VISUAL OPTICS II PROBLEM SHEET

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VISUAL OPTICS II PROBLEM SHEET Powered By Docstoc
					 ANSWERS TO VISUAL OPTICS PRACTICE PROBLEMS November 2007
                                                                                                                                          Page
Chapter 1         Paraxial Optics – Less than Meets the Eye .................................................................2
Chapter 2         Schematic Eyes and Introduction to Ametropia; Entoptic Phenomena .....................3
Chapter 3         Retinal Image Quality .................................................................................................9
Chapter 4         Emmetropia and Ametropia ......................................................................................15
Chapter 5         Introduction to Astigmatism; the Cornea; Astigmatic Eyes, Lenses & Image
                  Formation ..................................................................................................................16
Chapter 6         Astigmatism and Subjective Refraction....................................................................21
Chapter 7         The Pupil ...................................................................................................................23
Chapter 8         Accommodation and Near Vision in Corrected Ametropia ......................................25
Chapter 9         Intraocular Lenses and Refractive Surgery ...............................................................28
Chapter 10 Ophthalmic Prisms .................................................................................................. NA
Chapter 11 Retinal Image Size and Aniseikonia .........................................................................29
Chapter 12 Blur Circles and Depth of Focus ...............................................................................31
Chapter 13 Reflection in Visual Optics .......................................................................................33
Chapter 14 Binocular Indirect Ophthalmoscopy and Retinoscopy..............................................35
Chapter 15 Polarization ............................................................................................................ NA




                                                                       1
    Chapter 1 Answers

1   B
2   C
3   B
4   A
5   A
6   D
7   C
8   B
9   D
10 C
11 A




                        2
Chapter 2 Answers

1.   B
2.   C   Keratometry requires separation of the anterior cornea from the rest of the eye (Exact
         Eye).
3.   B   A thick lens consists of more than one refracting surface and at least two different
         refractive indices (SSE). The reduced eye is NOT a thick lens (single refracting surface).

4.   D
5.   A
6.   B
7.   C
     A  Femm  Fe  Femm  A  Fe   3.75  62.5   58.75 D  Femm
               n              n      1.333
     Femm           ax                      22.70 mm  ax
              ax             Femm    58.75 D


8.   D
          n       1.333
Femm                           61 D         A  Femm  Fe   61  59   2 D (hyperopia)
         ax   21.86  10 3 m
Based on the +60 D standard emmetropic reduced eye, this eye has +1 D axial hyperopia (1 D too
short), and +1 D refractive hyperopia (1 D too “weak”).

9.   A         LMR = A




                                                3
10. D

               n        1.333
     Femm                             55 D
              ax    24.24  10 3 m

     A  Femm  Fe   55  53   2 D  FO (ocular correction)

    Using the effective power formula :

               FO                2                               1        1
     FS                                     1.95 D   f S                   51.4 cm
            1  d FO       1  0.014  2                        FS    1.95 D


By focal lengths




                                                  4
11. A    The reduced surface sits 1.67 mm behind the cornea so the correct “vertex distance” to
         convert between spectacle correction and far point vergence is 16 mm:
                             FS              3.50
           LMR  FO                                          3.31 D
                          1  d FS   1  0.016  ( )3.50 


By focal lengths




                                              5
                      LMR           7.50
12.   B    FS                                   6.83 D
                   1  d LMR   1  0.013  7.50


By focal lengths




                                         1        1
           FS   3.75 DS       f S                   266 .7 mm
                                         FS    3.75 D
                                          1       1
13.   C    FO   3.55 DS        
                                fO                     281 .7 mm
                                         FO    3.55 D
           vertex distance  281 .7  266 .7  15 mm




                                               6
                             FS             8.00
           A  FO                                        8.89 D
                          1  d FS   1  0.0125  8.00
14.   A    A  Femm  Fe                 Femm  A  Fe   8.89  54   62.89 D
                      n      1.333
           ax                       21.20 mm
                     Femm    62.89 D

By focal lengths




15.   D
16.   B
17.   C
18.   C
19.   A
20.   A
                                         450  10 9
21.   A            sin  max1                   6
                                                       0.045    2.58
                                  d       10  10
22.   A
23.   C

                                                    7
Chapter 3 Answers
1.   B    coherence length increases with decreasing bandwidth
2.   C
3.   A    geometry of the interference set-up relates to spatial coherence only
4.   D
5.   B    for example, at the bottom of the slit, a point on the wavefront propagates a spherical
          wavefront, part of which travels downward

6.   A    after the first order minimum, the greatest intensity corresponds to an entire crest (or
          entire trough) that is not canceled (BC = 1.5  ). All other phase differences across
          the slit (between BC =  and BC = 2 ) will produce a greater degree of destructive
          interference.




