Optics

1. Optics Oral exam For 1st Part Master

2. What is the name of this test?

3. On examination, the following results are reported by your patients: Patient 1 Patient 2 Patient 3 What is the diagnosis in:a. Patient 1 ?b. Patient 2 ?c. Patient 3 ?

4. Answer a. Right suppression b. Left suppression c. Exotropia (crossed diplopia )

5. Q1 • Your patient needs +2.00 D to see distance clearly. However, he can tolerate up to +4.00D without gettingblurred distance vision. • His cycloplegic refraction is +6.00D sphere. What are the values in dioptres of his: a. ِ absolute hypermetropia ? b. manifest hypermetropia ? c. facultative hypermetropia ? d. latent hypermetropia ?

6. Answer • Absolute hypermetropia  = + 2.00D (absolute hypermetropia is defined as the least amount of plus lenses needed for clear vision without cycloplegia) • Manifest hypermetropia = + 4.00D (manifest hypermetropia is defined as without cylcoplegia, the most plus correction that can be  tolerated without blurring of vision) • Facultative hypermetropia = + 2.00D (facultative hypermetropia is defined as the difference between absolute and manifest hypermetropia     ( + 4.00D - + 2.00D = + 2.00 D) • Latent hypermetropia = + 2.00 D (latent hypermetropia is defined as the difference between manifest hypermetropia and hypermetropia measured with cycloplegia (+ 6.00D - + 4.00D = + 2.00D)

7. Q • Convert the following prescriptions to + or - cylinder notation and state the type ofastigmatism which is present in each a. +4.00 / -1.50 x 70 b. +1.25 / -3.00 x 90 c. PL / +1.50 x 45 d. -2.00 / +2.00 x 50 e. - 1.75 / -2.00 x 135

8. Answer a. + 4.00 / - 1.50 x 70 = + 2.50 / + 1.50 x 160 ; compound hypermetropic astigmatism. b. + 1.25 / - 3.00 x 90 = - 1.75 / + 3.00 x 180 ; mixed astigmatism. c. PL / + 1.50 x 45 = +1.50 / - 1.50 x 135 ; mixed astigmatism. d. - 2.00 / + 2.00 x 50 = PL / - 2.00 x 140 ; simple myopic astigmatism. e. - 1.75 / - 2.00 x 135 = - 3.75 / + 2.00 x 45 ; mixed astigmatism.

9. Q • A 15 year-old emmetropic boy has an amplitude of accommodation of 30 D. a. What is his range of accommodation? • Another 20 year-old man has a +10 D hypermetropia with a 20 D amplitude of accommodation. b. what is his near point and the range of accommodation when he is not wearing the spectacle correction? c. What is his near point and the range of accommodation when he is wearing his +10.00 dioptres?

10. Answer a. Infinity to 3.3 cm. (The range of accommodation is the linear distance between the far point and the nearpoint. In the absence of accommodation, the boy is in focus at infinity. With maximalaccommodation, he is in focus at 1/30 = 0.033 m = 3.3 cm) b. 10cm(To be in focus in infinity, the man needs to use up +10.00D of his accommodation amplitude. As a result, he has only 20 - 10 = 10 D for near vision. His near point is:     1/10 m = 0.1 m = 10 cm c. The near point is 5 cm and the range of accommodation is infinity to 5 cm. (With hypermetropic correction he does not need to use up any of his accommodation and can therefore use all the 20 D for near vision. The near point is therefore:     1/20 m = 0.05 m= 5 cm. The range of accommodation is therefore from infinity to 5 cm)

11. Q • The following are the results of the retinoscopy on a 20 year-old man usinga streak retinoscope (assuming a working distance of 2/3 metre). • OD With the streak in the vertical meridian and sweep horizontally, the retinareflex is neutralized with +0.50D.With the streak in the horizontal meridian and sweep vertically, the retinareflex is neutralized with -1.00D. • OS  With the streak orientated at 30ºand sweep at 120º, the retina reflexis neutralized with +1.00D.With the streak orientated at 120ºand sweep at 30º, the retina reflexis neturalized with -0.50D a. Draw the power cross (taking into account the working distance) b. Write down the glasses prescription for this patient (using negative cylinder)

