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Optometry Board Practice Test for the NBEO^{®} Part 1 Full Length Test #1 – Ophthalmic Optics This test is comprised of 21 items. This is PRACTICE mode. There is no countdown timer and answers are shown after each question. 
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A concave glass surface of index 1.50 has a power of 12.00D in air. What is the dioptric power of the glass surface if it is placed under water (n = 1.33)?
B. 4.08 D. You must first determine r by using the equation [Fair = 12.00 D = (1.50 – 1) / r]. You determine that r = 0.0417m. Using the new equation [Fwater = (1.50 – 1.33) / 0.0417m].
B. 4.08 D. You must first determine r by using the equation [Fair = 12.00 D = (1.50 – 1) / r]. You determine that r = 0.0417m. Using the new equation [Fwater = (1.50 – 1.33) / 0.0417m].
An insect (in air) is hovering 40cm above the surface of a pond (index 1.33). How far above the water does the insect appear to a frog beneath the surface?
A. The insect appears to be 53.20cm above the surface. To solve, first draw a diagram with the insect above the water with a line acting as the surface of the pond. Now use the equation (l’/n’) = (l/n). It should read (l’/1.33) = (40/1); l = 53.20
A. The insect appears to be 53.20cm above the surface. To solve, first draw a diagram with the insect above the water with a line acting as the surface of the pond. Now use the equation (l’/n’) = (l/n). It should read (l’/1.33) = (40/1); l = 53.20
A lens clock in air is calibrated for glass of index 1.53. The clock reads +6.00 D when it is placed on a spherical convex surface with an index of 1.66. What is the true dioptric power of the surface in air?
C. The true power is +7.47 D. To solve, first determine r using the clock reading: [Flc = +6.00 D = (1.53 – 1) / r; r = +0.08833m. Now solve for true with Flens = (1.66 – 1) / r = 0.66/0.08833 = +7.47 D)
C. The true power is +7.47 D. To solve, first determine r using the clock reading: [Flc = +6.00 D = (1.53 – 1) / r; r = +0.08833m. Now solve for true with Flens = (1.66 – 1) / r = 0.66/0.08833 = +7.47 D)
Which of the four Purkinje images is real and inverted?
D. Purkinje image IV is the only image that is real and inverted. The remaining three are virtual and upright.
D. Purkinje image IV is the only image that is real and inverted. The remaining three are virtual and upright.
A patient presents with the following glasses prescription: +2.00 3.00 x 090. What is the power in the 180 meridian?
D. The power in the 180 meridian is 1.00. On an optical cross, 2.00DS falls on the 90 meridian. Subtract the astigmatic difference of 3.00 to find the resulting 1.00DS.
D. The power in the 180 meridian is 1.00. On an optical cross, 2.00DS falls on the 90 meridian. Subtract the astigmatic difference of 3.00 to find the resulting 1.00DS.
A child looks at an insect embedded in glass 3 cm below the surface. What is the apparent depth of the insect?
A. The apparent depth is 2.0 cm. Apparent depth is determined using the index of refraction of both mediums and the given depth. Therefore, n2/l’ = n1/l → (1.0/l’ = 1.5/0.03m = 50 cm) → 1/50 = l’ = 2.0cm.
A. The apparent depth is 2.0 cm. Apparent depth is determined using the index of refraction of both mediums and the given depth. Therefore, n2/l’ = n1/l → (1.0/l’ = 1.5/0.03m = 50 cm) → 1/50 = l’ = 2.0cm.
What is the power of a thin lens made of Crown glass in air with a front radius of curvature of 4 cm (with a convex surface) and back radius of curvature of 2 cm (with a concave surface)?
C. The power of the thin, Crown glass lens (n = 1.523) is 13.08 D. First draw a picture of this image (should look like two “C”s). First find F1 = (n2 – n1)/r, then F2 with the same equation. Done correctly will result in F1 = +13.075 D; F2 = 26.15 D; F = F1 + F2 → F = 13.075 + (26.15) = 13.075 D.
