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Optometry Board Practice Test for the NBEO® Part 1 Test #2 – Ocular Motility & Binocular Vision This test is comprised of 54 items, which must be completed within 54 minutes. |
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A patient comes to your office with a spectacle prescription of -8.50-1.00×180. His chief complaint is blurry vision at distance with his spectacles and notices an improvement with what you observe is a pantoscopic tilt of the spectacles. What change in his refractive error would you expect according to his chief complaint?
E. Increase myopia and with the rule astigmatism. Remember panto tilt increases the power of the 180 axis
E. Increase myopia and with the rule astigmatism. Remember panto tilt increases the power of the 180 axis
A Prescription for an aphakic patient reads:
OD: +16.00 DS
OS: +16.00 DS
Refracting distance of 14 mm.
This patient’s far point is located approximately
C. vertex the power of the eye. This will help you find the patients far point F=16.00, L={16/(1-(0.014)(16)} = 48.5 mm behind the corneal apex.
C. vertex the power of the eye. This will help you find the patients far point F=16.00, L={16/(1-(0.014)(16)} = 48.5 mm behind the corneal apex.
Myopia, hyperopia, and astigmatism are considered what radial order using zernike polynomials?
B. second order
B. second order
If a person with high minus spectacle prescription switches to contact lens, the power on the contact lens would be which of the following when compared to the power of the spectacle lens?
A. less than. As you move a minus les closer to the eye, the effective power increases; this mean it gains more plus power. We compensate for this effective power by prescribing less power as we get closer to the eye which is why contact lenses are more plus compared to the glasses prescription.
A. less than. As you move a minus les closer to the eye, the effective power increases; this mean it gains more plus power. We compensate for this effective power by prescribing less power as we get closer to the eye which is why contact lenses are more plus compared to the glasses prescription.
A lens clock measures: Front surface @090 =+8.00D. Front surfaces @180 =+7.00 D. Ocular surface = -5.00D. What would the approximate power of the lens be if the lens were made from plastic n=1.60?
D.
D.
A 45 year old patient comes to your office with a chief complaint of blurry vision at distance at the end of the day. His current spectacle prescription has a back vertex power of -7.00 DS OD, OS with an add of +1.50 D. He explains he notices an improvement of symptoms if his spectacles are placed further from his eyes. According to his chief complaint and objective findings, the practitioner may expect what findings after finalizing the glasses prescription?
B. less myopia.
By increasing the vertex distance of the lenses, he is effectively inducing more plus power. Since the chief complaint is mainly for distance, we know presbyopia isn’t the best answer in this case as these patients have difficulties seeing at near. Given his age, its not a new finding since his habitual prescription has an add already built in.
B. less myopia.
By increasing the vertex distance of the lenses, he is effectively inducing more plus power. Since the chief complaint is mainly for distance, we know presbyopia isn’t the best answer in this case as these patients have difficulties seeing at near. Given his age, its not a new finding since his habitual prescription has an add already built in.
Which of the following aberrations is used as part of the refraction in visual exam?
A. Chromatic aberration.
A. Chromatic aberration.
An ocular lens +4.00 +1.00 x180 is ground in minus cylinder form with the flattest of the two back surfaces having a power of -3.00 D. What is the base curve?
C. +8.00
Don’t let the plus cyl format fool you. Transpose this power into minus cyl form = +5.00 – 1.00 x 090. In minus cyl format, the base curve is the most plus curve. In plus cyl format, the base curve is the most minus curve. The flatter of the two meridians is the most plus of the two.
The base curve is the front surface of the lens and gives the lens its shape.
C. +8.00
Don’t let the plus cyl format fool you. Transpose this power into minus cyl form = +5.00 – 1.00 x 090. In minus cyl format, the base curve is the most plus curve. In plus cyl format, the base curve is the most minus curve. The flatter of the two meridians is the most plus of the two.
The base curve is the front surface of the lens and gives the lens its shape.
You are performing duochrome test in the refraction. The lens placed in the phoropter at the beginning of the test is a -4.00 DS lens. If the patient reports seeing the letters on the green side darker and better than the red side; what does this mean?
B. RAM GAP. Red add minus because the lens is over plused or under minused. Green add plus because the lens is under plused or over minused
B. RAM GAP. Red add minus because the lens is over plused or under minused. Green add plus because the lens is under plused or over minused
A person with 20/80 VA is just able to read a letter 11 meters in front of him. What is the size of the letter?