12. AnswerRemember that :                              i) the power of the cylinder is 90º to the axis                             ii) 2/3 metre is equivalent to -1.50 which has to be taken off)a.   b. OD  -1.00 / -1.50 X 90    OS  -0.50 / -1.50  X 120

13. Q • a. A 18 year-old university student presents to the optician one week before his final examination complaining of intermittent double vision and poor distance vision. • He was seen one month earlier and the refraction was emmetropic. • Qs i.  What is the most likely diagnosis ?     ii. What treatment would you prescribe?

14. Answer i. Accommodation spasm. When the symptoms occurs the patient is likely to have constrictedpupils and esotropia. The spasm can be broken with cyloplegia. ii. The following treatment is useful: •     Frequent looking up from work or •     Prescription of reading glasses 

15. Q • A 40 year-old myopic woman is recently prescribed soft contact lenses for the first time. • She returned two weeks later and complains that her reading vision is not as good as with her glasses. • Retest shows her visual acuity to be 6/6 in both eyes with the contact lenses and the lenses were of the right prescription and well-fitted. • Q. Why does she have problem reading with her contact lenses but not with her glasses?

16. Answer • The patient is pre-presbyopic. • Myopes require less accommodation with glasses thancontact lenses. • In addition, the prismatic effect (base-in prism) offered by the concave glasses assist convergence during reading.

17. Q • Eight weeks after a left extracapsular cataract extraction and implant, your patient has the following keratometry:  40.00 @4044.00 @140 Q. • What is the power and axis of cylinder required to correct the post-operative astigmatism? b. If a tight radially placed suture is present, in which meridian would you most likely to find it?

18. Answer a. There is a 4D difference between the two meridians. • Therefore, the required cylinder  correction is either:      - 4.00 X 40 or      + 4.00 X 140 b. The tight suture is at 140º and need to be removed.

19. Q • While performing a right cataract operation, you inadvertently inserted a 25.00 D IOL instead of a 20.50D (aiming for -0.05D) into the capsular bag. a. Would the patient become more myopic or hypermetropic in the right eye? • The patient was unable to tolerate the anisometropia and found contact lens difficult to use. You performed a lens exchange and insert the new lens in the sulcus. Post-operatively, thepatient's refraction was -0.50D. However, he returned two weeks  later, complaining ofdecreased vision. Repeated refraction showed that he had developed a 3.50D astigmatism. b. What is the most likely cause for this change of refraction?

20. Answer • More myopic. b. Dislocation of the implant. Astigmatism can occur due to tilting of the lens. 

21. Q • a. The following diagram represents the reduced schematic eye.     Write down the values for:      a ( in mm ), b ( in mm ), c ( in mm )      the total eye power      the refractive index of the eye • b. A 20 cm object is 1 metre away from the eye. What is the image size on the retina?

22. Answer • a.     a = 17 mm    b = 17 mm    c = 22.5 mm    total lens power = + 60 D    refractive index of the eye = 1.33

23. b. • The nodal point is 17 mm anterior to the retina and all refraction occurs at the plane through the nodal point. • The size of the retina image can be calculated as: Object height / retinal image height = distance from nodal point / 17 mmretinal image height = object height  X 17 mm/ distance from nodal pointretinal image height = 200 X 17 / 1000retinal image height = 3.4 mm

24. Q • Calculate the image location and state if the image were upright or inverted when an object is placed: i)   50 cm ii)  20 cm to the left of a +5.00D convex lens.