C. The power of the thin, Crown glass lens (n = 1.523) is 13.08 D. First draw a picture of this image (should look like two “C”s). First find F1 = (n2 – n1)/r, then F2 with the same equation. Done correctly will result in F1 = +13.075 D; F2 = 26.15 D; F = F1 + F2 → F = 13.075 + (26.15) = 13.075 D.
A corneal topographic map shows a “Figure 8” pattern with noted horizontal steepening. What does this represent?
D. This CT pattern is indicative of againsttherule astigmatism. WTR astigmatism is characterized by a bow tie pattern with vertical steepening. Keratoconus would present with inferior steepening. Pellucid marginal degeneration scans are known for a kissing dove or crab claw appearance.
D. This CT pattern is indicative of againsttherule astigmatism. WTR astigmatism is characterized by a bow tie pattern with vertical steepening. Keratoconus would present with inferior steepening. Pellucid marginal degeneration scans are known for a kissing dove or crab claw appearance.
You measure a pair of spectacles using a manual lensometer. Your first reading is +1.00 D at the 90 meridian. Your second reading is 2.00 D. What is the final prescription?
D. The correct answer is +1.00 3.00 x 090. Although cylinder lines are cleared at 2.00 D, the difference between both meridians is the DC of the prescription.
D. The correct answer is +1.00 3.00 x 090. Although cylinder lines are cleared at 2.00 D, the difference between both meridians is the DC of the prescription.
Transpose the following prescription: 2.00 2.00 x 180
B. 4.00 +2.00 x 090. To transpose, first move the total cylindrical value to the spherical portion. In this example, 2.00 + (2.00) = 4.00. Once you have the sphere, reverse the symbol of the DC and flip the axis of the original equation.
B. 4.00 +2.00 x 090. To transpose, first move the total cylindrical value to the spherical portion. In this example, 2.00 + (2.00) = 4.00. Once you have the sphere, reverse the symbol of the DC and flip the axis of the original equation.
Which of the following is defined as the optical image of a physical aperture stop formed by all the lenses in front of it?
A. The entrance pupil is defined as the image of the aperture stop formed by all lenses in front of it. The exit pupil is the image of the aperture stop formed by all lenses behind it. An entrance port is the image of the field stop formed by all lenses in front of it. The image of the field stop formed by lenses behind it is known as the exit port.
A. The entrance pupil is defined as the image of the aperture stop formed by all lenses in front of it. The exit pupil is the image of the aperture stop formed by all lenses behind it. An entrance port is the image of the field stop formed by all lenses in front of it. The image of the field stop formed by lenses behind it is known as the exit port.
A patient presents with acute angle closure glaucoma. You are able to reduce intraocular pressure in the office and refer for laser peripheral iridotomy. What type of laser is most commonly used in this procedure?
C. YAG laser is used for LPI, PCO, and SLT. Argon and Krypton are used for PRP, ALT, retinal tears, and in rare instances LPIs. Excimer/LASIK laser is used for LASIK and PRK. Helium neon is used to illuminate the retina to view the fundus.
C. YAG laser is used for LPI, PCO, and SLT. Argon and Krypton are used for PRP, ALT, retinal tears, and in rare instances LPIs. Excimer/LASIK laser is used for LASIK and PRK. Helium neon is used to illuminate the retina to view the fundus.
Which of the following is FALSE regarding BIO when compared to the direct ophthalmoscope?
C. When compared to a direct ophthalmoscope, the BIO has a larger field of view, larger depth of focus, SMALLER magnification, and an inverted image.
C. When compared to a direct ophthalmoscope, the BIO has a larger field of view, larger depth of focus, SMALLER magnification, and an inverted image.
The ANSI Z87.1 Safety Standard for high mass impact is measured using which of the following? (PICK 3)
A,C,D: There are two definitions within the ANSI Z87.1 Safety Standard for Impact: high mass impact and high velocity impact. High mass impact is the drop ball test. This test requires a pointed projectile that weighs 500g dropped from 50 inches. The high velocity impact is assessed using a 0.25 inch diameter steel ball fired at 150 feet per second.
A,C,D: There are two definitions within the ANSI Z87.1 Safety Standard for Impact: high mass impact and high velocity impact. High mass impact is the drop ball test. This test requires a pointed projectile that weighs 500g dropped from 50 inches. The high velocity impact is assessed using a 0.25 inch diameter steel ball fired at 150 feet per second.
