Questions

1. Which of the following regarding the wave properties of light is correct?

a. Wavelength is determined by the distance between crests of the wave.

b. The wavelength of visible light is between 200 and 500 nm.

c. Frequency is the maximum value attained by the electric field.

d. The speed of light in a vacuum is 6 × 10⁸.

2. When light interacts with matter, individual quanta of energy (photons) are emitted or absorbed. Which of the following is not true regarding the particle or photon characteristics of light?

a. The amount of energy (E) per photon is equal to Planck’s constant multiplied by the frequency.

b. A photon of red light has greater energy than a photon of blue light.

c. The light emitted through fluorescence has a longer wavelength than the excitation light.

d. Planck’s constant is equal to 6.626 × 10^–34 J/s.

3. Antireflection films prevent reflection of light by what mechanism?

a. constructive interference.

b. absorption of light photons.

c. polarization.

d. destructive interference.

4. The laser interferometer utilizes the principles of all of the following except

a. constructive interference.

b. destructive interference.

c. high coherence.

d. polarization.

5. The Haidinger brush phenomenon is due to which special characteristic of light transmission?

a. interference.

b. polarization.

c. diffraction.

d. scattering.

6. Which of the following concerning diffraction is not true?

a. It is responsible for a limit on pinhole acuity of approximately 20/25.

b. Diffraction is responsible for the blue color of the sky.

c. It is a limiting factor for visual acuity with pupils smaller than about 2.5 mm.

d. Long wavelengths are diffracted more than short wavelengths.

7. Why does the sky appear blue?

a. Because blue light is scattered more than other light given its longer wavelength.

b. Because blue light is scattered less than other light given its longer wavelength.

c. Shorter wavelengths are scattered more.

d. coherence.

8. The features of laser light that enhance its intensity include all of the following except

a. polychromaticity.

b. directionality.

c. coherence.

d. polarization.

9. Which of the following laser–tissue interactions are matched correctly?

a. Photodisruption: selective absorption of light energy and conversion of energy to heat with thermally induced structural change in the target.

b. Photocoagulation: uses pulsed lasers to ionize the target and rupture the surrounding tissues.

c. Photoablation: high-powered ultraviolet light exceeds covalent bond strength of target protein precisely removing a submicron layer.

d. Photodisruption: primary type of laser tissue interaction used in LASIK surgery.

10. In a simple thin lens system, the object is upright with a height of 5 cm, the image is inverted with a height of 10 cm, and the object is 2 cm from the center of the lens. What is the distance of the image from the center of the lens?

a. 1 cm.

b. 2 cm.

c. 4 cm.

d. 8 cm.

11. Which of the following concerning linear magnification is true?

a. Magnification is equal to the ratio of object size to image size.

b. A magnification value < 0 implies inversion of the image relative to the object.

c. Magnification is equal to the image vergence divided by the object vergence.

d. A magnification value >–1, and <1 implies that the image is greater than the object.

12. All of the following concerning refraction of light at interfaces are true except

a. Light will bend toward the surface normal as it enters a medium of higher index of refraction.

b. The index of refraction of any given substance is greater for longer wavelengths.

c. Total internal reflection renders the anterior chamber angle invisible by a slit lamp.

d. Refractive index is the ratio of the speed of light in a vacuum to the speed of light in the medium.

13. You have gone fishing and see a fish in the water. You do not have a fishing rod. The only equipment that you have is a spear to catch the fish. Where do you throw the spear?

a. in front of the fish.

b. behind the fish.

c. directly at the fish.

d. It is not possible to hit the fish as it is a virtual image.

14. All of the following is true of total internal reflection except

a. Total internal reflection occurs when light travels from a low-index medium to a highindex medium and the angle of incidence exceeds a certain critical angle.

b. Total internal reflection makes it impossible to view the eye’s anterior chamber angle without the use of a contact lens.

c. The critical angle is the angle of incidence that produces a transmitted ray 90° to the surface normal.

d. It may be possible to view the anterior chamber angle when the cornea is ectatic as in keratoconus.

15. What is the power of a +20 D lens under water?

a. 6.2 D.

b. 6.5 D.

c. 7.3 D.

d. 8.0 D.

16. Which is a true statement regarding vergence?

a. If light rays converge to a point, the vergence is negative.

b. Vergence is directly proportional to distance from the object point to the image point.

c. As light travels away from an object point or toward an image point, its vergence stays the same.

d. Vergence is the reciprocal of distance.

17. What is the vergence of an object 50 cm away from a lens? What is the vergence of an object 25 cm away from the lens?

a. –4 D; –8 D.

b. –2 D; –4 D.

c. 2 D; 4 D.

d. 4 D; 8 D.

18. What is the vergence of an image 50 cm away from a lens? What is the vergence of an image 100 cm away from the lens?

a. –4 D; –2 D.

b. –2 D; –1 D.

c. 2 D; 1 D.

d. 4 D; 2 D.

19. Consider an object 10 cm in front of a 5 D convex thin lens in air. What is the vergence after the light leaves the lens?

a. –5 D.

b. 5 D.

c. 10 D.

d. –10 D.

20. How far away is the image from the lens in question 19? Is the image virtual or real?

a. 20 cm; real.

b. 20 cm; virtual.

c. 10 cm; real.

d. 10 cm; virtual.

21. Consider an image is 50 cm in front of a 5 D convex thin lens in air. What is the vergence after the light leaves the lens?

a. 1 D.

b. –1 D.

c. 3 D.

d. –3 D.

22. How far away is the image from the lens in question 21? Is the image virtual or real?

a. 11 cm; real.

b. 11 cm; virtual.

c. 33 cm; real.

d. 33 cm; virtual.

23. Consider an object 25 cm in front of a 4 D concave thin lens in air. What is the vergence after the light leaves the lens?

a. –8 D.

b. 8 D.

c. –2 D.

d. 2 D.

24. How far away is the image from the lens in question 23? Is the image virtual or real?

a. 12.5 cm; real.

b. 12.5 cm; virtual.

c. 50 cm; real.

d. 50 cm; virtual.

25. Where will the image be formed for an object placed 25 cm in front of an 8 D convex lens that is separated from a 1 D concave lens by 50 cm?

a. 20 cm to the left of the concave lens.

b. 20 cm to the right of the concave lens.

c. 30 cm to the left of the concave lens.

d. 30 cm to the right of the concave lens.

26. An object is 25 cm to the left of a convex lens with a focal point of 10 cm. Is the image inverted or upright? Is the image to the left or right of the lens?

a. inverted; left of the lens.

b. upright; left of the lens.

c. inverted; right of the lens.

d. upright; right of the lens.

27. An object is 5 cm to the left of a convex lens with a focal point of 10 cm. Is the image inverted or upright? Is the image to the left or right of the lens?

a. inverted; left of the lens.

b. upright; left of the lens.

c. inverted; right of the lens.

d. upright; right of the lens.

28. An object is 25 cm to the left of a concave lens with a focal point of 10 cm. Is the image inverted or upright? Is the image to the left or right of the lens?

a. inverted; left of the lens.

b. upright; left of the lens.

c. inverted; right of the lens.

d. upright; right of the lens.