25. Answer • This is calculated using the formula:  U + D = V   U = object vergence in dioptres                       D = lens power in dioptres                       V = image vergence in dioptres) i) The image is 0.333 m to the left of the lens and is inverted.  U = 1/50 cm = 1/0.5 m = - 2D the sign is negative because the vergence of the object is divergence     U + D = -2 + 5 = +3     V = 3  and therefore the image vergence = 1/3 = 0.333m)  ii) The image is at infinity and erect. U = 1/0.2 m = -5 D , negative sign because the vergence of the object is diveregence       U + V  = -5 + 5 = 0        zero vergence means that the image is at the infinity.

26. Q • A patient wears -10.00 contact lens fitted on K. • If the patient is fitted with appropriate spectacle lenses worn at a vertex distance of 10mm. What is: • a. the required power of the spectacle lense? • b. the percentage change in spectacle magnification? 

27. Answer • a. The required spectacle power is calculated using the formula for lens effectivityDo = Do / (1-sDo) = -10 / ( 1-0.1) = -11.11D • b. The change in spectacle magnification is = retinal image size with the contact lens / retinal image size with the spectacle = power of the contact lens / power of the spectacle = -10.00 / -11.11 = 0.9 The percentage spectacle magnification change is           =  (0.9 - 1) 100           =   -10%

28. Q • a. Draw the ray diagram for a Galilean telescope using two lenses with focal lengths+10.00DS and -20.00DS. • b. What is the length of the constructed Galilean telescope in a. • c. What is the magnification achieved with the above telescope.

29. Answer Fo = objective lens  fo = focal length of the objective lens  Fe = eye piece        fe = focal length of the eyepiece lens 

30. b. 5cm.  (The length of the Galilean telescope is equal to the focal length of the objective lens minus the focal length of the eyepiece lens) • c. Magnification of the Galilean telescope : = W' / W  = Fe / Fo = fo / fe = 2

31. Q • A patient who is about to undergo cataract surgery has the following results during herpre-assessement. • Axial length = 23.00 mm • K1 = 42.00 DS • K2 = 44.00 DS • A-constant of the lens to be used = 118.0 • Calculate the lens power needed to achieve emmetropia. • During the operation, the patient has a large ruptured posterior capsule and youdecide to fit the lens in the sulcus. Would you use a lens (same A-constant) witha higher or lower power compared with a. 

32. Answer • a. Using the IOL formula= A - 2.5 (axial length) - 0.9 (average K reading )  = 118 - 2.5 (23) - 0.9 (43)  = 21.8 D ( As the lens come in step of 0.5 D, the one used would be 22.0D) b. Moving the lens forward increases the power of the lens and therefore a weaker lens is needed. This isusually 0.5D less than in the bag IOL

33. Q • Draw a ray diagrms to illustrate the formation of the images by aconcave mirror of an object: • a. between the centre of curvature and the principal focus. • b. inside the principal focus.

34. a.

35. b. F =  principal point, O = object, I = image

36. Q • Calculate the spherical equivalent of the following prescriptions and draw the powercross for each. • a. +2.50 / -1.50 X 80 • b. -4.00 / + 6.00 X 90 • c. +1.50 / -3.50 X 45

37. a. +2.50 / -1.50 X 80 = +1.75 (spherical equivalent) • b. -4.00 / + 6.00 X 90 = -1.00 (spherical equivalent) • c. +1.50 / -3.50 X 45 = -0.25 (spherical equivalent) (To draw the power cross remember that the power of the cylinder is 90 degrees to the axis. The spherical equivalent is calculated by adding the value of the sphere and half the value of the cylinder.)

38. Q • A 65 year-old jeweller is prescribed the following glasses for close work.Right eye +5.00DS andLeft eye +5.00DSAt work, he sees 10 mm below the optical centres of each lens. • a. What is the induced prism in each eye? • b. How much would the image of a piece of diamond be displaced if it were held20 cm from the glasses? • c. If he were to pick up the diamond with a pair of forcep, would he reach above or below it?