A. 6.4cm : 80/20(8.73)(11/6)= 64 mm which is equivalent to 6.4 cm
A. 6.4cm : 80/20(8.73)(11/6)= 64 mm which is equivalent to 6.4 cm
A patient has a minimum angle of resolution at far and near of 3 feet, what is this patients VA at 40 cm?
B. 1.2 M ; 3’=M/0.4
B. 1.2 M ; 3’=M/0.4
A patient can read the line labeled 1.5M at 40 cm. What is the equivalent Snellen Acuity. (SELECT 4)
A.B.C.D 1.5/4=3.75 =x/20->20/75 ; 20/75=6/x
A.B.C.D 1.5/4=3.75 =x/20->20/75 ; 20/75=6/x
What is the nominal magnification of a 3 inch focal length lens?
A. 3.28X ; 3in(0.0254/1in)=0.0762 m; (1/0.0762m)= (13.12/4) = 3.28X
A. 3.28X ; 3in(0.0254/1in)=0.0762 m; (1/0.0762m)= (13.12/4) = 3.28X
When a lensometer reads plano, the target should be where:
A. Primary focal length coincides with the far point of a lens. If the far point of a lens is at infinity then the power of the lens is plano.
A. Primary focal length coincides with the far point of a lens. If the far point of a lens is at infinity then the power of the lens is plano.
If a keratometer is calibrated for a corneal index of 1.3376, with 76mm diameter mire, 3mm diameter of the image, what is the K reading if the mire is located 76.66 mm aways from the cornea?
D. +55.75D
D. +55.75D
While performing a slit lamp biomicroscopy, at what distance must the object be from the objective lens such that an objective lens has a focal power of 22.75 mm and the ocular lens has 200 mm focal lens separated by a distance of 500 mm. The OTL is 277.25 mm. Assume an emmetropic optometrist
C.
C.
A patient is a 3D uncorrected hyperopia and examiner is an 3D uncorrected myope. Assuming accommodation fully relaxed on the examiner and patient, what lens should be used in the peephole of the direct ophthalmoscope to view the fundus?
D. +3+-3=0D
D. +3+-3=0D
Performing retinoscopy on a 3D myope, a +1.50-3.00×090 cylinder lens is placed behind the phoropter. What will be your retinoscopy finding?
D.
D.
On a lensometer, what is the main device that detects parallel light?
D. Telescope.
D. Telescope.
An aphakic patient was refracted at 14 mm and found +16.50 D. If the spectacles are going to be worn with a vertex of 8mm; what back vertex power must the lenses have?
E. +18.37 DS.
E. +18.37 DS.
What type of astigmatism would a patient with soft contact lens prescription of : -2.00+2.00×100 have?
B. Simple myopic with the rule
B. Simple myopic with the rule
A reduced eye with index 1.333 is corrected with spectacle -5.25DS lens placed 12 mm in front of the eye. What is the power of this eye assuming standard axial length?
B. 67D ; (-5.25)/[1-(0.012)(-5.25)] = -4.93D; 1.333-1.000/-4.93= 67.54D
B. 67D ; (-5.25)/[1-(0.012)(-5.25)] = -4.93D; 1.333-1.000/-4.93= 67.54D
While doing retinoscopy neutralize the 045 meridian with +6.00D lens at 40 cm. You then neutralize the 135 meridian with a +4.25 D lens at 67 cm. What is the prescription you would prescribe to this patient?
C.
C.
Given the index the cornea to be 1.376 and index of the aqueous to be 1.336 of a schematic eye, what’s the complete equivalent power of the cornea when the distance between the front and back surfaces of the cornea is 0.5 mm (rfront= 7.7mm, rback= 6.8mm)
B.+43.056D
B.+43.056D
While working on a lensometer with a standard +25D badal lens, you moved the dial 5mm closer to the badal. You see a clear neutral target in the lensometer, what is the back surface power of the lens? How many mm of movement correspond to 1D?
C. +3.125D, 1.6mm
C. +3.125D, 1.6mm
You perform near point of accommodation on a 2.00D myope wearing contact lens for full correction and determine it to be 8 cm in front of the cornea. What is his/her amplitude of accommodation?
C. 10.5D. 100/8=12.5D far point; Near point 2D. Aof A is near – far point : -12.5-2=10.5D
C. 10.5D. 100/8=12.5D far point; Near point 2D. Aof A is near – far point : -12.5-2=10.5D
Angle between pupillary axis and nodal axis is referred to as?