29. All of the following are true of a Galilean telescope except

a. The Galilean telescope produces an upright image.

b. The objective lens is positive and has a low power, whereas the eyepiece is negative and usually has a high power.

c. The two lenses in the Galilean telescope are separated by the sum of the focal lengths.

d. The Galilean telescope is often used as a lowvision aid or surgical loupes.

30. All of the following are true of Keplerian telescopes except

a. The Keplerian telescope produces an inverted image.

b. The two lenses in the system are separated by the sum of the focal lengths.

c. The Keplerian telescope consists of a low-power objective and a high-power ocular, with both lenses being positive.

d. In the Keplerian telescope, some of the light collected by the objective is lost.

31. What is the angular magnification of a Keplerian telescope whose eyepiece is +20 D and whose objective is +4 D.

a. –5×.

b. 5×.

c. –10×.

d. 10×.

32. An afocal telescope is constructed with –10 D and +4 D lenses. What is the distance between the lenses?

a. 5 cm.

b. 15 cm.

c. 20 cm.

d. 25 cm.

33. Which of the following is true in regard to ophthalmic prisms?

a. The power in prism diopters is the number of centimeters light is displaced over a 100 cm distance.

b. The image formed by a prism is virtual.

c. Images created by prisms are deviated toward the apex.

d. Prism power decreases as the distance from the optical center in a lens increases.

34. Which of the following is true regarding mirrors?

a. Plane mirrors add negative vergence.

b. Concave mirrors add plus vergence.

c. Light rays reflected off of a convex mirror are convergent.

d. Convex mirrors form real images to the left of the mirror.

35. All of the following are true regarding mirrors except

a. Convex mirrors form virtual images on the opposite side of the object.

b. The focal point of a convex mirror is to the left of the mirror.

c. The focal length is equal to half the radius of curvature of the mirror.

d. The reflecting power of a mirror is equal to the reciprocal of the focal length.

36. What is the reflecting power of a concave mirror whose radius of curvature is 50 cm?

a. +2 D.

b. –2 D.

c. +4 D.

d. –4 D.

37. What is the reflecting power of a convex mirror whose radius of curvature is 100 cm?

a. +2 D.

b. –2 D.

c. +4 D.

d. –4 D.

38. Consider an object 2 m to the left of a concave mirror with a radius of curvature of 100 cm. Where is the image?

a. 33 cm to the left of the mirror.

b. 33 cm to the right of the mirror.

c. 66 cm to the left of the mirror.

d. 66 cm to the right of the mirror.

39. Consider an object 2 m to the left of a convex mirror with radius of curvature of 100 cm. Where is the image?

a. 20 cm to the left of the mirror.

b. 20 cm to the right of the mirror.

c. 40 cm to the left of the mirror.

d. 40 cm to the right of the mirror.

40. If a man is 6 ft tall, what is the minimal length of a plane mirror for the man to see himself from head to toe?

a. 1 ft.

b. 3 ft.

c. 6 ft.

d. 12 ft.

41. If an object is 50 cm in front of a plane mirror, where is the image in relation to the mirror? Is the image virtual or real?

a. 50 cm from the mirror; virtual.

b. 50 cm from the mirror; real.

c. 100 cm from the mirror; virtual.

d. 100 cm from the mirror; real.

42. In the reduced schematic eye, what is the approximate focal length in air?

a. 17 mm.

b. 17 cm.

c. 22 mm.

d. 22 cm.

43. Using the dimensions from the reduced schematic eye, what is the image size on the retina of a Snellen letter that is 60 mm in height 20 ft (6,000 mm) from the cornea?

a. 17 mm.

b. 0.17 mm.

c. 22 mm.

d. 0.22 mm.

44. In a patient with astigmatism, all of the following are true of myopia and hyperopia except

a. In simple myopic astigmatism, one focal line lies in front of the retina and the other is on the retina.

b. In compound myopic astigmatism, both focal lines lie in front of the retina.

c. In simple hyperopic astigmatism, both focal lines lie behind the retina.

d. In mixed astigmatism, one focal line lies in front of the retina and one lies behind the retina.

45. A patient wears a –10 D spectacle lens at a vertex distance of 20 mm for distance correction. What power contact lens will be required for proper distance correction?

a. –8.25 D.

b. –9.00 D.

c. –12.50 D.

d. –11.50 D.

46. A patient wears a spectacle correction of +10.0 D at vertex distance of 20 mm. Where should a +3.0 D lens be placed to correct this patient for distance?

a. 10 cm in front of the eye.

b. 25 cm in front of the eye.

c. 33 cm in front of the eye.

d. 35 cm in front of the eye.

47. A patient with a corneal scar is carefully refracted. Best corrected visual acuity is 20/40. With a pinhole over his correction, his acuity improves to 20/25. The best explanation for this is

a. spherical aberration.

b. myopic astigmatism.

c. cataract.

d. irregular astigmatism.

48. An examiner sits 50 cm from a patient being refracted by retinoscopy. With the streak oriented in the horizontal meridian (sweeping vertically), a +3.00 D sphere neutralizes the reflex; with the streak oriented in the vertical meridian (sweeping horizontally), a +5.00 D sphere neutralizes the reflex. What is the patient’s final retinoscopic refraction?

a. +3.00 +2.00 × 090.

b. +3.00 +2.00 × 180.

c. +1.00 +2.00 × 090.

d. +1.00 +2.00 × 180.

49. The angular magnification of a retinal image afforded by direct ophthalmoscopy in an emmetrope is approximately

a. 5×.

b. 10×.

c. 15×.

d. 20×.

50. A 5-year-old child is noted to have a 20 prism diopter esotropia that increases to 45 prism diopters while reading at 20 cm. The patient’s pupillary distance is 50 mm. With the child reading through his distance correction, a +3.00 D lens placed over each eye decreases the esotropia at near to 20 prism diopters. The patient’s accommodative convergence to accommodation (AC/A) ratio, as determined by the gradient method, is

a. 5:1.

b. 8:1.

c. 10:1.

d. 15:1.

51. For the child described in the previous question, what would the AC/A ratio be, as determined by the heterophoria method?

a. 5:1.

b. 8:1.

c. 10:1.

d. 15:1.

52. A patient wearing new glasses comes in complaining of diplopia. The prescription is +3.00 –1.00 × 180 OD and –2.50 –1.00 × 90 OS with a +2.00 D add OU. You note a tropia present when the patient is reading at a comfortable distance. While reading, the visual axis is 2 cm nasal and 2 cm inferior to the distance optical center and 0.5 cm nasal and 1.2 cm above the optical center of the add in each lens. What is the induced prismatic effect for this patient?

a. 3 prism diopters base up OD and 1 prism diopter base out OD.

b. 3 prism diopters base up OD and 1 prism diopter base in OD.

c. 9 prism diopters base down OD and 1 prism diopters base in OD.

d. 9 prism diopters base up OD and 1 prism diopter base out OD.

53. Image jump is most troublesome for bifocals of the

a. round-top type.

b. flat-top type.

c. ribbon type.

d. progressive type.

54. How far from a +24 D camera lens should an object be placed to be focused onto film 5 cm to the right of the lens?

a. 25 cm to the left of the lens.

b. 50 cm to the right of the lens.

c. 50 cm to the left of the lens.

d. 25 cm to the right of the lens.

