Optics Supplement. Christopher S. Wolfe, OD, FAAO

Transcription

1 Optics Supplement Christopher S. Wolfe, OD, FAAO One of the most consistently difficult sections of the NBEO Part 1 exam has been optics. We feel that in order to truly master this material one must: 1. Have a working knowledge of what optics principles to use and when to use them. 2. Be able to answer questions even when equations can not be used because questions are more theoretical. 3. Be able to apply a knowledge of optics to various clinical situations. To that end, we have created a series of questions that builds on fundamentals from the pre-course flashcards and the KMK Textbook and stretches the knowledge solidified in the essential post-course flashcards. As you study these concepts, we encourage you to continue to think of different ways each question can be asked. We feel this exercise will be extremely beneficial to your exam preparation, but more importantly, we hope it will help you excel as an optometric physician. The following questions are based solely on our opinions and interpretation of the material presented on the NBEO s detailed matrix which can be found at Question 1: Which (3) of the following is true if a near point card is placed 12 cm in front of a 5 diopter converging lens? a. The image would be located 30 cm behind the lens b. The image would be located 7.5 cm in front of the lens c. The image is considered real d. The image is considered virtual e. The optics of this system are similar to a handheld (collimating) magnifier f. The optics of this system are similar to a stand (non-collimating) magnifier g. The optics of this system should create no magnification effects Answer 1: In this case we are placing an object (near point card) in front of a lens and we are asked to locate the image and understand something about the optics of the 1

2 system. The object is 0.12 m which corresponds to a vergence of 8.33 D. The exiting vergence would then be 3.33 D which corresponds to an textbfimage location of 30 cm in front of the lens. If an image is on the left side of the lens system it is textbfconsidered virtual. Since the object is placed closer to the lens than the primary focal point of the lens, then this system creates divergent exiting light and thus textbfthe optics are similar to that of a stand or non-collimating magnifier. Answers A, D and F are correct. Question 2: A Franklin bifocal has a base curve of D and a segment surface of The ocular surface of the lens measures 3.00 D when the clock is placed horizontally and 7.00 D when the clock is placed vertically. What is the prescription of the lens? a x 090 Add b x 180 Add c x 090 Add d x 180 Add e x 090 Add Answer 2: The lens clock is telling you what the surface powers of the front and the back of the lens are and remember that if they are asking you about the total power without giving you a thickness or specifying a refractive index then you can simply add the front and back surfaces together in order to get the total power in a thin lens. So the power in the vertical portion of the lens is 2.00 D and the power of the horizontal portion of the lens is this gives a distance power of x 090 since the front surface curve of the segment is 1.50 diopters more plus than the base curve or the distance front surface curve then the textbfadd power is Answer C is correct. Question 3: Which (2) statements are most likely true in a patient with Fuch s endothelial dystrophy? a. More hyperopic in the morning b. More myopic in the morning c. More presbyopic in the morning d. The refracting power of the eye is more convergent in the morning e. The refracting power of the eye is less convergent in the morning 2