                               587.6  10 9 m
7.   A    sin min                                 min  2.87 
                        d       11.75  10 6 m




                          587.6  10 9 m
8.   C    sin min                                 min  11.53
                       d   2.94  10 6 m




                        1.22    1.22  587.6  10 9 m
          sin m in                                          m in  0.014
                           d           3  10 3 m
                        x ( source separation)
9.   C    tan m in                            x   tan m in  6 (m) tan 0.014
                         (observer distance)

          x  1.45  10 3 m  1.45 mm

                                                  8
                              x retina (retinal Airy disc separation)   x
             tan min                                                 retina
                              NF  (distance nodal point to retina)      fe

10.   B            x retina   f e tan min   16.67 mm tan 0.014

             x retina  4.02 m




11.   C Above 1.3 mm, aberrations also affect resolution


                                                  1.22         1.22  587 .6  10 9 m
          sin  m in  sin  radius Airy Disc                                          7.16  10 7
                                                  d objective            1m
12.   C
           m in  4.11  10 5   0.148 sec arc
                                  




13.   B Must magnify 0.148 seconds of arc to 49 seconds of arc, which is ~ 300
14.   A Both have the same minimum angle of resolution
15.   B
16.   D
17.   B
18.   B
19.   E same in all four cases (interferences fringes equally spaced)



                                                           9
20.   D Petzval’s surface is the surface on which the lens (free of oblique astigmatism) focuses
        parallel light for all possible directions of incidence. The Far Point Sphere is the surface
        for which the retina is focused for all possible directions of gaze. This answer is the
        “all-directions” equivalent of the definition of spectacle correction of ametropia: “place
        the second focus of the spectacle correction at the patient’s far point.” Petzval’s surface
        is “all the possible locations of the second focus of the spectacle correction for all
        possible directions of incidence of parallel light” and the Far Point Sphere is the surface
        for which the retina is focused for all possible directions of gaze.

21.   B
22.   A see answer to question 20
23.   D see answer to question 20
24.   A Oblique astigmatism is quantified as the (dioptric) separation of tangential and sagittal
        focal lines. It decreases to zero when the tangential and sagittal focal planes (and
        therefore focal lines) coincide.

25.   C 517642 means lens index 1.517 and Abbe Number 64.2 (low disperion). This should be
        combined with a high dispersion negative element to provide high dispersion but lower
        net power. The most appropriate choice is 620364 (Abbe Number 36.4, high
        dispersion).

26.   A due to the abrupt change in contour at the edge of the ablation zone


                         
27.   B sin  min 
                         d


                               x (distance on screen )
28.   A tan  m in 
                          L (distance from slit to screen )




                                                  1.22         1.22  500  10 9 m
          sin  m in  sin  radius Airy Disc                             3
                                                                                      2.03  10 3
                                                  d objective       0.3  10 m

29.   B  m in  0.1165           This is Airy Disc radius. Double to get diameter

          Diameter of Airy Disc  0.1165  2  0.233 


                                                           10
                                   x (distance on screen )
          2  tan  min 
                              L (distance from slit to screen )
30.   C
           x  L  2 tan  min  7.5 m  tan 0.233  3.05 cm




                                                 1.22    1.22  587.6  10 9 m
          sin  min  sin  radius Airy Disc                         3
                                                                                  2.39  10 3
                                                    d          0.3  10 m

31.   C  min  0.137

                        x
          tan  min       x  L tan  min  10 (m) tan 0.137   2.39 cm
                        L


32.   D SA is the main reason; CA is next


                                                      TSA1   TSA2
          TSA  y 3  k y 3               k           3
                                                              3
                                                       y1     y2
33.   C
                                   3
                                 y2              2.5 3
          TSA2  TSA1                0.35 mm         0.684 mm
                                 y13              23




                                                       LSA1   LSA2
          LSA  y 2  k y 2                k           2
                                                               2
                                                        y1     y2
34.   D
                                    2
                                  y2           2.5 2
          LSA2  LSA1                 100%         156.25% or 56.25% increase
                                  y12           22


35.   D
36.   A SA is a longitudinal spread of foci as a function of aperture diameter. TSA allows us to
        capture the image on a screen and evaluate the resulting image degradation




                                                         11
37.   D Positive spherical surfaces focus more peripheral rays further and further left. With a
        larger pupil, a patient with little peripheral corneal flattening will have a large spread of
        foci and the best form of the image (waist of least aberration) will be shifted
        significantly to the left of the paraxial focus, making the patient effectively myopic.


38.   C Theory of aperture “shells” and their corresponding comatic circles
39.   B Remember to divide the new tangential coma value by 3 to get sagittal coma
40.   A Greater thickness to traverse; reduced effective aperture (steeper incidence)
41.   D Same theory as Q40
42.   C “Angle of obliquity”-dependent aberrations
43.   D
44.   B Microscopes are large aperture systems
45.   D
      D D  h   k h 
                    3          3
46.

47.   C   LSA  y 2 ; TSA  y 3            LCA aperture  independent; TSA  y


48.   E SA, coma, and OA all degrade image points; curvature of field degrades image quality
        of object planes (curved image planes); distortion produces no degradation of image
        quality.