55. A crystal ball with an opaque rear surface sits on a pedestal. Its internal radius of curvature is 50 cm. Its index of refraction is 3.00. What is the refractive power of the crystal ball?

a. +2.00 D.

b. +4.00 D.

c. +8.00 D.

d. +10.00 D.

56. Which of the following is a characteristic of handheld magnifiers?

a. greater working distance.

b. greater ease of use for patients with poor manual dexterity.

c. smaller range of magnifying powers.

d. wider field of view.

57. The main advantage of telescopic aids for near work is

a. decreased convergence requirement.

b. wider field of view.

c. greater depth of focus.

d. greater working distance.

58. Which of the following is true regarding accommodation and contact lenses?

a. In a myopic patient, contact lenses decrease the accommodative demand compared to spectacles.

b. In a hyperopic patient, contact lenses increase the accommodative demand compared to spectacles.

c. In a myopic patient, contact lenses increase accommodative demand compared to spectacles.

d. There is no difference noted in accommodation when wearing contact lenses.

59. A patient requesting rigid gas-permeable (RGP) contact lens has a best refraction in the right eye of –4.00 +1.25 × 90. Keratometry is 44.0 D at 90° and 42.5 D at 180°. The conversion to radius of curvature is 7.70 mm at 90°, 7.95 mm at 180°. The only posterior curve available is 7.80 mm. The contact lens power that should be prescribed for best vision is

a. –2.00 D.

b. –2.75 D.

c. –3.00 D.

d. –3.50 D.

60. Which of the following is true regarding rigid-gas permeable (RGP) contact lenses?

a. Most lenses are made of PMMA.

b. RGP lenses are useful in correcting astigmatism.

c. RGP lenses generally have a shorter adaptation period compared to soft contact lenses.

d. RGP lenses are easier to fit than soft contact lenses.

61. Which of the following is true regarding examination lenses?

a. Goldmann fundus lens produces an inverted image.

b. Images generated by the Goldmann contact lens are “real” images.

c. The Hruby lens generates a virtual image.

d. The Hruby lens generates an inverted image.

62. Goldmann applanation tonometry readings

a. are not affected by corneal astigmatism.

b. are not affected by scleral rigidity.

c. are affected by surface tension of the tear film.

d. are not affected by the corneal thickness.

63. Which one of the following concerning direct ophthalmoscopy is false?

a. The linear magnification is 15×.

b. The optic disc of a myopic eye appears larger than that of an emmetropic eye.

c. The image is a virtual, upright image.

d. The dial on a direct ophthalmoscope is intended to neutralize both the examiner’s and patient’s refractive error.

64. Which of the following regarding indirect ophthalmoscopy of an emmetropic eye is false?

a. The image the examiner observes is a real, inverted image in the focal plane of the condensing lens.

b. Conjugate planes include the patient’s and the examiner’s retinas and the patient’s and examiner’s pupils.

c. If the examiner uses a 20 D condensing lens, the lateral magnification is approximately 3× and the axial magnification approximately 2.25×.

d. A 30 D condensing lens provides greater magnification and a larger field of view than a 20 D condensing lens.

65. Which of the following concerning keratometry is false?

a. Corneal curvature is measured by using the cornea’s power as a convex mirror.

b. A central image is doubled to negate the effect of eye movement.

c. Conventional keratometry measures the curvature of the central 6 mm of the cornea.

d. The refractive power of the average cornea equals 337.5 divided by its radius of curvature (in mm).

66. Which of the following is not a complication of ACIOLs?

a. anterior uveitis.

b. hyphema.

c. chronic cystoid macular edema.

d. inability to achieve good vision.

67. Which IOL calculation formula is most accurate for shorter eyes (<24.5 mm)?

a. Holladay 1.

b. SRK.

c. Hoffer Q.

d. SRK/T.

68. Which IOL calculation formula is most accurate for longer eyes (>26.0 mm)?

a. Holladay 1.

b. SRK.

c. Hoffer Q.

d. SRK/T.

69. Which of the following is false?

a. A-scan ultrasound speed is higher in the lens and cornea than in the aqueous and vitreous.

b. Axial length in staphylomatous eyes may be better measured with B-scan.

c. Applanation A-scan measurements tend to give longer axial length readings.

d. Immersion A-scan is more accurate than applanation A-scan.

70. A 60-year-old man with prior history of LASIK for myopia in both eyes presents to you for cataract evaluation. You perform your standard exam using refraction, potential acuity, axial length, and keratometery readings. Surgery is uncomplicated. That evening, you remember that you did not take into account the patient’s prior refractive surgery into IOL calculations. What type of refractive error will the patient most likely have as a result?

a. myopia.

b. hyperopia.

c. astigmatic error.

d. none.

71. Which of the following refractive surgeries do not change the index of refraction of the cornea?

a. LASIK.

b. LASEK.

c. RK.

d. PRK.

72. Which of the below answers is true regarding IOL calculations in eyes with previous retinal surgery and silicone-oil implantation?

a. A-scan velocity is faster in eyes with silicone oil versus vitreous.

b. A-scan velocity is slower in eyes with silicone oil versus vitreous.

c. Silicone oil acts as an intraocular positive lens when IOL’s are placed.

d. B-scan must be done to determine axial length.

73. Multifocal IOLs

a. can induce monocular diplopia.

b. improve night vision.

c. improve contrast sensitivity.

d. can contain only two zones of focus.

74. Which of the following is a third-order refractive aberration?

a. tetrafoil.

b. spherical aberration.

c. vertical coma.

d. secondary astigmatism.

75. What is true regarding the direct ophthalmoscope?

a. A hyperopic eye has a higher magnification than an emmetropic eye.

b. A myopic eye has a higher magnification than an emmetropic eye.

c. An emmetropic schematic eye has a refractive power of about 80 D.

d. Refractive status of an eye does not affect magnification when using a direct ophthalmoscope.

76. How can one improve fundus visualization using the indirect ophthalmoscope in a patient with small pupils?

a. increasing the distance of the ophthalmoscope mirror and the observer.

b. decreasing the distance between the ophthalmoscope eyepieces.

c. moving closer to the patient’s eyes.

d. decreasing the observer’s interpupillary distance.

77. Which of the following is true regarding the slit-lamp biomicroscope?

a. The slit lamp contains both astronomical and Galilean telescope components.

b. The slit lamp contains only astronomical components.

c. The slit lamp contains only Galilean components.

d. The slit lamp is a monocular viewing system.

78. The 78-D fundus lens when used in conjunction with the slit lamp produces an image that is

a. virtual.

b. real.

c. minified.

d. upright.

79. Which of the following is not a method used to accurately measure corneal thickness?

a. optical focusing.

b. optical doubling.

c. slit-lamp measurement.

d. ultrasound.

80. Which of the following is true regarding laser interferometry?

a. It requires one clear area of the cataractous lens.

b. The fringe spacing directly correlates with visual acuity.

c. It is a good test for optic nerve function.

d. It is based on the principle that cataractous lenses have some clear spaces.