A. Angle Kappa. Angle lambda is the angle between pupillary axis and line of sight. Angle alpha is the angle between optic axis and visual axis. Angle gamma is the angle between fixation axis and the optical axis.
A. Angle Kappa. Angle lambda is the angle between pupillary axis and line of sight. Angle alpha is the angle between optic axis and visual axis. Angle gamma is the angle between fixation axis and the optical axis.
Which purkinje image is used when measuring the cornea on a keratometer?
A. I. Purkinje image 1 is focused on the anterior aspect of the cornea. II is the posterior aspect of the cornea. III is the anterior aspect of the lens. IV is the posterior aspect of the lens as well as the only image that is inverted and real
A. I. Purkinje image 1 is focused on the anterior aspect of the cornea. II is the posterior aspect of the cornea. III is the anterior aspect of the lens. IV is the posterior aspect of the lens as well as the only image that is inverted and real
An astigmatic reduced eye has a power of +60 D @ 090 and +58 D @ 180 with standard axial length. What is the ocular refraction?
B. +2.00-2.00X180 without doing any math we know there is 2.00D of cylinder on this patient. So options A and D are eliminated. WTH astigmatism means there is more power on the 090 corneal axis. ATH astigmatism means there is more power on the 180 corneal axis. In this case the higher power is on the 090 so the patient is WTH astigmatism. The standard plano power of an eye is 60D so the 090 axis will have plano power and -2.00 on the 180 axis.
B. +2.00-2.00X180 without doing any math we know there is 2.00D of cylinder on this patient. So options A and D are eliminated. WTH astigmatism means there is more power on the 090 corneal axis. ATH astigmatism means there is more power on the 180 corneal axis. In this case the higher power is on the 090 so the patient is WTH astigmatism. The standard plano power of an eye is 60D so the 090 axis will have plano power and -2.00 on the 180 axis.
The far point of a patient was determined to be a real object placed 50 cm from the eye’s principal point while wearing a +4.00DS contact lens. What amount and type of ametropia does the patient have?
B. 2D of hyperopia. Real object vergence is -2.00D hitting a +4.00D len . -2.00D-+4.00D=+2.00D
B. 2D of hyperopia. Real object vergence is -2.00D hitting a +4.00D len . -2.00D-+4.00D=+2.00D
You determined the near point of accommodation (push up method) to be 40 cm in front of point P when fully corrected. You repeated the NPA without correction and found it to be 67 cm in front of point P. Assuming full accommodation on both procedures; the amount and type of ametropia is?
C. 1D hyperopia. Patient FP vergence is -2.50D corrected. Patients FP vergence uncorrected is -1.50. Comparing the difference between these numbers will give you the total ametropia of the patient
C. 1D hyperopia. Patient FP vergence is -2.50D corrected. Patients FP vergence uncorrected is -1.50. Comparing the difference between these numbers will give you the total ametropia of the patient
According to Snell’s Law, if light enters a material from air at 25 deg, what will be the new angle of refraction, if the material has an index of 1.53?
C. 16 deg. We can solve this without any sin inverse or calculator. When light hits an object with a higher index of refraction the light is going to bend more creating less of a degree. 16 is the only number that is smaller than 25. Keeping this theory in mind will help you solve any problem like this.
C. 16 deg. We can solve this without any sin inverse or calculator. When light hits an object with a higher index of refraction the light is going to bend more creating less of a degree. 16 is the only number that is smaller than 25. Keeping this theory in mind will help you solve any problem like this.
What is the object height of a system where the inverted real +5 mm image is located 25 cm behind a +5D lens?
C. Real erect 1.25 mm object. Real objects create real images. We first have to find the object vergence ( L’=L+F ; (100/25)=L+(5)-> 4=L+5 -> L=-1 ; then we find the magnification of the system ; M= obj/img ; M=(-1/4)-> M= -0.25 ; M=obj/image ; 0.25=obj/5 -> 1.25 mm
C. Real erect 1.25 mm object. Real objects create real images. We first have to find the object vergence ( L’=L+F ; (100/25)=L+(5)-> 4=L+5 -> L=-1 ; then we find the magnification of the system ; M= obj/img ; M=(-1/4)-> M= -0.25 ; M=obj/image ; 0.25=obj/5 -> 1.25 mm
A 1.53 index prism with an apical angle of 15 deg used at 25 cm, produces an image displacement of how much in cm?
A. 7.95 degrees ; d=A(n-a) -> d= 15(1.53-1)
A. 7.95 degrees ; d=A(n-a) -> d= 15(1.53-1)
What is the latency for optokinetic nystagmus reflex?
