optics optoprep

Question 1

How is the segment of a fused bifocal lens constructed compared to the segment of a one-piece bifocal lens, and what material is a fused bifocal made from?

Change of curvature, glass material

Change of curvature, plastic material

Change of index of refraction, plastic material

Change of index of refraction, glass material

Answer 1

Change of index of refraction, glass material

Question 2

An equiconcave thin lens has a power of -10.00 D, an index of refraction of 1.49 and a radius of curvature of 9.8 cm. If the back surface of the lens is coated to create a reflective surface, what would be the resulting power of the lens-mirror combination?

+20.40 D

• 40.40 D
• 30.50 D
• 10.00 D

Answer 2

-40.40 D

The total power of a lens-mirror combination can be determined using the formula Dlm = 2 D1 + P(n’), where Dlm= the power of the lens-mirror combination, D1= the front surface power, and n’= the index of refraction of the lens. However, the power of the mirror must first be determined using the formula P= -2n/r, where P= the power of the mirror in diopters, n= the index of refraction and r= the radius of curvature of the mirror in meters. Solve for P = -2(1.49)/0.098 = -30.40 D. Solve for Dlm= 2(-5.00) + -30.40 D = -40.40 D. It is important to note that because the lens is equiconcave, the power of the front surface is equivalent to the back surface D1=D2. Because the total power of the lens is -10.00 D, we know that the front surface power is -5.00 D. Also, when determining the power of the mirror, you must be sure to watch your signs!! The radius of curvature is positive and the power of the mirror is negative because the mirror is convex and would diverge light.

Question 3

Which 2 of the following are TRUE in regards to ANSI standards for progressive addition lenses (PALs) and single vision or bifocal lenses? (Select 2)

The tolerance for add power is higher for PALs as compared to bifocal lenses

The tolerance for sphere power is higher for PALs as compared to single vision or bifocal lenses

The tolerance for cylinder power is higher for PALs as compared to single vision or bifocal lenses

The tolerance for cylinder axis is higher for PALs as compared to single vision or bifocal lenses

Answer 3

The tolerance for sphere power is higher for PALs as compared to single vision or bifocal lenses

The tolerance for cylinder power is higher for PALs as compared to single vision or bifocal lenses

General tolerances for single vision (SV) and multifocal lenses (MF; these include bifocals and trifocals) are slightly different than progressive addition lenses (PALs), according to ANSI standards.

Sphere power

• SV/MF: For powers up to +/- 6.50, tolerance is +/- 0.13D
• PALs: For powers up to +/- 8.00, tolerance is +/- 0.16D
• Powers higher than this have a tolerance of +/-2% for both SV/MF and PALs

Cylinder power
-Tolerances are higher for PALs at all cylinder power ranges

Cylinder axis
-Tolerance ranges are the same for SV/MF and PALs

Add power
-Tolerance ranges are the same for SV/MF and PALs

Others

• Unmounted prism, vertical prism imbalance, and horizontal prism imbalance tolerances are the same for SV/MF and PALs
• Vertical segment height (fitting point height) and vertical segment difference (fitting point difference) tolerances are the same for MF and PALs
• Horizontal segment location (fitting point location) tolerance is less for PALs than MF (1.0 mm vs. 2.5 mm)

Question 4

A polycarbonate lens measures 6 mm thick at its nasal edge and 8 mm thick at its temporal edge. The power of the lens is -3.50 D and it measures 45 mm horizontally. Given the lens parameters, calculate the amount of horizontal prismatic effect that is present within the middle of the lens.

1. 55 prism diopters
2. 62 prism diopters
3. 13 prism diopters
4. 22 prism diopters
5. 89 prism diopters

Answer 4

2.62 prism diopters

Using the formula P=100g(n-1)/d, where P= the power of the prism, g= the difference in thickness between the apex and the base, n= the index of the lens, and d= the distance between the apex and the base, one is able to calculate the power of the prism that is located in the middle of the lens. Input the appropriate values into the equation and solve for P. P= 100 (2) (1.59-1)/45, P=2.62. Be sure to memorize the index of refraction for polycarbonate (1.59), and CR-39 (1.50) because these values may not always be given. Also, be aware of distractors such as extra information that is not needed but is provided in the question, as in this case where the power of the lens was not required to calculate the correct answer.