81. Legal blindness as described by the World Health Organization (WHO) is defined as

a. best-corrected vision of 20/100 or worse in the better eye.

b. best-corrected vision of 20/200 or worse in the better eye.

c. best-corrected vision of 20/70 or worse in the better eye.

d. best-corrected vision of 20/400 or worse in the better eye.

82. Per WHO definition, what is the visual field limit of legal blindness in the better eye?

a. 15°.

b. 20°.

c. 25°.

d. 30°.

83. The Kestenbaum rule

a. is used to measure horizontal deviation.

b. is used to refract for distance vision.

c. is used to determine the “add” needed for bifocals.

d. is used to determine predicted prism needed for glasses.

84. What is true regarding stand magnifiers as low-vision aids?

a. The stand magnifiers usually cause an unstable image.

b. Maintaining lens working distance is difficult.

c. They are useful for patients with limitation of hand dexterity.

d. The magnifiers require no specific positioning of reading material.

85. What is the most favorable size for the pinhole in an occluder to provide the best vision?

a. 1.0 mm.

b. 1.2 mm.

c. 1.5 mm.

d. 1.7 mm.

86. When testing for Jaeger near acuity, how far is the near card held from the patient?

a. 10 inches.

b. 12 inches.

c. 14 inches.

d. 16 inches.

87. Which type of testing below requires no participation from the patient?

a. Teller acuity cards.

b. visual evoked potential.

c. Vernier acuity cards.

d. optokinetic nystagmus drum.

88. Which of the following is true regarding the Ferris-Bailey distance visual acuity chart?

a. The letters used are of equal viewing difficulty.

b. Each line has six letters.

c. The letter spacing varies between letters on the same line.

d. It is a poor vision acuity chart.

89. Which of the following is a cause of acquired myopia?

a. orbital tumor.

b. central serous chorioretinopathy.

c. intravitreal silicone oil.

d. congenital glaucoma.

90. All of the following are causes of acquired hyperopia except

a. Adie’s tonic pupil.

b. posterior lens dislocation.

c. phenothiazines.

d. posterior staphyloma.

91. Which of the following drugs can cause acquired myopia?

a. chloroquine.

b. sulfonamides.

c. phenothiazines.

d. benzodiazepines.

92. Acquired astigmatism may be caused by all of the following except

a. ptosis.

b. pellucid marginal degeneration.

c. macular degeneration.

d. ciliary body mass.

93. Which of the following is false regarding aphakic glasses?

a. The lenses magnify images.

b. A “pincushion” effect may be present.

c. The lenses can cause a ring scotoma.

d. The lenses do not affect depth perception.

94. According to the Sanders-Retzlaff-Kraff formula, a 1 mm error in axial length causes what magnitude of diopter error for final IOL power?

a. 1.3 D.

b. 2.5 D.

c. 5 D.

d. 7.7 D.

95. Pantoscopic tilt relates to which of the following?

a. surgical microscope position.

b. slit-lamp position.

c. contact lens fitting.

d. spectacles.

96. Which of the following is true about closed circuit television (CCTV) used as a low-vision aid?

a. CCTV is useful for patient with manual dexterity limitations.

b. CCTV is relatively cheap.

c. CCTV is easily portable.

d. CCTV provides high magnification.

97. The Geneva lens clock measures

a. radius of curvature.

b. prismatic effect.

c. trifocal segment size.

d. time.

98. The most important feature of the slit lamp is

a. illumination.

b. magnification.

c. common focus point.

d. portability.

99. Which one of the following regarding diurnal refractive fluctuations following incisional refractive surgery is false?

a. It is more common with incisional keratotomy than photorefractive keratoplasty.

b. Fluctuations in the second eye generally closely parallel those of the first eye.

c. Diurnal fluctuations have been measured up to 11 years after refractive surgery.

d. The most frequently seen pattern is a hyperopic shift from morning to evening.

100. Which of the following will yield artificially high IOP measurements during applanation?

a. applanation over corneal edema.

b. applanation against a soft contact lens.

c. applanation over a corneal scar.

d. applanation after LASIK surgery.

Helpful Optics Equations

Transverse magnification = image height / object height = image distance / object distance

Snells’s Law:

n_i sin θ_i = n_t sin θ_t

Critical angle:

n_i sin θ_c = n_t sin 90°

The power of a thin lens immersed in fluid can be given by the equation:

D_air / D_fluid = (n_lens – n_air ) / (n_lens – n_fluid )

Angular Magnification for a telescope:

M_A = –D_e / D_o

Reduced schematic eye calculations:

Object height / Image height = Distance from nodal point to object / 17 mm

Accomodative Convergence to Accomodation Ratio Gradient method:

AC/A = [(deviation with lens – deviation without lens) / (lens power)]

Heterophoria method:

AC/A = [(deviation at near – deviation at distance)/ (accommodation at near) + PD]

Prentice’s Rule Δ = hD

Angular magnification: (distance of 25 cm) M = D/4

Refracting power of a spherical surface: D_s = (n₂ – n₁) / r

Answers

In optics, the term magnification refers to making images larger or smaller than the object. If an image is inverted, it is negative and if it is upright it is positive. Object and image distances are negative when they point to the left of the lens and positive when they point to the right. Object distance is measured from the anterior nodal point to the object, and image distance is measured from the posterior nodal point to the image. For a simple thin lens immersed in a uniform medium such as air, the nodal points overlap in the center of the lens. Therefore, the image distance and object distance are measured from the center of the lens. In this problem, we use the equation:

Transverse magnification = image height/object height = image distance/object distance

–10 cm/5 cm = image distance/–2 cm Image distance = 4 cm to the right of the lens.

In this problem, 10 cm is negative because the image is inverted, and the distance of the object from the lens is negative because it is to the left of the lens. The image distance is positive. Therefore, it is to the right of the lens.

11. b. Magnification is equal to the ratio of image size to object size (i/o). Magnification < 0, a negative value, implies an inverted image. Magnification <1 (or if negative, >–1) implies an image smaller than the object.

12. b. An imaginary line perpendicular to the optical interface is called the surface normal. Light will bend toward the surface normal as it enters a medium of higher index of refraction and away from the surface normal as it enters a medium of lower index of refraction. This can be shown by Snell’s law:

n_i sinθ_i = n_t sinθ_t

where n_i is the refractive index of the medium, n_t is the refractive index of transmitted medium, θ_i is the angle of incidence, and θ_t is the angle of transmission.

The index of refraction for any medium is greater for shorter wavelengths. In any medium except for a vacuum, short wavelengths travel

more slowly than long wavelengths. This is called dispersion. Chromatic dispersion in the human eye leads to chromatic aberration, where yellow wavelengths will be focused on the retina, blue light will be focused in front of the retina, and red light behind the retina. This chromatic aberration can be helpful in refraction when using the duochrome, or red-green, test. When the image is clearly focused on the retina, the green symbols will be focused in front of the retina and the red symbols behind the retina. Some clinicians use the RAM-GAP mnemonic (Red Add Minus, Green Add Plus) to help them with refraction.

13. a. It is true that the fish you see is a virtual image, but it is possible to hit the fish. Remember that the light coming from the fish is going from a medium of higher index of refraction (water) to a medium of lower index of refraction (air). Therefore, the light from the fish will be going away from the surface normal, which is perpendicular to the air–water interface. Because of this, the image of the virtual fish, which is what you see, will be behind the actual fish. Yo u must aim in front of this virtual fish to hit the actual fish.