49.   A This is the Resting State Theory of Accommodation. Overaccommodation at distance
        means the paraxial focus is in front of the retina, so longer wavelengths (from the object
        plane) are focused on the retina. Progressively “increasing” underaccommodation for
        near vision moves the paraxial focus progressively further behind the retina, resulting in
        shorter and shorter wavelengths focusing on the retina.

50.   C LCA is the only one of the listed aberrations that is a property of light, not the eye
51.   B Coma is one of the two transverse Seidel aberrations (the other is distortion). Coma
        produces variable transverse magnification with incident height (how far out in the
        aperture the ray is incident). Dividing a lens into “aperture shells”, the smallest shell
        produces the smallest comatic circle; the largest (most peripheral) aperture shell
        produces the largest comatic circle, and is used to quantify the resulting comatic pattern.
        Note that the magnification produced by distortion varies with paraxial object/image
        height, NOT incident height. Distortion is affected by aperture POSITION, not
        diameter.
                                                 12
                                                       k  I scat  4  I scat 1 1  I scat 2 4
                                1           k
                 I scat                                                            4

                                           
                                   4            4                                                    2

52.   (I)   D
                                                    1
                                                     4
                                                               5554
                I scat 2    I scat 1              3.5        10.67 cd
                                                    4
                                                     2         4204




                                                       k  I scat  4  I scat 1 1  I scat 3 3
                                1           k
                 I scat                                                            4               4

                                  4
                                               4

      (II) A
                                              1
                                               4
                                                        5554
                I scat 3    I scat 1    4  3.5         1.55 cd
                                              3        680 4


53.   B
54.   D Mie scatter is directional and weakly wavelength-dependent
55.   (a) Red (b) Green (c) BlueShorter wavelength deviated through larger angle


                                                           d green            5.25
      B d  n  1                                                             10
56.
                                                     n   green    1        0.523


57.   A
58.   A
59.   B Note that the 587.6 nm term allows us to ignore chromatic aberration, not Seidel
        aberrations

60.   D x0 is often expressed in terms of field angle (OA and curvature of field)
61.   B Tangential and sagittal astigmatic error are equal and opposite meaning that the circle of
        least confusion is falling on the Far Point Sphere. This corrects curvature of field, but
        leaves residual OA

62.   A One curve, showing progressive undercorrection of oblique error. This means OA is
        fully corrected; curvature of field is not (Petzval’s Surface flatter than FPS).

63.   D No oblique error; one line. OA fully corrected; Petzval’s surface matches the FPS, so
        curvature of field is also fully corrected.

                                                                     13
Chapter 4 – Ametropia Answers
1.    B     Moderate to high hyperopic refractive errors tend to remain stable until middle age
2.    C     High myopia tends to be progressive at least through the school years
3.    A     Progressive high myopia is believed to result from habitual underaccommodation
            in near vision. The focused image sits behind the retina, stimulating axial growth
            “toward the focused image.” To prevent this, a reduced myopic correction could
            be prescribed, moving the image forward to the retina. This is the theory behind
            progressive addition lenses in progressive myopes (latest COMET data, 2004,
            suggests that PALs may work in an “underaccommodating subset of myopes”).
4.    D     Both genetic factors (myopic parent(s)) and environmental factors (e.g. “excessive
            reading”) are important, but their relative contributions are unknown.

5.    D     Myopia lower in magnitude than 5 D (correlation myopia) tends to be a non-
            predictable mix of axial and refractive components.
6.    D     The patient is hyperopic in both eyes, therefore a myopic crescent (conus) would
            not be expected. All other tests could help determine refractive and axial
            components of ametropia.

7.    D     Applanation biometry, especially in the hands of an inexperienced operator tends
            to underestimate ocular axial length due to indentation of the cornea.




                                             14
Chapter 5 – Answers

1      C
F90  F sin 2   F sin 2 50    2  0.587   1.17 D

       ( measured from cylinder axis)




2      A
F30 (cyl )  F sin 2 60   3  0.75   2.25 D
F30 (total)   2.25 D  3.00 DS   0.75 D



Shortcut: +3.00 3.00 axis 90 transposes to Plano +3.00 axis 180  +3.00 DC axis 180
For  3.00 DC axis 180 :    F30  F sin 2 30    3  0.25   0.75 D


3.     B Astigmatism (in terms of Fe ) almost identical to ocular surface astigmatism.




                                              15
4.   B Index difference between tear film and cornea is the same as that between cornea and
       aqueous. Posterior corneal power (5.88 D) is low, but NOT negligible.

5.   D At age 10, with-the-rule total ocular astigmatism in the range of 0.50 to 1.00 D is very
       common (anterior cornea steeper vertically, lower [atr] physiological astigmatism).
       Wtr total ocular astigmatism decreases, but remains predominant, until the mid 40s,
       after which against-the-rule total ocular astigmatism becomes more common. A
       patient who has 2 D wtr astigmatism at age 10 (higher than average) is likely to still
       have wtr, but of lower magnitude, at age 45 years).
6.   A At age 55, the average astigmat has with-the-rule corneal and against-the-rule total
       ocular astigmatism.