Question 5

What it the spherical equivalent of -2.50 -3.00 x 097?

• 2.50 DS
• 5.50 DS
  
  • 4.00 DS
• 3.50 DS

Answer 5

-4.00 DS

The spherical equivalent is calculated by adding half of the amount of the cylinder power to the sphere power. For the above prescription, it would be calculated as follows: -2.50 + (-3.00/2) = -2.50+ -1.50= -4.00 DS.

Question 6

The power of a thin lens in air can be found by using which of the following equations?

The radius of curvature subtracted by the primary focal length

The reciprocal of the secondary focal length in meters

The square of the sagittal height of the lens divided by the chord length

The reciprocal of the radius of curvature of the lens in meters

Answer 6

The reciprocal of the secondary focal length in meters

The power of a thin lens in air can be determined using the formula P= n/f’ where n=the index of refraction of the surrounding medium, which in the above case is air (1.00) and f’=the secondary focal length.

Question 7

Which of the following terms describes the vertical measurement taken from the top of the reading segment on a bifocal to the deepest part of the frame or lens?

The segment fusion

The segment width

The segment height

The segment depth

Answer 7

The segment height

The segment height is measured as the distance between the top of the segment to the deepest portion of the frame (this measurement is frame dependent). The segment depth is measured as the longest vertical distance from the top of the seg to the bottom of the seg. The seg width is determined by measuring the widest horizontal portion of the seg. The segment fusion is fictional.

Question 8

Your patient states that he requires safety glasses for work. According to the American National Standards Institute (ANSI), what is the minimum center thickness allowable for high-impact prescription safety lenses made from polycarbonate?

1. 0mm
2. 0mm

There is no minimum thickness

4. 0mm
5. 0mm

Answer 8

2.0mm

ANSI requirements state that in order for a prescription lens to be deemed as high-impact, it cannot measure less than 2.0mm thick at its thinnest point and must pass the high-velocity impact test. Currently, only materials made from polycarbonate resins adhere to both of these requirements. Previously, lens materials had to be at least 3.0mm thick but with the introduction of polycarbonate, safety lenses can now be made thinner.

Question 9

Base-in prism can be induced by which of the following means?

Decentering the optical center of a minus powered lens inwards (towards to patient’s nose)

Decentering the optical center of a minus powered lens downwards
Decentering the optical center of a plus powered lens inwards (towards the patient’s nose)

Decentering the optical center of a plus powered lens upwards

Answer 9

Decentering the optical center of a plus powered lens inwards (towards the patient’s nose)

Rather than specifically ordering prism, low amounts can sometimes be induced by decentering the optical center of the ophthalmic lens from the patient’s pupillary distance (PD), causing the patient to look through a different area of the lens other than the optical center. If a minus powered lens is decentered inwards (nasally), the induced prism will be will be base out; if the same lens is decentered outwards (temporally), the induced prism will be base in. A plus powered lens that is decentered nasally induces base in prism, while temporal decentration induces base out prism.

Question 10

Reflection of incident light at the front and back lens surfaces will be the greatest for which of the following lens materials?

CR-39

Trivex

Crown glass

Polycarbonate

Answer 10

Polycarbonate
Reflection from a lens surface is proportional to the lens index. Polycarbonate (n=1.59) has the highest index of refraction of the lenses listed followed by Trivex (n=1.53), Crown glass (n=1.52) and CR-39 (n=1.50).

Question 11

How much image jump will be created by a +2.50 D round 28 mm segment add with a carrier lens of +3.25 DS?

1. 25 prism diopters
2. 5 prism diopters
3. 1 prism diopters
4. 0 prism diopters
5. 5 prism diopters

Answer 11

3.5 prism diopters

Image jump is created by the vertical prismatic effect when looking through the reading addition of a bifocal lens. When the patient looks from the distance portion of their lenses into the reading area, the viewed image will appear to move or jump. The greater the distance between the reading add optical center and the bifocal line, the greater the image jump experienced by the patient. The total amount of image jump depends on the reading add and the distance between the optical center of the reading add and the segment line. The optical center of a round segment is located in the center of the segment, therefore one must divide the diameter of the segment by 2. For the above example, 28/2=14 mm. Next apply the Prentice rule to solve this problem: prism diopters(pd) =d*F where d is equal to the distance from the optical center in centimeters and F= the power of the add. Pd= 1.4(2.50) = 3.5 prism diopters.