14. a. Total internal reflection occurs when light travels from a high-index medium to a low-index medium. The critical angle is the angle of incidence that produces a transmitted ray 90° to the surface normal. The critical angle θ_c can be calculated from Snell’s law using the equation:

n_i sinθ_c = n_t sin90°

15. c. The power of a thin lens immersed in fluid can be calculated using the equation:

16. d. Vergence is the reciprocal of distance. Vergence is inversely proportional to distance from the object point or the image point. As light travels away from an object point or toward an image point, its vergence constantly changes. If an object point moves closer to the lens, the light is more divergent and when the object point is farther from the lens the light is less divergent. Similarly, if the lens is closer to an image point, light is more convergent and when the lens is farther from the image point, the light is less convergent.

17. b. Vergence is measured in diopters (D). Vergence is the reciprocal of the distance in meters: 1/distance in meters. Both answer choices are negative because the light rays are diverging from the object. The answers can be found by using the calculations: 1/0.5 m and 1/0.25 m. Therefore, the answers are –2 D and –4 D, respectively. As you can see by these calculations, the vergence is greater as the object moves closer to the lens.

18. c. Both answer choices are positive because the light rays are converging on to the image point. The answers can be found by using the calculations: 1/0.5 m and 1/1.0 m. Therefore, the answers are 2 D and 1 D, respectively. As you can see by these calculations, the vergence is less as the image point moves farther away from the lens.

19. a. In this problem, we must use the vergence formula:

U +D = V

U is the object vergence, D is the lens power, and V is the image vergence.

Remember that the light rays are diverging from the object, so the object vergence is negative.

The object vergence can be determined by taking the reciprocal of the distance in meters, which is 1/0.10 m. Thus, the object vergence is –10 D. Convex lenses add plus power and concave lenses add minus power. Therefore, the lens gives a power of +5 D.

In order to determine the vergence leaving the lens, we must use the equation, U + D = V.

–10 D+(+5 D) = V = –5 D.

20. b. The vergence leaving the lens is –5 D. Because the vergence leaving the lens is negative, this means that the light leaving the lens is still diverging and does not converge to form an image to the right of the lens. Instead, imaginary lines from the diverging light rays leaving the lens can be drawn to form an image to the left of the lens. Because the image is to the left of the lens, and is not formed by the true light rays, the image is virtual. To find the distance of the image, we need to find the reciprocal of the vergence leaving the lens, which is (1/5 D) = 0.2 m = 20 cm to the left of the lens.

21. c. Again, in this problem we must use the vergence formula:

U +D = V

U is the object vergence, D is the lens power, and V is the image vergence.

Remember that the light rays are diverging from the object, so the object vergence is negative. The object vergence can be determined by taking the reciprocal of the distance in meters, which is 1/0.50 m. Therefore, the object vergence is –2 D. Convex lenses add plus power and concave lenses add minus power. Therefore, the lens gives a power of +5 D.

In order to determine the vergence leaving the lens, we must use the equation, U + D = V.

–2 D + (+5 D) = V = +3 D.

22. c. The vergence leaving the lens is +3 D. Because the vergence leaving the lens is positive, this means that the light leaving the lens is now converging and will form an image to the right of the lens. Because the image is to the right of the lens, and is formed by the true light rays, the image is real. To find the distance of the image, we need to find the reciprocal of the vergence leaving the lens, which is (1/3 D) = 0.33 m = 33 cm to the right of the lens.

23. a. Remember that the light rays are diverging from the object, so the object vergence is negative. The object vergence can be determined by taking the reciprocal of the distance in meters, which is 1/0.25 m; the object vergence is –4 D. Convex lenses add plus power and concave lenses add minus power. Therefore, the lens gives a power of –4 D.

In order to determine the vergence leaving the lens, we must use the equation, U + D = V.

–4 D+(–4 D) = V = –8 D.

24. b. The vergence leaving the lens is –8 D. Because the vergence leaving the lens is negative, this means that the light leaving the lens is still diverging and does not converge to form an image to the right of the lens. Instead, imaginary lines from the diverging light rays leaving the lens can be drawn to form an image to the left of the lens. Because the image is to the left of the lens, and is not formed by the true light rays, the image is virtual. To find the distance of the image, we need to find the reciprocal of the vergence leaving the lens, which is (1/8 D) = 0.125 m = 12.5 cm to the left of the lens.

25. a. When solving a problem with multiple lenses, it is important to locate the first image. The position of this first image is then used to determine the vergence entering the second lens. For the first lens, we will use the equation:

U₁ +D₁ = V₁

U₁ is the object vergence, D₁ is the lens power, and V₁ is the image vergence for the first lens. Remember that the light rays are diverging from the object, so the object vergence is negative. The object vergence can be determined by taking the reciprocal of the distance in meters, which is 1/0.25 m. Therefore, the object vergence is –4 D. Convex lenses add plus power. Therefore, the lens gives a power of +8 D.

In order to find the vergence leaving the lens, we must use the equation, U₁ + D₁ = V₁ –4 D + (+8 D) = V₁ = +4 D. The distance is the reciprocal of the vergence, which is 1/4 = 0.25 m = 25 cm.

The location of the image formed by the first lens becomes the object location of the second lens. Remember that the object distance used for the second lens must be the distance from the second lens, not the first lens. The total distance between the two lenses in this problem is 50 cm. The image formed is 25 cm to the right of the first lens. Thus, this image must be 25 cm to the left of the second lens.

For the second lens, we will use the equation: U₂ +D₂ =V₂

U₂ is the object vergence, D₂ is the lens power, and V₂ is the image vergence for the second lens.

Remember that the light rays are diverging from the object, so the object vergence is again negative. The object vergence can be determined by taking the reciprocal of the distance in meters, which is 1/0.25 m. Therefore, the object vergence is –4 D.

Concave lenses add minus power. Therefore, the lens gives a power of –1 D.

In order to find the vergence leaving the lens, we must use the equation, U₂ + D₂ = V₂ –4 D + (–1 D) = V₂ = –5 D. The distance is the reciprocal of the vergence, which is 1/5 = 0.20 m = 20 cm to the left of the second lens.

26. c. The image is inverted and to the right of the lens. Light coming from the anterior focal point (F_a) exits the lens and comes to a focus at plus optical infinity. Light coming from minus optical infinity images to the posterior focal point (F_p). Three rays can be drawn through a thin lens to locate a corresponding point in the image. Only two rays are needed. The first two rays can pass through F_a and F_p and the third ray, which is known as the central ray, pass through the nodal points. For a thin lens, the nodal points overlap at the optical center of the lens. For a convex lens, when the object lies at a distance from the lens greater than the focal point, the image is inverted, real, and to the right of the lens. This can be shown through ray tracings.

27. b. When the object distance from the lens is less than the focal point of a convex lens, the image is magnified, upright, and virtual, and is located to the left of the object and lens. This can be shown through ray tracings.