7.   C Similar lens power ranges could be covered with front or back surface toric lenses

8.   C




                                             16
9.    B
                                                      n       1.333
      F170   9.50 D     F80   3.50 D    Femm                           56 D
                                                     ax   23.81  10 3 m
      A  Femm  Fe        Fe  Femm  A
      Fe (170)   56  (9.5)   65.5 D            Fe (80)   56  (3.5)   59.5 D
                         n       1.333
      f ' e (170)                         20.36 mm
                      Fe (170)    65.5 D
      Position of Anterior FL : (23.81  20.36)  3.45 mm in front of retina
                                       F (170)  Fe (80)            65.5  59.5 
      Anterior FL Diameter : h170  y  e
                                                          4 mm  
                                                                                    0.366 mm
                                            Fe (170)                    65.5    

10.   D (Focal line and COLC diameters all vary in size with pupil diameter)




                                            17
11.   A Optimum reading vision at any distance for an uncorrected astigmat occurs when a
        vertical focal can be placed on the retina. A young “simple” astigmat who cannot
        place a vertical focal line on the retina in distance vision must be a simple myopic
        astigmat (young hyperopic astigmats should have plenty of accommodation to bring
        the posterior focal line onto the retina). The choice is between the myopic astigmats.
        Which one has a vertical anterior focal line? This corresponds to greater ocular power
        in the horizontal meridian, which defines against-the-rule astigmatism. In near vision,
        the against-the-rule myopic astigmat could simply vary reading distance to place the
        vertical focal line on the retina.




                                             18
12.   B




          19
13.   C

                     n        1.333
           Femm                            57 D       A  Femm  Fe
                    ax   23 .39  10 3 m

           A90   57  58   1 D myopia           A180   57  61   4 D myopia

          A90 = 1 D; A180 = 4 D. At a 50 cm viewing distance, horizontal parts of letters
          would be clearer (focused closer to retina)




14.   C A characteristic of uncorrected astigmatic vision is blur at all distances.




                                               20
Chapter 6 - Astigmatism and Subjective Refraction - Answers

1. B With the JCC handle parallel to the trial cylinder axis, cylinder axis is being refined
     (obliquely crossed cylinders). Chasing the minus, for the “second” view the negative JCC
     axis is rotated toward 90 from the 45 negative trial cylinder axis. Therefore, rotate the
     trial cylinder axis toward 90 (55).



                 n        1.333
       Femm                            57 D     A  Femm  Fe
                ax    23.39  10 3 m
2. B   A70     57  58   1 D myopia       A160   57  61   4 D myopia
                A70  A160   1  4
       BVS                          2.5 DS
                    2          2


3. D The patient’s response indicated that less negative trial cylinder power was preferred.
     This corresponds to the positive JCC axis aligned with the negative trial cylinder axis at
     90. Therefore, the negative JCC axis must have been oriented at 180.

4. D This question makes the point that the COLC is a non-orientation-dominant image, while
     focal lines are orientation-dominant. The safest way to assure that patients do not base
     their JCC responses on one or other focal line (which can happen if the COLC is NOT on
     the retina) is to use circular viewing targets (or groups of dots).

5. D Reducing the cylinder power in the patient’s correction from 5.5 DC to 3.0 DC amounts
     to removing 2.5 diopters of cylinder. The equivalent sphere to 2.5 DC is 1.25 DS.
     Because this has been removed in the reduced astigmatic correction, it must be
     incorporated into the sphere component. The new, partial astigmatic correction is +3.25
     3.0  30.

6. D You are refining axis (obliquely crossed cylinders between TCA and JCC negative axis).
     With TCA 80 and JCC negative axis 35, the resultant will be between 80 and 35.
     Therefore, rotate the TCA toward 180

7.     Patient’s current Rx in minus cylinder form is 1.50 2.00  115. New Rx will be
       0.50 2.50  105.

       (A) Axis refinement: JCC handle at 115, patient wants axis rotated toward 105.
           Therefore minus JCC axis 45 counterclockwise from handle = 70

       (B) Power refinement: JCC axes 105 and 15 (parallel and  to TCA). Patient wants
           more cylinder; therefore prefers minus JCC axis along 105
                                               21
8. D any letters that are predominantly parallel to or perpendicular to the patient’s PM’s will be
     parallel / perpendicular to their focal lines. With the COLC behind the retina, the “wrong”
     choice for cylinder power (increases interval of Sturm) may push the anterior focal line
     forward to the retina. Conversely, the posterior focal line can always be moved to the
     retina by accommodation (especially younger patients).

9. (A) True. The higher the cyl, the more the image degrades for a given cyl axis rotation
      (B) False. Either works fine
      (C) False. Better to have the COLC 0.125 D behind the retina and allow minimal
          accommodation to bring it forward to the retina. If the COLC is set 0.125 D in front of
          the retina, the patient cannot bring it “back” to the retina
      (D) False. With the COLC on the retina, reducing the interval of Sturm will give better
          vision regardless of which side of the retina the focal lines are located
      (E) True

10.    B Patient prefers JCC negative power meridian 135 (axis 45). The JCC PM’s are aligned
         with the patient’s PM’s  power refinement. Preference for negative JCC axis 45
         means there was too little negative cylinder in the phoropter (Less than 3.00 DC).