Question 12

Where is the base curve of a modern spectacle lens typically located?

The base curve is located on the same side as the cylinder power

The base curve is located on the front lens surface

The base curve is located on the back side of a modern lens

The base curve is located on the opposite side of where the bifocal segment is located

Answer 12

The base curve is located on the front lens surface

Base curves are located on the front or non-ocular side of a modern spectacle lens (minus cylinder form). Older lenses had the base curve on the back of the lens, with the cylinder power on the front side (plus cylinder form). The advantage to the minus cylinder lens form is that the cylinder is closer to the eye, minimizing marginal astigmatism when one views away from the optical center.

Modern bifocal lenses have the bifocal on the front of the lens; thus, the base curve is on the same side as the bifocal but on the opposite side of the cylinder power.

Question 13

A Keplerian telescope possesses an eyepiece with a corresponding power of +15.00 D and a +5.00 D objective lens. What is the tube length of the telescope?

5. 0 cm
6. 67 cm
7. 33 cm
8. 0 cm

Answer 13

26.67 cm

To determine the tube length of the telescope, the corresponding focal lengths of the lenses are added together which is found by taking the reciprocal of the dioptric powers of the ocular and objective lenses. For the above problem, the focal length of the ocular lens is (1/15.00 D)= 0.066667 or 6.67 cm and the focal length of the objective lens is (1/5.00 D)=0.2 or 20.0 cm. Adding the two values together yields a tube length of 26.67 cm (6.67 cm + 20.0 cm).

Question 14

A telescope has an objective lens diameter of 15.0 mm and an exit pupil diameter of 2.0 mm. What is magnification of the telescope?

1.3x

30x

15x

7.5x

Answer 14

7.5x

The magnification of the telescope can be determined by using the equation Mtel= objective lens diameter/exit pupil diameter. Solving for the magnification yields, Mtel=15.0 mm/2.0 mm, Mtel= 7.5x. The amount of light that may enter a telescope is limited by the size of the entrance pupil. For the majority of telescopes, the entrance pupil is also the objective lens. The exit pupil is the image of the entrance pupil (objective lens) as viewed through the ocular of the telescope. For a Galilean telescope, the exit pupil is virtual and is located inside the telescope. A Keplerian telescope possesses a real exit pupil.

Question 15

A patient is seen at your office complaining of distance blur with her current glasses. With her current prescription of -3.25 D in place, you determine that her far point is 70 cm from the spectacle plane for her right eye. Given this information, which of the following is the MOST appropriate spectacle prescription to obtain a clear retinal image when an object is viewed at optical infinity (rounded to the nearest 0.25 D)?

• 4.75 D
• 1.75 D
• 1.00 D
• 2.50 D
• 3.25 D

Answer 15

-4.75 D

With the current prescription the patient’s far point is 70 cm. The far point vergence at the spectacle plane necessary to obtain a clear image is the reciprocal of the far point in meters. 1/0.70= 1.43 D, rounded to the nearest quarter diopter yields 1.50 D. Therefore, to achieve a clear retinal image for an object focused at optical infinity requires -4.75 D at the spectacle plane.

Question 16

A 24-year old female wears soft contact lenses with a Dk/t of 175 and admits to sleeping in her lenses. She is very satisfied with both the comfort and the vision of her lenses. Biomicroscopy reveals mucin balls under her lenses bilaterally that leave impressions in her central corneas upon removal of her lenses. Which of the following actions would BEST help to eliminate the formation of mucin balls?

Instructing the patient to increase her blinking frequency

Maintaining the same lens material but changing to a steeper base curve

Changing her multi-purpose solution

Altering the power of the contact lens but maintaining the same lens material

Answer 16

Maintaining the same lens material but changing to a steeper base curve

Mucin balls appear as small, white, pearl-like debris that occur behind the posterior surface of contact lenses. They generally occur with silicone hydrogel lenses that are fit too flat and are used for extended wear purposes. Mucin balls do not actually pose a threat to vision and do not generally compromise the integrity of the cornea. However, if they are severe enough, there are several options available to clinicians to combat their formation. An easy way to decrease generation of mucin balls is to steepen the base curve of the lens. Alternatively, one can decrease the amount of extended wear or add re-wetting drops to the patient’s contact lens regimen. Upon removal, mucin balls will cause pooling of sodium fluorescein but will not cause staining of the cornea.