28. b. A ray of light directed through F_a will exit the lens parallel to the optical axis. A ray of light entering the lens parallel to the optical axis will pass through F_p. Using these same principles as used for the convex lens, three rays can be drawn through a concave lens to locate the corresponding image. No matter where a real object is placed in front of a minus lens, the resulting image is upright, minified, virtual, and to the left of the lens. This can be shown through ray tracings.

29. c. The lenses in a Galilean telescope are separated by the difference in focal lengths. Because the image produced by a Galilean telescope is upright, and the Galilean telescope is shorter than a Keplerian telescope, it makes it more ideal in surgical loupes and low vision aids.

30. d. In a Galilean telescope, some of the light is lost, not in a Keplerian telescope. Because of this, Keplerian telescopes use light much more effi-ciently than Galilean telescopes and as a result, are more ideal for astronomical observation.

31. a. The angular magnification (M_A) of both a Keplerian and Galilean telescope is equal to the power of the eyepiece (D_e) divided by the power of the objective (D_o). This is given by the equation:

M_A = –D_e /D_o

M_A = – (+20)/(+4) = –5×. The negative sign indicates that the image is inverted, which is true of a Keplerian telescope.

32. b. In an afocal telescope, the lenses must be placed such that the secondary focal point of the objective lens coincides with the primary focal point of the eyepiece. If both lenses are positive, as in an astronomical or Keplerian telescope, the distance between the lenses is the sum of the focal lengths. If the eyepiece is negative, as in a Galilean telescope, the distance between the lenses is the difference between the focal lengths. In this Galilean telescope, f_objective = 25 cm, and f_eyepiece = 10 cm. The difference between the two, 15 cm, is the length of the device.

33. a. The image formed by a prism is real and images are deviated toward the base of a prism, not the apex. Prism power increases as the distance from the optical center in a lens increases. This prismatic effect becomes important clinically in patients with anisometropia in the reading position, where it can cause vertical misalignment of the visual axis.

34. b. Concave mirrors add plus vergence as the light rays reflected off of the mirror is convergent. Convex mirrors add minus vergence as the light rays reflected off of the mirror is divergent. Plane mirrors add zero vergence.

35. b. The focal point of a convex mirror is to the right of the mirror and the focal point of a concave mirror is the left of the mirror.

36. c. Remember that a concave mirror adds positive vergence. The focal length (f) is equal to half of the radius of curvature (r) given by the equation: f =< r/2. The radius and focal length are in meters. The focal length in this problem is f = 0.50 m/2 = 0.25 m.

The reflecting power of a mirror is given by the equation: D_m = 1/f.

The reflecting power of concave mirror in this problem is D_m = 1/0.25 m = +4 D.

37. b. The focal length is found by using the equation f = r/2 = 1 m/2 = 0.50 m to the right of the mirror. The reflecting power of a mirror is given by the equation D_m = 1/f = –1/0.50 = –2 D. Remember that a convex mirror adds minus vergence.

38. c. To find the image distance from the mirror, we can use the equation U + D_m = V. U is the object vergence, which in this problem is U = –1/2 m = –0.5 D. This is negative because light is diverging from the object. The focal length is

f = r/2 = 1.0 m/2 = 0.50 m.

The reflecting power of the mirror is D_m = 1/f = 1/0.50 m = +2 D. Remember that concave lenses give plus power. Now we need to find the vergence of the image rays (V) using the equation U + D_m = V.

–0.5 D + (+2 D) = V = +1.50 D. Using the image vergence, we can find the image distance from the mirror, which is 1/1.50 D = 0.66 m or 66 cm to the left of the mirror. The image formed is real.

39. d. To find the image distance from the mirror, we can use the equation U + D_m = V. U is the object vergence, which in this problem is U = –1/2 m = –0.5 D. The focal length is f = r/2 = 1.0 m/2 = 0.50 m. The reflecting power of the mirror is

D_m = 1/f = 1/0.50 m = –2 D. Remember that convex lenses give minus power. Now we need to find the vergence of the image rays (V) using the equation U + D_m = V.

–0.5 D + (–2 D) = V = –2.50 D. Using the image vergence, we can find the image distance from the mirror, which is 1/–2.50D = 0.40 m or 40 cm to the right of the mirror. The image is virtual.

40. b. The minimal length of a plane mirror needed to view the entire body from head to toe is half the height of the person.

41. a. Remember that a plane mirror adds no vergence, D_m = 0 D. To find the location of the object, we can use the equation U + D_m = V. U = –1/0.50 m = –2 D. The vergence of the rays forming the image is therefore –2 D + 0 D = V = –2 D. The image location is 1/–2 D = 0.50 m or 50 cm to the right of the mirror. The image is virtual.

42. a. It is important to know the dimensions of the reduced schematic eye. In this simplified version of the eye, the power of the eye is +60 D. The nodal point of the eye lies within the eye 5.6 mm posterior to the cornea. This distance is often insignificant in calculations and often ignored. Without taking into account this distance, the anterior focal length in air is 17 mm in front of the cornea, and the posterior focal point, which lies on the retina, is 17 mm posterior to the nodal point. The refractive index for air is taken to be 1.0 and the simplified refractive index for the eye is 1.33.

43. b. Similar triangles can be used to figure out the height of the image. Remember that the distance from the nodal point to the retina is 17 mm. To figure out the height of the image, we can use the equation: Object height /Image height = Distance from nodal point to object /17 mm 60 mm/Image height = 6,000 mm/17 mm Therefore, the image height is 0.17 mm.

44. c. In a patient with myopia, the eye possesses too much optical power for its axial length and the image is focused in front of the retina. In a patient with hyperopia, the eye does not possess enough optical power for its axial length and the image is focused behind the retina. In an astigmatic eye, there is no single focal point, but rather a set of two focal lines. In simple hyperopic astigmatism, one focal line lies on the retina and the other focal line lies behind the retina. In compound hyper-opic astigmatism, both focal lines lie behind the retina.

45. a. The principle of neutralization of refractive errors is to create an image of an object at infin-ity at the eye’s far point. With accommodation relaxed, the far point and the retina are conjugate. Thus, the corrective lens’ secondary focal point must coincide with the eye’s far point. Because this eye is corrected with a minus 10 D spectacle lens, this eye must be myopic. In a myopic eye, the far point lies somewhere in front of the eye.

T o find the secondary focal point of the spectacle lens, we take the reciprocal of the diopters, which is 1/10 D = 0.1 m = 10 cm. The eye’s far point must be 10 cm in front of the spectacle lens. Given a vertex distance of 2 cm, the far point must be 12 cm from the cornea. Thus, the proper contact lens must have its secondary focal point 12 cm or 0.12 m in front of the cornea. The power in the contact lens is found by taking the reciprocal of the secondary focal point: 1/0.12 m = –8.33 D.

46. b. Again, the secondary focal point of a corrective lens must coincide with the eye’s far point. The far point of a hyperope lies somewhere behind the retina. This eye’s far point is 0.1 m (1/10 D) or 10 cm behind the spectacle plane. Given a 2 cm vertex distance, the far point is 8 cm behind the cornea. A +3.00 D lens must be placed with its secondary focal point at this same point. Because its secondary focal length is 0.33 m or 33 cm, it must be placed 25 cm in front of the cornea.