11.    C First pair with handle at 90                 135 is closer to 110 than is 45
         Second pair with handle at 45  90 is closer to 110 than is 180

12.    D Second view has negative JCC axis at 55. Patient wants more negative cylinder



13.    B To start the Fan and Block (Astigmatic Dial) Refraction, the posterior focal line should
         be moved to the retina. 20/60 vision with BVS (COLC on the retina) corresponds to 2 D
         predicted astigmatism. To move the posterior FL to the retina, add plus sphere to the
         BVS equal to one half predicted astigmatism: +2.75 (BVS) + 1.00 = +3.75 DS starting
         sphere. Note that the additional +0.50 D fog to move the posterior FL in front of the
         retina (and verify focal line “identity”) is a subsequent step and is not used for the first
         view of the Fan Chart.

14.    C The Fan and Block Method relies on placing the posterior focal line on the retina. If this
         is not done correctly, the method is likely to yield inconsistent or inaccurate patient
         responses. Recheck BVS power before continuing.

15.    D Equivalent to the Paraboline Arrow (Raubischeck Arrow) used to refine cyl axis.




                                                 22
Chapter 7 - The Pupil - Answers

1. B The entrance pupil is the image of the real pupil seen (by an observer) through the cornea.
     Because the pupil is imaged through the cornea, EnP position will be influenced by
     corneal power (and through the lateral magnification m = L/L) EnP diameter will be
     affected by corneal power. An astigmatic cornea has different corneal powers between
     PM’s; therefore EnP position and diameter will vary between PM’s. In highly astigmatic
     patients, the EnP will actually appear elliptical. The exit pupil is the image of the pupil
     through the crystalline lens. Its position and diameter is unaffected by the cornea.

2. B Image the pupil (in aqueous object medium) through the cornea into air (image medium).
                                                     naqueous             1.336
              3.2 mm ( AC depth)          L                                       417.5 D
                                                                      3.2  10 3 m
           L   L  Fcornea   417.5  43.08   374.42 D
                nair        1.000
                                     2.67 mm
                 L       374.42 D
           The EnP is closer to the cornea than the real pupil



3. C Use the Lateral magnification equation to find EnP diameter:
                  L     417.5               h   EnP diameter
           m                    1.115     
                  L    374.42              h    Pupil diameter
           h  h  m  4.0 mm  1.115  4.46 mm



4. A Image the pupil (in crystalline lens medium) through the posterior lens into vitreous:

                                                  ncr. lens           1.413
          4.0 mm (lens thickness)        L                                     353.25 D
                                                                  4.0  10 3 m
       L   L  F post . lens   353.25  15.40   337.85 D

              nvitreous      1.336
                                   3.95 mm from posterior lens (0.05 mm behind ant. lens)
                L         337.85 D

        ExP location : 3.2  0.05  3.25 mm right of cornea

       Because of minimal posterior lens power, the ExP is almost coincident with the actual
       pupil.
                                                23
5. B Use the lateral magnification to find ExP diameter:
               L     353.25              h   ExP diameter
         m                   1.046     
               L    337.85              h    Pupil diameter
         h  h  m  4.0 mm  1.046  4.182 mm


6. B

7. C

8. D

                                       
9. B axial length  Cornea  H eye  f eye


10. B     Focused image height for a distant object:
                              1   
                                    tan 0   
                                                     1 
         h  f e tan 0   
                             F                          tan 2.5   0.640 mm
                              e                 68.23 

          This equation finds focused image height for a distant object in all optical systems.


11.    C Changing pupil diameter has no effect on retinal image height. However, if the retinal
         image is blurred, the size of retinal blur circles will change in direct proportion to the
         change in pupil diameter.




                                                 24
Chapter 8 – Crystalline Lens, Accommodation, and Near Vision – Answers

1. D Both crystalline lens surfaces have positive power. Decreasing either radius increases
     surface power. Equivalent power for the eye varies with the distance between cornea and
     crystalline lens (actually H of the crystalline lens). If the crystalline lens moves closer to
     the cornea (decreased anterior chamber depth), equivalent power of the eye increases by
     decreasing the t/nF1F2 term that is subtracted from F1 plus F2 for the eye. Increasing the
     thickness of the crystalline lens increases the t/nF1F2 term that is subtracted from F1 plus
     F2 for the crystalline lens.


2. A Accommodation changes retinal conjugacy (focused object plane) from far point to near
     point:

        Amp Accom (O)  Far Point Vergence  Near Point Vergence  LMR  LMP

                  1           1
        LMR                        2.5 D
                 MR        40 cm

        LMP  LMR  Amp Accom (O)   2.5  4.0   6.5 D

                 1       1
        MP                   15.38 cm
                LMP    6.5 D

        ROCV from 15.38 cm (near point ) to 40 cm ( far point )


3. C Correcting an ametrope with an ocular correction (for distance) makes them the same as
     an emmetrope. Clear vision will be obtained from infinity (distance) into a near object
     plane that is the reciprocal of the ocular amplitude of accommodation:

                                            1              1
                LMP (corrected )                           20 cm
                                       Amp Accom (O )     5D

                LMR (corrected )  in finity ( distance)

Note that when uncorrected, this patient will see clearly only at distance (infinity) because ocular
amplitude of accommodation matches hyperopia.