3 Answer 3: Remember the underlying pathology in patients with Fuch s endothelial dystrophy is that they have a loss of endothelial cell function subsequently resulting in the a reduction in the ability of the cornea to maintain proper hydration. If these pumps are not functioning well they can not pump out extra fluid from the cornea, causing it to swell. This effect is typically increased during prolonged periods of eyelid closure due to the lack of corneal to the external environment. We can think of the cornea as a thick meniscus lens with 2 refracting surfaces (one convex and one concave),if the cornea swells we can think of it as getting thicker which allows us to use and equivalentlens equation to calculate the answer: F e = F 1 + F 2 t n 2 F 1F 2 (1) If we create variables where we know that the overall refracting power of the cornea is about 43 diopters and we hold F 1 (positive), F 2 (negative) and n 2 constant and increase t we will see that the equivalent power becomes more convergent (plus), making the patient more myopic in the morning (since the patient will have light focused more anterior to the retina under these conditions). B and D are correct. Question 4: Which (3) of the following would be true in a patient with Brown s syndrome on the left side? a. Hypertropia in primary gaze in the left eye b. Hypotropia in primary gaze in the left eye c. Hypotropia during ADduction of the left eye d. Hypertropia during ABduction of the left eye e. Widening of the palpebral fissure in the left eye on direct upgaze f. Narrowing of the palpebral fissure in the left eye on direct upgaze Answer 4: Brown s syndrome is a problem with the superior oblique muscle, tendon or the troclea that it passes through causing a restriction of the eye that can look similar to an inferior oblique palsy. Since there needs to be a balance in primary gaze between all 4 vertically acting muscles in each eye for the eyes to point straight ahead, and the eye can t elevate well, the affected eye will appear hypo in primary gaze. If the inferior oblique is the primary elevator during ADduction, the patient would likely be hypo in the effected eye when it is ADducted. If the patient looks directly up from the midline, the eye that is hypo will not move as much as the eye without the defect, however, the levator will still be stimulated on that side wich would cause the palpebral fissure to widen. B, C and E are correct. Question 5: Simulated K readings are as follows: OD: 015 // 105 OS: 075 // 165 In what meridian would you position the plus cylinder to correct the astigmatism? 3

4 a. OD: 015, OS: 165 b. OD: 015, OS: 075 c. OD: 105, OS: 165 d. OD: 105, OS: 075 Answer 5: If you look at the K readings you will see that this patient has a steeper vertical meridian in both the right and left eyes. In minus cylinder this would mean that the respective axes would be along the flatter meridian (OD: 015, OS: 165), however in plus cylander we would see the axes along the steeper meridian (OD: 105, OS: 075). D is the correct answer. Question 6: A patient has a prescription of x 090 in the right eye, however, the right lens was made incorrectly and he is wearing a prescription of x 180. If you present the astigmatic dial to him while he is wearing the new prescription which (2) of the following are true? a. Plus power needs to be added to place both line foci in front of the retina b. They would report the 3-9 o clock line as the darkest c. They would report the 6-12 o clock line as the darkest d. The interval of Sturm will be larger with the new glasses on than if the patient is uncorrected Answer 6: To answer this question we must first visualize what type of astigmatism we should be correcting for and then figure out which corneal meridian would be the steepest and finally find where the line foci fall while he is wearing the wrong prescription. So, if the patient is should wear a prescription of x 090 then this tells us that he is a compound hyperopic astigmat, both line foci fall behind the retina when he is uncorrected, and the horizontal line foci will fall behind the vertical line foci. When he is wearing the incorrect prescription that will cause the vertical meridian of his cornea to corrected by 1 diopter of excess plus and the horizontal meridian of his cornea to be corrected by 4 diopters of excess plus. This will cause both line foci to fall in front of the retina with the horizontal line falling behind the vertical line foci with a larger distance between these 2 lines than if the patient were uncorrected. With the line foci in this position, the patient would report that the 3-9 o clock line would be the darkest. B and D are correct. Question 7: A patient s distance refractive correction is x 172. If the examiner performs retinoscopy at 10 cm without a working lens in place, what lens combination will be in the phoropter when a neutral reflex is observed in all meridians? 4