47. d. The corneal scar may be producing irregular astigmatism, which cannot be corrected by sphe-rocylindrical spectacle lenses. The pinhole greatly minimizes nonaxial light rays, which require refraction to come into focus on the retina.

48. c. Remember that the working distance in this problem is 50 cm, so 2.00 D must be subtracted from the spherical portion, giving a net result of +1.00 +2.00 × 090.

49. c. The magnification of a simple plus lens is defined as the ratio of the angular size of the image produced by the lens to the angular size of the object viewed at 25 cm. The formula for angular magnification by a simple plus lens of power

(P) is M = P/4. Because the average power of an emmetropic eye is +60 D, magnification M = 60/4 = 15×.

50. b. Using the gradient method, AC/A (the ratio of prism diopters of accommodative convergence to diopters of accommodation) is calculated by placing plus lenses in front of each eye and determining the change in the deviation at near. This change, divided by the power of the plus lens used (amount of weakened accommodation), gives the number of prism diopters of accommodative convergence per diopter of accommodation. In this case, 45 – 20 = 25, divided by 3 is (approximately) 8:1.

Gradient method:

AC/A = [(deviation with lens – deviation without lens)/(lens power)].

In these calculations, esodeviations are positive and exodeviations are negative, by convention.

51. c. The heterophoria method makes use of similar reasoning, while taking into account the effect of interpupillary distance (PD), that is, larger PDs require greater convergence for fusion. Heterophoria method:

AC/A = [(deviation at near – deviation at distance)/(accommodation at near) + PD]

where PD = interpupillary distance in cm. The denominator represents the number of diopters of accommodation required for the near target (the reciprocal of the reading distance in meters). In this case, PD is 5 cm, the deviation at near is 45 prism diopters, and the deviation at distance is 20 D. Accommodation will be 5 D at 20 cm. Thus, AC/A = (45 –20)/5 + 5 = 10:1.

52. d. Solving this problem is facilitated by preparing diagrams marked with the powers in the 90° and 180° meridians, the visual axes, and the optical axes for the distance corrections and adds. Image displacement in prism diopters is by Prentice’s rule:

Δ = hD

where h is in centimeters and D is the power of the lens in diopters.

This needs to be done for both the distance corrections and the adds separately. After this, the results are combined, first for each side, and then for a net effect on both eyes as shown in the figure at the top of the next page:

Note that when determining a net prismatic effect for two eyes, vertical prisms of different base orientation (base up plus base down) are additive, whereas horizontal prisms of same orientation (base in plus base in, or base out plus base out) are additive. For the net effect over one eye, prism of the same orientation are additive (e.g., base up plus base up, base in plus base in).

53. a. Image jump occurs at the segmentation line when the optical center of the add is not at its upper edge. A round top bifocal segment has its optical center farther from its upper edge than other types and therefore causes the greatest image jump. These are commonly used only for aphakic spectacles to minimize image displacement.

54. a. We can use the formula U + D = V. Because we know the power of the lens and the distance of the image from the lens, we can solve for U and therefore the distance from object to lens.

D = +24 D and V = 1/0.05 m = +20 D (real image and converging light rays).

Thus, U = V – D, U = 20–24 = –4 D. Finally, to get the distance of the object from the lens, we must take the reciprocal of –4 D, which is 1/ (–4 D) = –0.25 m or 25 cm to the left of the lens.

55. b. The refracting power of a spherical interface separating two media of different indices of refraction is given by

D_s = (n₂ – n₁)/r, where D_s = power of the surface, n₁ = refractive index of the first medium that the light travels in, n₂ = refractive index of the second medium that the light travels in, and r = radius of curvature of surface in meters. n₂ = 3.00, n₁ = 1.00 (air), and r = +0.5 m. The sign of “r” is assigned by convention: r is positive if the center of curvature of the surface is on the opposite side of the origin of the incident light. If the center of curvature is on the same side of the interface as the origin of light, then r is negative. Thus, D_s = (3–1)/+0.5 = +4.00 D.

56. a. Hand-held magnifiers have a greater working distance (i.e., a greater eye to object distance). They are available in a range of powers from +3.00 to +68.00 D. However, they do have a smaller field of view than high adds and must be held in the hand.

57. d. Focal telescopes magnify without decreasing the working distance. Adds may be used to lessen the accommodative demand of these aids. However, they do have a small field of view and small depth of field, so the head must be positioned precisely.

58. c. Contact lenses also decrease the accommodative demand in a hyperopic patient.

59. d. When fitting rigid contact lenses, if there is greater than a 0.2 mm difference between the radii of the principal corneal meridians, then a lens of intermediate curvature, steeper than the flatter meridian by one-third to one-half the difference between meridians, can be used. This creates a positive tear film lens (add 0.25 D to every 0.05 mm difference in radius of curvature), so minus power must be added to the original power. The starting power is the sphere of the refraction written in minus cylinder form and corrected for vertex distance if necessary (powers greater than ±4.00 D). –4.00 +1.25 × 90 is equivalent to –2.75 +1.25 × 180 in positive cylinder form. A plus tear lens of +0.75 is created by the choice of 7.80 mm base curve (0.25 D for every 0.05 mm). Thus, –0.75 D must be added to –2.75 D. (Vertex distance is not important for powers <4.00 D.).

60. b. RGP lenses today tend to be made of silicone. They offer clearer vision and correct astigmatism. On the other hand, RGP lenses have a longer adaptation period and a greater difficulty of fit.

61. c. Both the Hruby lens (–55 D, planoconcave) and the Goldmann fundus contact lens (–64 D, planoconcave) essentially nullify the refractive power of the cornea and create an upright, virtual image of the fundus approximately 2 cm posterior to the lens.

62. c. Measurement of intraocular pressure (IOP) with applanation can be inaccurate if significant corneal astigmatism is present. An elliptical rather than a circular area will be applanated. Splitting the ellipse at a 43° angle to the major axis gives the best results. Alternatively, taking the mean of readings at 90° and 180° can reduce the error. With-the-rule cylinder causes underestimation (1 mm Hg/4 D) and against-the-rule cylinder causes overestimation of IOP. Forces of scleral rigidity and tear film surface tension cancel each other out.

63. a. The direct ophthalmoscope operates on the optical principle that light emanating from the retina of an emmetropic patient will be focused on the retina of an emmetropic observer. Lenses are used between the patient and observer to correct for nonemmetropic situations. In emmetropia, magnification obtained with the direct ophthalmoscope is approximately 15× because the patient’s eye acts as a simple plus magnifier of power 60 D (angular magnification—M = P/4 or 60 D/4 = 15×). In conditions of ametropia, the interposition of lenses creates a Galilean telescope. Therefore, if the patient is myopic, the retina appears more magnified, and if the patient is hyperopic or aphakic, it appears less magni-fied. Remember that in afocal systems, magnifi-cation must be considered in angular (not linear) terms.

64. d. A 30 D lens offers less magnification (2×) but a larger field of view than a 20 D lens. Although axial magnification is the square of linear mag-nification, it must be reduced by a factor of 4 for indirect ophthalmoscopy because the patient’s pupil is expanded by a factor of 4 for visualization by the examiner.

65. c. Conventional keratometry measures the curvature of only the central 3 mm of the cornea.

