
  • 80. 20 10Δ Risely Prism over OD BU y 0 ∆2 = ∆1 ∆1 φ φ2 = - φ1 1 BD ∆ 20 x

Editor's Notes

  1. This also makes the internal angles equal.
  2. This slide looks at the incident angle relationship to deviaiton
  3. Negative incident angle
  4. The 90 degree incident and emergent angles are only theoretical examples. Any incident angle less than 90 (in the case above) would create TIR at the second surface. Therefore, the example above demonstrates the case where maximum deviation occurs and is limited by the apical angle of the prism. The largest incident angle (90) creates a situation where we have the bests opportunity to refract light through the prism. Angles less than this are more likely to cause TIR. Therefore, we can use the critical angle its relationship to the apex angle to determine if a prism has any ability to refract light or if it will only be a reflecting prism. Alpha = 2 x Critical angle gives us this relationship
  5. Remember Prism Diopter is cm/m
  6. Remember Prism Diopter is cm/m
  7. Blue line represents a prism 10 cm away from the eye. The red line represents a prism 6 cm away from the eye. Notice that the further you move the prism away from the eye for any near object, the less effective the prism is.
  8. Clinically we may have a patient that requires 6 prism diopters of prism as the prism demand for the eye. If that patient picks up a book to read, they my become diplopic because the prism no not providing them 6 prism diopters but some amount less than that.
  9. Don’t over counter rotate the risley prism