5 a x 172 b x 172 c x 172 d x 082 Answer 7: Remember that even at radical working distances the convention of accounting for your working distance still holds true. In this case you would want to plot your answer on a power cross starting with the known prescription of x 172 and add 10 diopters from both meridians. This gives us a lens in place of x 172. A is correct. Question 8: A patient s distance prescription is x 060. If you perform retinoscopy at 50 cm without a working lens in place, describe the (2) reflexes that will be seen. a. Against motion when scoping the 060 meridian b. With motion when scoping the 150 meridian c. Against motion when scoping the 150 meridian d. With motion when scoping the 060 meridian e. Neutral motion when scoping the 060 meridian f. Neutral motion when scoping the 150 meridian Answer 8: The best way to think about this question is to draw a picture. The next step is to think of the eye as purely an optical system that is either too weak or too strong. In this case the patient is a mixed astigmat and when she is uncorrected the line foci corresponding to the 060 meridian falls 2 diopters behind the retina (too weak) and the line corresponding to the 150 meridian falls 1 diopter in front of the retina (too strong). Since we are scoping at a distance of 50 cm and not using a working lens, we would expect to further weaken the system by 2 diopters in all meridians, so now the line corresponding to the 060 meridian falls 4 diopters behind the retina (too weak) and the line corresponding to the 150 meridian falls 1 diopter behind the retina (too weak). If we remember that WEAK = WITH motion, then we will see with motion in both meridians. We could also say that since the 150 meridian is closest to the retina it will be a faster motion than that seen along the 060 meridian. B and D are correct. Question 9: While performing retinoscopy at 33 cm without a working lens in place, you observe WITH motion in all meridians which is neutralized by a 5 diopter lens. What is the distance prescription? 5

6 a b c d Answer 9: Again if you would like to use an equation to solve this problem we have: R x = F d w (2) Since each meridian shows with motion it means that at this distance the optical system is to weak (not convergent enough), if a 5 diopter lens is needed to perfectly strengthen this optical system, then it would need to be a D lens. If we subtract the working distance to find our refractive error we would see that F = +5.00D and d w = 3.00D which gives us a prescription of D. C is the correct answer. Question 10: Which (2) of the following findings would occur if an emmetropic patientšs accommodative system was not fully relaxed during distance retinoscopy at 50 cm with a retinoscopy lens in place? a. A more hyperopic prescription would be seen b. A more myopic prescription would be seen c. Against motion would be seen d. With motion would be seen Answer 10: In this case we have a neutralizing lens that accounts for our working distance so, we don t need to compensate for the divergence that occurs from a distance of 50 cm. So it is important to understand that accommodation works to increase the dioptric (converging) power of the eye, if this occurs we would bring light from infinity that should be focused on the retina (since the patient is an emmetrope) and place the focus anterior to the retina, similar to a myope. Remember, since we see against motion in systems that are too strong (convergent) we will see against motion. So a more myopic prescription would be found. B and C are correct. Question 11: What lens power is needed in the phoropter to obtain a neutral reflex on a cyclopleged emmetrope while performing retinoscopy at a distance of 20 cm? a b c

7 d Answer 11: A D lens must be in place to neutralize the amount of divergence that occurs at a distance of 20 cm in front of the spectacle plane. Since the patient is an emmetrope that can t accommodate their retina should be conjugate with infinity with a plano lens so the lens in the phoropter would be a D. B is the correct answer. Question 12: While performing retinoscopy with the correct working lens in place you observe with motion in all directions. Of the 2 principal meridians, the thinner reflection occurs when holding the streak vertically. Which of the following prescriptions is possible? a x 090 b x 180 c x 090 d x 180 Answer 12: Remember if we see with motion in all meridians with the correct working lens in place, that means that both line foci are falling behind the retina. So the patient is a compound hyperopic astigmat. We also know that the thinner (or slower) reflex comes from scoping the horizontal meridian using a vertical streak, which means that this is the more hyperopic of the 2 meridians. So on a power cross we could place along the horizontal meridian and along the vertical meridian and see that our answer would be x 180. D is the correct answer. Question 13: Which (3) of the following are true about a patient with the following prescription: x 065? a. The patient is a compound hyperopic astigmat b. The patient is a simple myopic astigmat c. The patient s prescription is against the rule d. The patient s prescription is with the rule e. The patient s prescription is oblique f. The patient s steepest corneal meridian is at 065 g. The patient s flattest corneal meridian is at 065 7