A patient’s pupillary distance (PD) is measured at 70 mm. The patient is fit into a frame with a PD of 74 mm. How much and in what direction must the right lens be decentered to fit correctly?
D. The lens will need to be moved 2 mm to place the optical center in front of the patient’s eye. Do not forget that TOTAL decentration is 4 mm but each lens only moves half of that. Easy rule of thumb – patient PD < frame PD = move lens in and vice versa
D. The lens will need to be moved 2 mm to place the optical center in front of the patient’s eye. Do not forget that TOTAL decentration is 4 mm but each lens only moves half of that. Easy rule of thumb – patient PD < frame PD = move lens in and vice versa
A patient presents with complaints of blurred vision while reading. You see he is wearing a pair of bifocals and that the seg height appears too low. What adjustment can you make to adjust the spectacles?
B. You can increase the seg height by any of the following: narrow the pads, move pads down, increase vertex distance, reduce pantoscopic tilt, or shrink the bridge. Consider the opposite in cases where the seg height appears too high.
B. You can increase the seg height by any of the following: narrow the pads, move pads down, increase vertex distance, reduce pantoscopic tilt, or shrink the bridge. Consider the opposite in cases where the seg height appears too high.
“Barrel” distortion of an image is created when looking through what type of lens?
C. A barrel distortion is created by a minus lens. A plus power lens creates a pincushion effect. The other two options are distractors.
C. A barrel distortion is created by a minus lens. A plus power lens creates a pincushion effect. The other two options are distractors.
Which of the following lens materials causes the most dispersion of light leading to increased chromatic aberrations?
B. Polycarbonate has the lowest Abbe value of all these options. Abbe value is inversely proportional to the level of chromatic aberrations. In other words, the higher the Abbe value, a lower amount of chromatic aberrations will be induced.
B. Polycarbonate has the lowest Abbe value of all these options. Abbe value is inversely proportional to the level of chromatic aberrations. In other words, the higher the Abbe value, a lower amount of chromatic aberrations will be induced.
A patient looks 4 mm to the right in a +2.00 D lens over the right eye. What is the resulting power and direction of the prism induced?
D. Total prism induced is 0.8 BI. First, recall Prentice’s law (Total prism = power of lens x decentration in centimeters). In this example, the answer is 0.8. Next, draw a picture of the lens (two triangles sharing one side). We can see that the patient is looking through the basein portion of the lens.
D. Total prism induced is 0.8 BI. First, recall Prentice’s law (Total prism = power of lens x decentration in centimeters). In this example, the answer is 0.8. Next, draw a picture of the lens (two triangles sharing one side). We can see that the patient is looking through the basein portion of the lens.
An image is projected on a screen 3 m away. A 3 D prism is placed in front of the projector with the base facing to the right. How far and in which direction is the image displaced?
D. The image is displaced 9 cm to the left. The equation for this example is PrismPower = image displacement (in cm)/distance to screen (in m). This is why the answer is in centimeters. Next, we note that a prism moves images towards its apex. When the base is positioned to the right, the image moves to the left.
D. The image is displaced 9 cm to the left. The equation for this example is PrismPower = image displacement (in cm)/distance to screen (in m). This is why the answer is in centimeters. Next, we note that a prism moves images towards its apex. When the base is positioned to the right, the image moves to the left.
Which of the following changes will decrease spectacle magnification of a plus lens?
B. In order to increase spectacle magnification of a plus lens, you can increase vertex distance, thickness, and/or base curve. To decrease SM of a plus lens, increase the index of refraction. Trends for minus lens are slightly different due to the shape of the lens. To increase SM, increase thickness or base curve. To decrease SM, increase vertex distance or index of refraction.
B. In order to increase spectacle magnification of a plus lens, you can increase vertex distance, thickness, and/or base curve. To decrease SM of a plus lens, increase the index of refraction. Trends for minus lens are slightly different due to the shape of the lens. To increase SM, increase thickness or base curve. To decrease SM, increase vertex distance or index of refraction.