4. A Calculations show an effectivity advantage for the spectacle-corrected myope. Less ocular
     accommodation is required to view a given near distance (4.32 D) compared to an
     emmetrope (4.87 D required). Therefore, this question is demonstrating that the spectacle-
     corrected myope would have a lower Amp Accom (O) than an emmetrope with the same
     “near” point (uncorrected of course for the emmetrope vs. spec-corrected for the myope).
                                              25
5. C Calculations show an effectivity disadvantage for the spectacle-corrected hyperope. More
     ocular accommodation is required to view a given near distance (3.3 D) compared to an
     emmetrope (2.89 D required).

6. B Spectacle-corrected hyperopia creates an effectivity disadvantage in near vision that makes
     the hyperope exert more ocular accommodation to focus a given near distance. The effect
     increases with magnitude of hyperopia, so the +8.25 D hyperope requires more ocular
     accommodation to focus 20 cm from the spectacle plane. Translation: the +8.25 D
     hyperope is younger (ocular amp Accom is higher in younger patients). Note the
     similarity to question 4.

7. False.     Myopes lose their effectivity advantage when they switch to contacts, bringing on
              presbyopia at an earlier age than with a distance spectacle correction.

8. False      Hyperopes and myopes corrected with contacts are very similar to emmetropes;
              negligible effectivity.

9. True.      Changing a hyperope from spectacles to contacts removes the effectivity
              disadvantage.


                                        1         1                   3
                 Amp Accom ( S )                    2 D           Amp Accom ( S )  1.2 D
                                        S      0.5 m                5
                      
                Want LS through near add   1.2 D
10.    C
                                                      1                1
                 LS from working distance                                    3.0 D
                                               working distance    33.33 cm
                               
                 FADD ( S )  LS  LS      3  (1.2)   1.8 D

11.   False.    More positive distance correction has a more remote near point

12.   True.     All spectacle distance-corrected astigmats have “reduced effective cylinder” at
                near with their distance correction (no near add).

13.   False     Contact lenses almost entirely remove the effectivity-induced imbalance.

14.    B        Effectivity “disadvantage” in the more hyperopic principal meridian.
15.    B        The main disadvantage of the ROCV method is variable illumination with distance
16.    A
17.    C

                                                26
18.   C   Total hyperopia is revealed with cycloplegic refraction. With accommodation
          paralyzed, optimum VA will be obtained only with the “total” hyperopic
          correction.
19.   D   Manifest hyperopia is what is seen during routine, non-cycloplegic refraction. The
          highest plus lens power that does not decrease optimum VA is equal to the
          manifest hyperopia (manifest refraction).

20.   B   Latent hyperopia is the difference between total and manifest hyperopia
21.   B   Absolute hyperopia is that part of the manifest hyperopia that cannot be overcome
          by accommodation. The minimum amount of plus power that must be added
          under non-cycloplegic conditions to give optimum VA is equal to absolute
          hyperopia (here 1 D).

22.   C   Facultative hyperopia is that part of the manifest hyperopia that can be overcome
          by accommodation (4  1 = 3 D facultative hyperopia).

23.   C   Fractions of ocular and spectacle amplitude are almost identical

24.   C   accommodative imbalance becomes worse as the actual amount of accommodation
          increases




                                           27
Chapter 9 - Intraocular Lenses - Answers

1. A

2. C

3. B

4. C

5. D

6. D      (sorry about the overlap)

7. B




                                           28
Chapter 11 - RI Height and Aniseikonia - Answers

Q1.   D    aniseikonia is underdiagnosed because symptoms are vague and non-specific, and
           therefore of limited diagnostic value.
Q2.   A    Uncorrected retinal image height (reduced eye) is the same in all refractive
           ametropes because axial length is the same. Chief ray path for a given object is
           therefore constant for all eyes with refractive ametropia.

                FS             5.25
      FO                                    4.891 D
             1  d FS   1  0.014  ()5.25
      A  Femm  Fe       Refractive ametropia : Femm   60 D  ax  22.22 mm
       
      hU   M CR  ax tan  0   0.75  22.22 mm tan 2.5   0.728 mm

Q3.   B   Negative lenses deviate light away from the axis. This produces a shallower incident
          (and therefore refracted) chief ray path.

Q4.   B   In axial anisometropia, the more hyperopic eye is shorter, producing a smaller
          uncorrected retinal image height. Contacts have only a slight effect on image heights,
          so the more hyperopic eye has a smaller retinal image when corrected (see VOO, pp.
          20-22).
Q5.   D   The spectacle magnification produced by spectacles in Knapp’s plane (Feye) exactly
          compensates for differences in uncorrected retinal image height (Knapp’s Law).