8 Answer 13: With this question it is important to draw a power cross of the correcting lens and translate that to possible corneal meridians that would correspond to this correction. We will see that the correcting lens power in the 065 meridian is and in the 155 meridian. Both meridians are hyperopic so the patient is considered a compound hyperopic astigmat. In minus cylinder notation the rx would be x 155 which is considered with the rule. If we created a cornea to correspond to this prescription we would see that the corneal meridian that needs the most plus would be the flattest and the meridian that needs the least plus would be the steepest. In this case the corneal meridian that requires the most plus power would be the 155 meridian so it is the flatter of the 2 principle meridians and the one that requires the least plus power would be the 065 meridian so it is the steepest meridian. A, D and F are correct answers. Question 14: A patient s distance prescription is x 090. If she is viewing through her spherical equivalent power which JCC lens choice will make the circles appear the most round? a x 180 b x 090 c x 090 d x 180 Answer 14: This patient s spherical equivalent would be plano. The prescription that would most closely correct the patient s ametropia and subsequently cause the circles to appear the most round would be the one that is in the same axis and closest to the required prescription. The first step is to convert the patient s prescription to plus cylinder form (or you could convert all answer choices to minus cylinder), and we will see that the prescription is x 180, the answer choice that is the closest to this is x 180. A is the correct answer. Question 15: Consider a patient with the following keratometry readings in the right eye: 180 // 090. If the patient is an uncorrected mixed astigmat; which (2) of the following are true? a. Both line images fall in front of the retina b. One line image falls in front of the retina and one falls behind the retina c. One line image falls on the retina and one falls in front of the retina d. The vertical line image will be in front of the horizontal line image e. The horizontal line image will be in front of the vertical line image 8

9 Answer 15: By definition a mixed astigmat will have one line foci in front of the retina and one line foci behind the retina, an example of a prescription that would correct for this would be x 090. For this patient the steepest meridian 180 and this will cause a vertical line focus that is in front of the horizontal line focus. B and D are correct answers. Question 16: After fogging a patient and finding the best monocular sphere the following prescription is in the phoropter: x 015. Cylinder refinement by 0.75 more plus is needed. What should the final prescription be? a x 105 b x 105 c x 105 d x 105 Answer 16: If we change the cyl component of the prescription by diopters, we would need to change the spherical component by (technically it would be -0.37) so the patients prescription would be x 015, which if we transpose into plus cylinder is x 105. C is the correct answer. Question 17: Which (2) of the following would you expect to find in a patient with disciform keratitis? a. Increased hyperopic prescription b. Decreased hyperopic prescription c. Stromal haze d. Branching epithelial defects e. Against the rule astigmatism f. With the rule astigmatism Answer 17: Disciform keratitis is a corneal endotheliitis secondary to herpes simplex. Just like fuch s endothelial dystrophy, we have a reduction in endothelial cell function leading to stromal haze and edema. And just like in the first example we would see a decrease in the amount of hyperopia (more myopic prescription). B and C are correct. Question 18: Which (2) of the following are true about a patient who is an uncorrected lenticular against the rule simple myopic astigmat? 9

10 a. The horizontal line foci is on the retina b. The vertical line foci is on the retina c. On stenopaic slit testing the patient would see most clearly with the slit oriented at 180 d. On stenopaic slit testing the patient would see most clearly with the slit oriented at 090 e. If a spherical lid attached RGP were placed on his eye it would most likely decenter nasally or temporally f. If a spherical RGP were placed on his eye it would likely perfectly correct for his refractive error Answer 18: Since the astigmatism is lenticular, the cornea would be mostly spherical so we would likely not see significant decentration laterally and and RGP that is spherical would not correct for any astigmatic component of the patient s refractive error. If we draw the picture of what an against the rule astigmatšs cornea looks like we will see that the line foci that will fall on the retina will arise from the vertical meridian of the cornea and be a horizontal line on the retina. So if we pinhole one meridian to allow the patient to see most clearly we would minimize the effects of the more myopic meridian of the cornea the 180 and the slit would be oriented at 090. A and D are correct. Question 19: A 4 diopter hyperopic examiner uses a direct ophthalmoscope to view a patient s retina, the lens in the ophthalmoscope is in order for the examiner to see a clear image of the retina, what is the patient s distance prescription (assuming no accommodation and both are uncorrected)? a DS b DS c DS d DS Answer 19: For this question you need to remember that the principal behind the direct ophthalmoscope is that the examiner s retina and the patient s retina must pe conjugate to each other in order to allow a clear image. In order to solve this you can think of the examiner as having an eye that is 4 diopters to weak ( 4.00), which is strengthened by 2 diopters of plus in the ophthalmoscope. This means that the optics of the examiner and the lens in the ophthalmoscope is a total of 2 diopters 2 weak ( 2.00). Thus the patient s eye must be 2 diopters too strong or a 2 diopter myope, so their distance prescription is Answer A is correct. Question 20: When performing lensometry on the right lens, you see the crosshairs intersect the first circle on your left and the second lower circle. What is the base direction in the right lens? 10