C. 140 millisec: Vestibulo ocular nystagmus is 16 millisec, and OKN is 140 millisec
C. 140 millisec: Vestibulo ocular nystagmus is 16 millisec, and OKN is 140 millisec
What term best describes the reflex eye movement that is seen following the termination of a barany chair clockwise stimulation in the dark?
B. Post rotational nystagmus. OKAN describes the reflex eye movement that is seen after the stoppage of a barany chair stimulation in ambient room lighting that is enriched with objects.
B. Post rotational nystagmus. OKAN describes the reflex eye movement that is seen after the stoppage of a barany chair stimulation in ambient room lighting that is enriched with objects.
Which of the following laws states that each SCC influences a yoked pair of extraocular muscles?
B. Sherrington law : when one set of muscles is stimulated, muscles opposing the action of the first are simultaneously inhibited.
B. Sherrington law : when one set of muscles is stimulated, muscles opposing the action of the first are simultaneously inhibited.
The line within panum’s space that connects the set of spatial points in space that are perceived as single while binocularly fixating on an object point is called?
B. horopter.
B. horopter.
If the objective angle of deviation in esotropia is 35 degrees and the angle of anomaly is 45 degrees esotropia, what form of correspondence has a patient in the absence of diplopia?
A. Paradoxical type 1 anomalous retinal correspondence. harmonious ARC, the objective angle is equal to the subjective angle. In unharmonious ARC, the subjective angle is less than the objective angle. If the localization of the subjective and objective angles is crossed or uncrossed it is called paradoxical ARC. paradoxical type 1 is when the angle of anomaly is greater than the objective angle. Paradoxical type 2 is when the objective deviation is less than the subjective deviation.
A. Paradoxical type 1 anomalous retinal correspondence. harmonious ARC, the objective angle is equal to the subjective angle. In unharmonious ARC, the subjective angle is less than the objective angle. If the localization of the subjective and objective angles is crossed or uncrossed it is called paradoxical ARC. paradoxical type 1 is when the angle of anomaly is greater than the objective angle. Paradoxical type 2 is when the objective deviation is less than the subjective deviation.
A patient walks into your office and performs a bruckner test. Upon illumination you notice the right eye has a dark/dull reflex and the left eye has a bright beautiful reflex. Which eye is the fixating eye and which is deviating?
A. OD and OS. Fixating the eye will absorb all light on the macula. Deviating eye will glow because light is focused on the non macula area.
A. OD and OS. Fixating the eye will absorb all light on the macula. Deviating eye will glow because light is focused on the non macula area.
Which of the following does not affect accommodative response?
E. All of the above affect the response
E. All of the above affect the response
Which of the following is not a non-strabismic anomaly? (SELECT 2)
A,D. Non-Strabismic are visual disorders that affect the subject’s binocular vision and visual performance, especially when performing tasks requiring near vision. They occur when the accommodative and/or vergence response of the visual system is defective. Strabismus is usually unilateral condition or tropia
A,D. Non-Strabismic are visual disorders that affect the subject’s binocular vision and visual performance, especially when performing tasks requiring near vision. They occur when the accommodative and/or vergence response of the visual system is defective. Strabismus is usually unilateral condition or tropia
You perform Hirschberg test on a patient and the test reveals: OD is centered and OS -1.5mm. What is the magnitude?
A. 1.5mm x 22=33PD
A. 1.5mm x 22=33PD
What do we measure for risley prisms for blur to break?
A. range of accommodative vergence. Blur is the first indication that accommodation has reached its max. When break occurs then accommodative vergence is broken thus indicating the end of accommodation.
A. range of accommodative vergence. Blur is the first indication that accommodation has reached its max. When break occurs then accommodative vergence is broken thus indicating the end of accommodation.
The following are the results for the DEM test: vertical =+1.03 and horizontal is -2.00. Which of the following is the MOST probable diagnosis?
A. Saccadic dysfunction. DEM test results indicated 4 different types of responses. Type 1 Essentially normal performance in Horizontal Time, Vertical Time, and Ratio. Type 2 Characterized as abnormally increased time to complete the Horizontal Test in the presence of normal performance on the Vertical Test. The Ratio would be abnormally high in this case. Type II behavior is characteristic of oculomotor dysfunction. Type 3 Typified as an abnormal increase in both the Horizontal Test and Vertical Test times, but with a normal Ratio. In the presence of a normal Ratio, the Horizontal Test time is influenced and increased because of an abnormal baseline. This represents a case of difficulty in automaticity in number calling skills, not an ocular motility deficit. Type 4 Increased Horizontal and Vertical Test times, and an abnormally high Ratio. This case is a combination of Type II and Type III behavior patterns. There are deficiencies in automaticity/oculomotor skills.