Q6.   B   Uncorrected retinal image heights are equal in refractive anisometropia. Spectacle
          lenses cause greater spectacle magnification in the more hyperopic eye, producing
          aniseikonia.
Q7.   C   Contacts produce slight changes from the equal uncorrected retinal image heights in
          refractive anisometropia.
Q8.   A   Uncorrected retinal image heights are equal in refractive anisometropia. Spectacle
          lenses cause greater spectacle magnification in the more hyperopic eye, producing
          aniseikonia.
Q9.   D   Aniseikonia can be induced by “unaware” practitioners




                                             29
Q10.   B

                 FS             16.25
       FO                                     13.07 D
              1  d FS   1  0.015  ()16.25
       A  Femm  Fe        Axial ametropia : Fe
                                                                   n     1.333
       Femm  A  Fe   13.07  60   46.93 D ax                             28.41 mm
                                                                  Femm    46.93
        
       hU   M CR  ax tan  0   0.75  28.41 mm tan 3   1.117 mm

Q11.   C

Q12    A   all refractive ametropes have equal uncorrected image heights (for a given object
           angle). Therefore, differences in corrected retinal image height in a refractive
           anisometrope must be the result of differences in spectacle magnification.

Q13    D   equal retinal image heights producing retinal iseikonia (satisfying classical theory)
           and falling on equal numbers of receptive fields producing cortical aniseikonia

Q14    A   equal retinal image heights falling on unequal numbers of receptive fields producing
           cortical aniseikonia

Q15    C   unequal retinal image heights (retinal aniseikonia), but falling on equal numbers of
           receptive fields to produce cortical iseikonia

Q16    B   unequal retinal image heights producing retinal aniseikonia, and falling on unequal
           numbers of receptive fields (RF size same OD and OS) producing cortical aniseikonia

Q17    A   Eikonometry gives a direct subjective measure of aniseikonia as perceived by the
           patient. This is cortical aniseikonia.

Q18    D   Same basis as Q17

Q19    B   If the goal is emmetropia, there would be no need for distance correction after
           surgery, so the comparison is between corrected retinal image height prior to surgery
           and “uncorrected” retinal image height after surgery.




                                              30
Chapter 12 - Answers

Q1.    B   The patient (with the same sensitivity to blur) focused at the longer distance from the
           eye, will have the greater linear depth of field
Q2.    C   Two patients with equal sensitivity to blur and identical eyes will have the same
           dioptric depth of focus.
Q3.    C    Two patients with equal sensitivity to blur and identical eyes will have the same
           dioptric depth of field.
Q4.    C   Two patients with equal sensitivity to blur and identical eyes will have the same
           linear depth of focus.
Q5.    B   Dioptric depth of focus varies inversely with pupil diameter. Large pupil means
           smaller depth of focus.
Q6.    D
Q7.    B   Hyperfocal plane has an object vergence equal to the dioptric depth of focus.
           Therefore, as the pupil constricts, and dioptric depth of focus increases, the object
           vergence of the hyperfocal plane increases. Translation: the hyperfocal plane moves
           closer to the eye.
Q8.    A   Human eyes do in fact exploit depth of focus exactly as described in this problem.
           For distance vision, the (emmetropic) eye focuses in the hyperofocal plane, giving
           clear vision back to infinity and maximizing linear depth of field. In near vision, the
           eye tends to underaccommodate, placing the focused image progressively further
           behind the retina, but within the depth of focus range. This is the resting point theory
           of accommodation; it allows maximum ROCV at distance and ensures optimum
           economy in accommodative effort.
Q9.    B
Q10. B (A would be correct for axial myopia)
Q11.   C
Q12.   C
Q13.   A
Q14.   A
Q15.   B
Q16. A
Q17.   A
Q18.   D             see answer to question 21.

                                               31
Q19.   D
Q20.   A
Q21.   C   this is why clinical measurement of near adds is valid.
Q22.   A
Q23.   C   accommodative imbalance becomes worse as the actual amount of accommodation
           increases




                                               32
Chapter 13 - Answers

Q1.     C      The keratometer equation “assumes” that the virtual corneal image sits at the focus
               of the corneal mirror, a distance x from the mire object. Because the mire is not at
               infinity, the actual virtual image sits in front of the mirror focus. No image is
               present at the corneal focus, so there is no way to set the distance x with the
               keratometer.

Q2.     A      the virtual reflected image is optically equivalent to a real object in air sitting
               slightly behind the corneal vertex plane.
Q3.     C      Prism further from the graticule at 90, so x (image displacement) greater at 90, 
               h (graticule image height) larger at 90  h (virtual reflected corneal image height)
               larger at 90  anterior corneal radius (r) longer at 90. Longer radius means flatter
               (lower power). The anterior cornea has lower power at 90 than 180, which is
               against-the-rule astigmatism

Q4.     A      With the graticule plane images separated in both PMs for the right cornea and
               prism positions still set for the left cornea, the right anterior cornea must be steeper
               in both meridians than the left. It therefore has greater refractive power in both
               PMs than the left. This cornea MAY be spherical, but the information provided
               does NOT tell us that it MUST be spherical.