39. Answer • a.  The induced prism in each eye is 5 X 1cm = 5 dioptres base up.(Prentice's rule states that the prismatic effect is equal to the point from the optical centre in cm multiply by the dioptric power of the lens) • b.  1cm  X  5D  X  0.2m / 1.0m = 1cm displacement for each eye.(This is calculated by remembering that 1 prism dioptre deviates an object placed 1 m away by 1 cm.) • c. Below it. (The image is displaced downward by a a base-up prism)

40. Q • A patient has a visual acuity of 6/6 in both eyes while wearing glasses with the following prescriptions: OD -1.00/-0.50 X 90ºOS  -2.25 / -1.75 X 180º The keratometry reveals the following results: OD 7.85 mm along 180º (43.00D)7.85 mm along 90º (43.00D) OS  7.80 mm along 180º (43.25D)7.50 mm along 90º (45.00D) 1. Which structure contributes to the astigmatism in the a. right eye? b. left eye?  2. If the patient wants to wear spherical hard contact lenses, which eye will see better?

41. Answer 1. a. the lens b. the cornea Total ocular astigmatism = corneal astigmatism + lenticular astigmatism. In the R. eye, the keratometry shows no corneal astigmatism and therefore, the astigmatism found in the glasses is derived from the lens. In the L. eye, the keratometry shows the corneal astigmatism corresponds to that in the glasses, both in magnitude and axis, therefore, the astigmatism is derived from the cornea.

42. 2. The left eye. • A hard CL can neutralize about 90% of the corneal astigmatism.Therefore, the L.eye will see better than the R. because the astigmatism is entirely cornea.

43. Q • What is the stimulus to accommodation for an emmetropic subject in the following situations ? a. reading a book at a distance of 40 cm while     wearing +1.50D lenses. b. reading a book at a distance of 40 cm while     wearing -2.00D lenses

44. Answer • a. 1.0 D At 40 cm, the stimulus to accommodation for an emmetrope while reading a book is 2.5 D. As the patient is wearing +1.50 D, the final stimulus to accommodation is equal 2.50 – 1.50 = 1.0 D

45. b. At 40 cm, the stimulus to accommodation for an emmetrope while reading a book is 2.5 D. As the patient is wearing -2.50 D lenses, an extra 2.0 D is needed for accommodation. The final stimulus to accommodation is: 2.50 + 2.0 D = + 4.50 D

46. Q • Given a lens of  +3.00 / -1.00 X 90 with a diameter of 60mm and a pointobject at a distance of 1m in front of the lens, calculate the following: • a. the location of the line foci. • b. the location of the circle of least confusion. • c. the diameter of the circle of least confusion.

47. Q • The following patients have 6/6 vision (with spectacle correction where needed) and have no cataract. The results of the refraction and keratometry are as shown below:  Patient A:  Refraction:      OD   -2.50D       OS   plano Average Keratometry:    OD   44.50D        OS   44.50D Patient B:   Refraction:    OD    -2.50D         OS   plano Average Keratometry:     OS    44.50D        OS   42.00D a. What types of myopia does i) patient A  and ii) patient B have? b. Which patient is likely to get aniseikonia if the myopia were corrected with glasses?

48. Answer • Answer to Q a: patient A has axial myopia & patient B has refractive myopia. In patients A & B, the R. eyes are myopic, but the L. eyes are emmetropic. In patient A, the keratometry reading in both eyes are similar, and therefore, the myopia is axial. Whereas, in patient B, the keratometry reading in both eyes are different, and this difference is large enough to account for the myopia, therefore, the myopia is refractive.

49. Answer to Qb: Patient B. The use of glasses in refractive myopia is associated with diminusion of the retinal image. Whereas, in axial myopia, as in patient A, the size of the image is not changed.

50. Q • Using the heterophoria method, calculate the AC/A (accommodative convergence / accommodation) ratio using the following values (assuming the patient is emmetropic): ● Interpupillary distance     =  60 mm ● Alternate cover test at the distance (6 metre)  = 36 prism dioptres esophoria at near  (33cm)             = 30 prism dioptre esophoria