11 a. 1 BI, 2 BD b. 1 BO, 2 BU c. 1 BI, 2 BU d. 1 BO, 2 BD Answer 20: The crosshairs will point to the direction of the base, so if you are looking at the right lens and the crosshairs are decentered to the left that would be base out and the down would be base down, so the answer is 1 BO and 2 BD. Answer D is correct. Question 21: Which (2) of the following are true if you place a penlight 50 cm in front of a patient s eye? The patient s K s are 180 // a. The Hirschberg reflex appears to be behind the cornea by 10 mm b. The Hirschberg reflex appears to be behind the cornea by 4 mm c. The Hirschberg reflex appears to be 10 mm in front of the cornea d. The Hirschberg reflex appears to be 4 mm in front of the cornea e. The Hirschberg reflex is also known as Perkinje image 2 because it is behind the corneal surface f. The Hirschberg reflex is also known as Perkinje image 3 because it is located in the lens g. The Hirschberg reflex is virtual because it appears to be to the right of the interface h. The Hirschberg is virtual because it appears to be to the left of the interface Answer 21: We know that the Hirschberg reflex, is also known as Perkinje image 1, and is due to a reflection off of the anterior cornea. Since the K values give us the dioptric power of the cornea as a lens, we need to first find out the radius of curvature to calculate the dioptric power of cornea as a mirror since we need to view the reflection. So assuming the cornea has a refractive index of we will see that through the power formula that the radius of curvature is m. If seen as a mirror the cornea is a diverging mirror so the power is negative. Then using the mirror power equation, we see that the reflecting power of the cornea is D. If the object (penlight) is placed at 50 cm in front of the reflecting surface of the cornea we will see that it has a vergence of 2.00 D. So the outgoing vergence of the reflection is D, and the image appears to be located 3.8 mm (4 mm) to to the right of the cornea or behind the cornea and since it is located on the right side of the interface it is considered virtual. Answer B and G are correct. Question 22: Which of the following would improve the frame adjustment if a patient complains of the left lens being higher on the face than the right lens? 11

12 a. Bring the right temple IN b. Bring the right temple OUT c. Bring the right temple DOWN d. Bring the right temple UP Answer 22: If the left lens is higher than the right lens the 2 things that can be done are to bend the left temple up or bend the right temple down. Answer C is correct. Question 23: What is the total prismatic effect at the reading level in the following lens: Distance power: 2.00 DS Add power: Seg drop: 4 mm Seg type: Executive Seg height: 14 mm Reading level: 11 mm a BD b BD c BU d BD e BU Answer 23: When calculating total prismatic effect we need to know the prismatic effect when looking away from the optical center of the carrier lens AND the prismatic effect when looking away from the optical center of the add portion. In this case we are looking at a reading level of 11 mm so the prismatic effect from looking down in the distance lens is 2.2 BD and when we calculate the distance away from the near optical center we must know that we are reading at 7 mm below the top of the seg (11 mm - 4 mm for the seg drop) we also have to know that in an executive lens the near optical center is right at the seg line. So the prismatic effect induced from looking 7 mm below the near optical center is 1.05 BU, for a total prismatic effect of or 1.15 BD. Answer A is correct. Question 24: Which would be the best initial contact lens choice for a patient with the following findings in the right eye: K readings: 180 // 090 Spec Rx: x 090 a. Spherical rigid gas permeable b. Back toric rigid gas permeable c. Bitoric rigid gas permeable d. Front toric rigid gas permeable 12