66. d. Good vision, including 20/20 vision, can be attained with proper ACIOL placement. Many complications related to ACIOL can be attributed to improper size of implant. These complications include uveitis, glaucoma, hyphema, dislocation of ACIOL, spinning of ACIOL, chronic CME, and Uveitis-Glaucoma-Hyphema (UGH) syndrome. Refractive error can also be a complication if not adjusted for during biometry should ACIOL placement be required.

67. c. The Hoffer Q formula is shown to be better for eyes with axial length <24.5 mm.

68. d. SRK/T is shown to be better for longer eyes. For eyes in between the two ranges, the Holladay 1 formula has been shown to be very accurate.

69. c. Applanation A-scan tends to give SHORTER axial length readings due to inadvertent indentation of the cornea during the procedure. Using immersion A-scan tends to be more accurate than applanation A-scan. Sound does indeed travel faster in the lens and cornea when compared to the aqueous and vitreous. In eyes with long axial length (>25.0 mm) B-scan and IOLMaster may be useful in determining an accurate axial length.

70. b. Refractive surgery for myopia flattens the central cornea. Most keratometers/corneal topographers miss this flattened central cornea. Keratometers take the readings from a 3.2 mm zone of the central cornea. Topography overestimates the power of the cornea. This can lead to a hyperopic error.

71. c. RK proportionately flattens both the anterior and posterior surface of the cornea. The index of refraction of the cornea is based on the association of the anterior and posterior curvature. PRK, LASIK, and LASEK change this ratio and thus the index of refraction of the cornea.

72. b. A-scan velocity through silicone oil is much slower than vitreous, 980 m/s (oil) and 1,532 m/s (vitreous). Silicone oil acts as a negative lens in the eye, thus IOL power usually needs to be increased to offset this action.

73. a. Disadvantages of multifocal IOLs include monocular diplopia, decrease in image clarity, decrease in contrast sensitivity, and decrease in night vision. Multifocal IOLs come in bifocal and multiple zone focus.

74. c. Vertical coma is a third-order refractive aberration. Others are listed below. Second-order aberration: Astigmatism, Defocus Third-order aberration: Trefoil, Horizontal coma Fourth-order aberration: Tetrafoil, Secondary astigmatism, Spherical aberration.

75. b. The schematic emmetropic eye has about 60 D of power. The eye is a simple magnifier. At 25 cm, magnification = power/4 for the schematic eye. Therefore, the emmetropic schematic eye has a magnification of 15×. The more hyperopic the eye, the less magnification and the more myopic an eye the higher the magnification when using a direct ophthalmoscope.

76. d. Small pupils pose a challenge when using the indirect ophthalmoscope. The following can improve the ability to examine the fundus:

1. Decreasing interpupillary distance.

2. Decreasing the distance of the ophthalmoscope mirror and the observer.

3. Increasing distance of eyepieces.

4. Moving farther away from the patient.

77. a. The slit lamp optical components include astronomical telescope, inverting prism, Galilean telescope, objective lens, illumination system, and binocular viewing system. The astronomical component has two lenses that are both convex. The Galilean telescope has two lenses also, one convex and one concave.

78. b. The 60 D, 78 D, and 90 D lenses all produce images that are magnified, real, and inverted when examining the retina.

79. c. Optical focusing uses a specular microscope focused on the corneal endothelium. Optical doubling uses an image-doubling prism. Ultrasound and anterior segment ultrasound can be used also.

80. d. Laser interferometry is a good test for macular function. It is based on the principle that the cataractous lens has some clear spaces. For this test, one beam is optically split into two. This creates fringes on the retinal surface that can be appreciated. The fringes are not correlated to visual acuity.

81. b. The WHO defines legal blindness as best-corrected vision of 20/200 or worse in the better eye. The next question will address the second part of the requirement; visual field.

82. b. 20° or worse in the better eye. The combined definition is 20/200 or worse in the better eye or a visual field of 20° or worse in the better eye.

83. c. The Kestenbaum Rule is a starting point for determining the add needed for reading 1 M print. It is based on distance vision. For example, if a patient is 20/100, then take 100/20 = 5. This patient may benefit from a +5 D lens for viewing at 1/5 m. This is a good STARTING point for determining add.

84. c. The main advantage for stand magnifiers is the stability they can provide for patients with hand tremors or limitations of dexterity. These can provide a stable magnified image with little effort upon the patient. This allows the working distance to remain stable if the object is placed in a stable position.

85. b. 1.2 mm is the best size for a pinhole occluder. This allows a balance between diffraction and reducing inherent refractive error.

86. c. When testing near vision (Jaeger, Point type, or Reduced Snellen acuity), the conventional distance is usually 14 inches.

87. b. Visual evoked potential requires no response from the patient.

88. a. The Ferris-Bailey chart has five letters per line. Each space between letters is equal to the size of the letter on that specific line. Three line increases or decreases signify doubling or halving in the visual angle, respectively. This chart was used in the ETDRS and many subsequent studies.

89. d. Congenital glaucoma can cause acquired myopia due to increased axial length and increased corneal power. The others listed are causes of acquired hyperopia.

90. d. Posterior staphyloma causes acquired myopia since the eye wall is posterior to the point of focus. The other listed answers are causes of acquired hyperopia.

91. b. Sulfonamides can cause acquired myopia. Other medications that can cause acquired myopia include miotics, chlorthalidone, tetracycline, and carbonic anhydrase inhibitors. Acquired hyperopia can be caused by chloroquine, pheno-thiazines, benzodiazepines. Antihistamines may cause both acquired hyperopia and myopia.

92. c. Retinal disease may cause hyperopia and myopia but generally do not cause acquired astigmatism. Disorders that cause astigmatism are more anterior in nature.

93. d. Aphakic spectacles have several disadvantages. The lenses cause magnification, affect depth perception, cause a ring scotoma, and cause pincushion distortion. The spectacles can also be heavy on the face and quite expensive.

94. b. The original SRK formula is P = A – 2.5 L – 0.9 K. This states that there is a 2.5 D/ 1 mm error for axial length and about a 1 to 1 diopter error for keratometry. Intraocular lens position is also important, and has about a 1 mm error = 1.0 D change in power ratio.

95. d. Pantoscopic tilt is used to decrease the amount of astigmatism of oblique incidence. This type of astigmatism is caused by tilting of a spherical lens. This adds the same power and cylinder as the original lens in the axis of the tilt. For example, tilting forward a +5 D lens will induce more plus spherical power AND plus cylinder power in the 180 axis. Most spectacles are tilted to give a compromised lens position for distance and near work.

96. d. CCTV is difficult for patients to use if they have manual dexterity issues. It is also not cheap or portable. It does, however, provide high levels of magnification for reading.

97. a. The Geneva lens clock measures radius of curvature of a spectacle lens. It usually is calibrated for crown glass.

98. c. The common pivot point or focus point is the most important feature of the slit lamp. It allows illumination and examination to be at the same point of focus.

99. d. Gradual steepening of the cornea throughout the waking hours leads to progressive myopia. Some patients require myopic correction as the day progress.

100. c. Applanation over a corneal scar will yield artificially high IOP measurements. All of the other answers listed will yield artificially low IOP measurements. Increased corneal thickness may also yield an artificially high IOP measurement during applanation.

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