Q5.     B      With-the-rule means greater power vertically, therefore steeper radius and smaller
               h and h vertically. The image will be a horizontally elongated ellipse. The
               scissors appearance disappears as the instrument rotates into alignment with the
               PMs.

Q6.     A      Base-down single half-field prism deviates half the image rays down. ¾ of the
               composite image is below axis.

Q7.     C      One-position keratometers are of no use in evaluating irregular astigmatism
               because the doubling systems are locked 90 apart.


                      ncal  1     1.3375  1
             K 90                                45.61 D
                        r90       7.4  10 3 m
Q8     A
                      ncal  1     1.3375  1
             K180                                43.27 D
                        r180      7.8  10 3 m

      Javal' s Rule : AS  e AC  AP  1.25 (2.34 DC axis 90)  0.50 DC axis 90
      AS   2.92 DC axis 90  0.50 DC axis 90   2.42 DC axis 90   2.42 DC axis 180
                                                 33
Q9.    B   shape of the reflected mire pattern “mirrors” anterior corneal topography, larger
           images indicating longer (flatter) radius of curvature, and increasing separation of
           adjacent rings indicating progressive corneal flattening toward the periphery.
           Vertically elongated ellipses mean flatter radius in the vertical meridian (against-
           the-rule astigmatism). Increasing ring separation from center to periphery means
           progressive corneal flattening, which keeps spherical aberration at normal levels,
           lower than predicted for a true spherical refracting surface.

Q10.   D   PMs 80 apart, therefore irregular astigmatism.

Q11.   C   Topcon uses ncal = 1.3375. K values are +40.42 D @ 135 and +41.67 D @ 45.
           Mean K (mean of total corneal power estimates for the two PMs) = (+40.42 +
           41.67)/2 = +41.05 D (+41.04 to more decimal places).




                                            34
Chapter 14 – BIO and Retinoscopy – Answers

1.    A     Lower power means higher magnification

2.    D     BIO angular magnification and axial magnification are dependent on condenser
            power, patient ocular constants, and distance between the condenser and the
            patient’s eye. Angular mag and axial mag are both independent of BIO PD and
            observer PD. Stereoscopic magnification is a function of BIO reduced PD (axial
            mag  BIO PD/observer PD).

3.    B

4.    C     Same rationale as question 2.

5.    D     The function of the PD reduction system is to image the practitioner’s pupils in the
            patient’s pupil. The larger the BIO “reduced” PD, the larger the patient’s pupil
            must be to fit both of the practitioner’s pupils.




                                             35
6.   D   The main determinant of condensing (BIO) lens FOV is proximity of the
         condenser to the eye. Higher power condensers are held closer to the eye and
         therefore provide greater FOV (tracing a chief ray from the condenser margin into
         the eye provides the angular FOV).

         Greater FOV is partially offset by the smaller diameter of higher power
         condensers, but the “distance” effect is by far the more important factor.




7.   A


                                   1       1
            0.075 m       L                   13.33 D
                                       0.075 m
8.   D   L  L  F   13.33  16   2.66 D
              1        1
                           0.375 m   375 mm
              L    13.33 D
                                        36
9.    A   Narrow, dim, slow reflex means high refractive error. “With” means hyperopia.
          The patient therefore has high hyperopia.

10.   B   The reflex movement described is neutralization. However, no working distance
          lens is in place. For 2/3 meter, a +1.50 D WD lens is required to compensate.
          Therefore, neutralization with zero power is equivalent to a +1.50 D working lens
          and a 1.50 D myopic correction. The patient is about 1.50 D (therefore < 2 D)
          myopic.

11.   C   With the +1.50 D working distance lens, the patient’s uncorrected 1.50 D myopia
          shows up as a fairly rapid against movement.

12.   C   Neutralize the more hyperopic (slower with) or less myopic (faster against)
          meridian first. If the 90 meridian is more hyperopic, the 180 meridian must be
          less hyperopic (more myopic). This meridian requires a more negative correction
          than the 90 meridian, which means negative cylinder axis 90 (power 180).

13.   B   A “skew” reflex indicates astigmatism with principal meridians NOT parallel to
          the meridian being scoped OR the perpendicular meridian (assuming regular
          astigmatism.

14.   C   Adding more minus when a rapid against movement is seen (close to
          neutralization) pushes the negative correction to excess minus (overminused).
          Therefore, the reflex movement will be “with”, indicating “add plus” (decrease
          minus).

15.   B   Moving in closer, the required compensating lens power is more positive
          (therefore, “with” movement is seen). The opposite is true if the clinician moves
          back from the “reversal” position.

16.   C   Adding plus at reversal induces “against” (need to add minus to return to reversal);
          adding minus at reversal has the opposite effect.

17.   D   The concave mirror effect changes “with” to “against” and “against” to “with”, but
          does not change the appearance at reversal.

18.   C

19.   A

20.   C



                                           37

				
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