13 Answer 24: Since most of the patients refractive cylinder is internal ( 2.00 x 090) based on the keratometry values ( 0.25 x 180) and spectacle prescription ( 1.50 x 090), that means that a spherical RGP would likely give us significant ATR over refraction, both the bitoric and back toric are unnecessary because the physical fit of a spherical lens should be adequate based on the small amount of corneal toricity. Therefore the front toric is the best initial option as it allows us to correct for any residual astigmatism on the front surface of the contact while maintaining a spherical back surface, a soft toric would also be a good option. Answer D is correct. Question 25: What is the effective power for the patient if a trial lens of x 180 rotates 10 degrees to the left? a x 170 b x 180 c x 010 d x 170 e x 010 Answer 25: The key to this question is to understand what EFFECTIVE power the patient sees through the lens when it is in a rotated position. In LARS we add power when the lens is rotated to the left to compensate for the effective power that the patient will see. So in this question a lens that is rotated to the left would effectively be applying the toric power of the prescription axis 170 instead of 180. So the answer would be x 170. If we compensated for this rotation by ordering the lens with axis 010 (based on LARS) we would see that when that lens is rotated by 10 degrees to the left it would apply the power exactly where we wanted it, axis 180. Answer A is correct. Question 26: A monocular patient who needs to use their right eye for eccentric viewing would have most difficulty with which of the following tasks? a. Scanning up a vertical row of letters b. Scanning down a vertical row of letters c. Saccading from left to right to find the next word d. Saccading from right to left to find the beginning of the next line 13

14 Answer 26: If the patient only has the right eye and the have to eccentrically view in order to read will have the most difficulty when moving into the temporal field of their right eye because it will always be a blindspot (due to the optic nerve head). They may also have difficulty with the others but the most difficult for them will be saccading from left to right to find the next word on the page. Answer C is correct. Question 27: An emmetropic patient would like to use his 5x telescope with a +20 D ocular lens and an adjustable tube to read at 40 cm. What is the distance that must separate the 2 lenses in this telescope in order for him to read with no accommodation? a. 24 cm b. 72 cm c. 42 cm d. 55 cm Answer 27: Since this is a Keplarian telescope (ocular is positive) we must first understand that the goal of a telescope is to view distant objects so we have parallel incoming light and parallel outgoing light. If we look at a near target we will have divergent incoming light and divergent exiting light, we must offset this by either using a reading cap, accommodation or by changing the tube length such that light still leaves the system parallel. To achieve this we must place the focus of light leaving the objective lens at the primary focal point of the ocular lens. To find the power of the objective lens we must use the equation: M tel = Foc F ob (3) We will see that the power of the objective lens is D. Now we simply calculate the distance behind the objective lens that the image will be from an object that is 40 cm in front of the lens. By calculating this out we find that the image location is 66.7 cm behind the objective lens, we also know that this image will become the object for the ocular lens and must be placed at the primary focal point of the ocular for parallel light to exit so we would place the ocular lens 66.7 cm +5 cm behind the objective lens so that the telemicroscope now has a tube length of 71.7 cm. Answer B is correct. Question 28: A patient would like to do needle work at a distance of 20 cm. With their current add power of D, they have a near point of 35 cm. What new add power should be prescribed so that the patient remains comfortable at their new working distance? a b c