A. Saccadic dysfunction. DEM test results indicated 4 different types of responses. Type 1 Essentially normal performance in Horizontal Time, Vertical Time, and Ratio. Type 2 Characterized as abnormally increased time to complete the Horizontal Test in the presence of normal performance on the Vertical Test. The Ratio would be abnormally high in this case. Type II behavior is characteristic of oculomotor dysfunction. Type 3 Typified as an abnormal increase in both the Horizontal Test and Vertical Test times, but with a normal Ratio. In the presence of a normal Ratio, the Horizontal Test time is influenced and increased because of an abnormal baseline. This represents a case of difficulty in automaticity in number calling skills, not an ocular motility deficit. Type 4 Increased Horizontal and Vertical Test times, and an abnormally high Ratio. This case is a combination of Type II and Type III behavior patterns. There are deficiencies in automaticity/oculomotor skills.
Which of the following are associated with a motor response? (SELECT 3)
A,C,E
A,C,E
You perform duochrome and ask your patient which side is more clear; and patient says that both sides are equal. What could be the problem?
D. Room illumination. Room illumination is important when doing duochrome. If the illumination is too bright then you can cause a pinhole effect on the patient thus creating a false reading.
D. Room illumination. Room illumination is important when doing duochrome. If the illumination is too bright then you can cause a pinhole effect on the patient thus creating a false reading.
Which of the following schematic eyes has a two surface lens?
D. Gullstrand 2 is also known as the exact eye model. This means there are 6 refractive areas 2 of which are lens surfaces
D. Gullstrand 2 is also known as the exact eye model. This means there are 6 refractive areas 2 of which are lens surfaces
A patient with astigmatism was refracted at 10 mm and found the following spectacle correction; +2.50+2.50×180. Where are the far points located for this patient?
H. 19 and 39 cm behind the eye.
H. 19 and 39 cm behind the eye.
When light travels from a smaller index material to a higher index material, what happens to the vergence of the light?
B. Light always bends towards the higher index
B. Light always bends towards the higher index
A patient with a 35PD exotropia deviation has surgical correction. Afterwards, they are neutralized at 1PD exotropia with the unilateral cover test. What will the result be on W4DOT?
B. Uncrossed diplopia. Atypical UHARC paradoxical type 1: Pt will have opposite objective and subjective results
B. Uncrossed diplopia. Atypical UHARC paradoxical type 1: Pt will have opposite objective and subjective results
What percent of anisometropic amblyopes improves with best rx only?
C. 30% According to the Amblyopia treatment studies 30% of anisometropic amblyopes can be corrected by just Best rx only.
Chen AM, Cotter SA. The Amblyopia Treatment Studies: Implications for Clinical Practice. Adv Ophthalmol Optom. 2016 Aug;1(1):287-305. doi: 10.1016/j.yaoo.2016.03.007. PMID: 28435934; PMCID: PMC5396957.
C. 30% According to the Amblyopia treatment studies 30% of anisometropic amblyopes can be corrected by just Best rx only.
Chen AM, Cotter SA. The Amblyopia Treatment Studies: Implications for Clinical Practice. Adv Ophthalmol Optom. 2016 Aug;1(1):287-305. doi: 10.1016/j.yaoo.2016.03.007. PMID: 28435934; PMCID: PMC5396957.
You perform a Brock string test on a patient and they report that the strings cross in front of the center dot. What type of deviation do they have?
A. Eso deviation. Eso deviation has a nasal eccentric fixation point thus creating a crossing of the lines closer than normal
A. Eso deviation. Eso deviation has a nasal eccentric fixation point thus creating a crossing of the lines closer than normal
When your patient is doing a left head tilt, the respective action of the left inferior rectus and the right superior rectus can be explained by the following principle
A. Sherrington’s law of innervation is concerned with the contraction and relaxation of the muscles in the same eye. Hering’s law is the muscle innervation between 2 eyes. Remember sherring the same eye.
A. Sherrington’s law of innervation is concerned with the contraction and relaxation of the muscles in the same eye. Hering’s law is the muscle innervation between 2 eyes. Remember sherring the same eye.