15 d Answer 28: Remember we want to leave half of their amplitude of accommodation in reserve when prescribing add powers. In this case with a D add the patient has a near point of 35 cm or about 2.85 D. If her add power is giving her an advantage of 1.25 diopters, her amplitude of accommodation is 1.60 D. If we allow her to use half of that amplitude to see comfortably at near, then she can use 0.80 diopters. The working distance requires 5 diopters to offset the vergence from the near target placed at 20 cm so the add power we would prescribe would be +4.25, since she can also use about 0.80 diopters comfortably. Answer B is correct. Question 29: A patient with thyroid disease and hypertension presents with double vision that is worse when the patient looks to her left. Which (2) of the following are possible? a. She has a negative forced duction test showing a restricted left medial rectus b. She has a positive forced duction test showing a restricted left medial rectus c. She has a negative forced duction test showing a paralytic right lateral rectus d. She has a positive forced duction test showing a paralytic right lateral rectus e. She has a negative forced duction test showing a paralytic left lateral rectus f. She has a negative forced duction test showing a restricted right medial rectus Answer 29: In a patient with hypertension (a possible cause of cranial nerve 6 palsy) and thyroid disease (a possible cause for a medial rectus restriction) we must know what each finding will give us and tell us. Since the double vision is greatest when the patient looks to her left, we know the problem is either a paralytic left lateral rectus or a restricted left medial rectus. A positive forced duction test means that the muscle is restricted, while a negative forced duction test means that the eye complies and the diplopia is likely due to a paralytic muscle. Answer B and E are correct. Question 30: An elderly patient presents with complaints of double vision after falling and hitting his head 3 weeks ago. He has a small hypertropia in the right eye in the primary position of gaze. When he walks into your office he has a head turn left and a head tilt left. Which (3) of the following are most likely true. a. He has a right superior oblique palsy b. He has a left inferior oblique entrapment c. He likely has no torsional complaints of his diplopia d. He likely has torsional diplopia 15

16 e. He will likely have normal vertical vergence ranges f. He will likely have large vertical vergence ranges Answer 30: This patient presents placing himself in a position to minimize his diplopia. So a head turn right (gaze left) will maximize his diplopia as will a head tilt right. If we do a Parks 3 step we will see that the impacted muscle is the right superior oblique. Since the onset is new, he will likely have torsional diplopia and normal vertical vergence ranges, these are both characteristic of acquired fourth nerve palsies. Congenital CN 4 palsies are more likely NOT to have torsional diplopia, but can have large vertical vergence ranges, you may also see a fuller face on one side due to the longstanding nature of the head tilt. Answers A, D and E are correct. Question 31: A 3 diopter hyperopic patient is 1 exo at 40 cm when looking through his manifest refraction. With a add his is 15 exo. What would you predict his phoria to be at 40 cm if he was not wearing his prescription. a. 27 eso b. 20 eso c. 1 exo d. 22 exo e. 29 exo Answer 31: From the given phoria information we see that he becomes 14 diopters more exo with a add, this means his ACA ratio is 14/2, or 7/1. So if he were uncorrected he would have to accommodate by 3 diopters more than if he has his correction on, so he would be 21 diopters more eso than he was with correction. So we would measure his phoria to be about 20 eso. Answer B is correct. Question 32: Which (3) of the following findings are consistent with the diagnosis of convergence excess? a. Normal positive relative accommodation b. Normal negative relative accommodation c. High amplitude of accommodation d. Normal monocular accommodative facility e. Reduced base out ranges at near f. Excess accommodative lag on near retinoscopy g. Significant esophoria at distance 16

17 Answer 32: Remember in a patient who has convergence excess we expect to have more eso at near and about ortho at distance. These patients have a high AC/A ratio and we would expect that these patients would have more plus findings on their near retinoscopy findings, this is due to the fact that they will try to relax their accommodation in order to reduce their eso posture. Since negative fusional vergence ranges are typically reduced but positive fusional ranges are large, these patients will have low positive relative accommodation findings but normal negative relative accommodation findings. Since fusional vergence is eliminated, patients will typically have normal monocular accommodative facility results. Answers B, D and F are correct. 17