Question 17

A ray of light is deviated 5.50 cm by a prism made of crown glass located 7.0m away. Which of the following equations will CORRECTLY determine the total prism power?

P=(100)(0.55/700)

P=(100)(700/0.55)

P=(100)(5.50/700)

P=(100)(5.50/7.0)

P=(100)(700/5.50)

P=(100)(7.0/5.50)

Answer 17

P=(100)(5.50/700)

Prism power is found by using the equation P=(100)(x/d), where P= power of the prism (in prism diopters, pd), x=the total distance that a ray of light is deviated, and d=the total distance from the prism to where the deviation is measured. For the above question, P= (100)(5.50 cm/700 cm)=0.786 pd. Key: both the distances must be in either meters or centimeters. If the units of the distances are not the same then 100 must be dropped from the formula. For example, P=5.50 cm/7.0 m=0.786 pd.

Question 18

The presence of small, iridescent, fleck-like opacities scattered throughout the crystalline lens (also known as “Christmas tree cataracts”) are associated with which of the following systemic diseases?

Atopic dermatitis

Neurofibromatosis

Down’s syndrome

Wilson’s disease

Correct answer Myotonic dystrophy

Answer 18

Myotonic dystrophy: associated with multi-colored opacities known as “Christmas tree” cataracts (close to 90% of patients develop these cataracts)

• Atopic dermatitis: associated with shield-like, dense anterior subcapsular plaques
• Neurofibromatosis: associated with posterior subcapsular or posterior cortical cataracts
• Wilson’s disease: associated with green “sunflower” cataracts
• Down’s syndrome: low association with Cerulean or “blue-dot” cataracts

Question 19

A 20-year old patient is complaining of double vision when reading. He is wearing a spectacle prescription of -5.00 DS OD and -2.00 DS OS. If he reads 8 mm below the optical centers of the lenses, what is the induced vertical prismatic imbalance when reading?

4 2.4 BD

2. 4 BU
3. 24 BU
4. 24 BD

Answer 19

2.4 BD

Prism can be induced when a patient is looking above/below or lateral to the optical centers of a spectacle lens. The induced prismatic effect from a spectacle lens is a function of the lens vertex power (Fv) and the distance between the patient’s line of sight and the optical center of the lens (d). This prismatic effect can be calculated using Prentice’s Rule: prism = d(Fv). It is important to remember that the unit for d is always in centimeters.

When dealing with vertical prismatic effects, if the induced prism values over each eye have the same base direction, the prism values are subtracted. If prism values have opposite base directions, then the prism values are added.

The base direction is dependent on whether the lens is a plus or minus powered prescription. Since a plus powered lens has a greater center thickness, looking above the optical center will induce base down (BD) prism, and looking below the optical center will induce base up prism (BU). A way to visualize this is to picture two prisms stacked together base to base, creating a thick center. The opposite is true for a minus lens, since the thickness of the lens is greatest at the edge. Looking above the optical center of a minus lens induces BU prism, and looking below the optical center induces BD prism. A minus lens can be visualized as two prisms stacked apex to apex, creating a thin center and thick edge.

To determine the vertical prismatic imbalance in the question, calculate the induced prism effect of each eye:
Right Eye: Prism = (0.8cm)(5D) = 4.0 BD
Left Eye: Prism = (0.8cm)(2D) = 1.6 BD
Each eye has the same base direction, so the values will be subtracted to determine the vertical prismatic imbalance.
Imbalance = 4.0 - 1.6 = 2.4 BD over the right eye

Question 20

Which of the following lenses does not possess an optical axis?

A bi-concave lens

A bi-convex lens

A meniscus lens

A plano-convex lens

A plane parallel lens

Answer 20

A plane parallel lens

The optical axis of a lens system can be thought of as an imaginary line that serves to join the centers of curvature. Light transmitted through a lens will pass through its center of curvature and is said to travel along the optical axis of the lens. A surface that does not possess any curvature will not have a natural optical axis.

Question 21

A patient walks into your office because he recently moved and lost his glasses in the process. He remembers that he had prism in his glasses but he is unsure how much. Your refraction reveals OD: -0.25 -7.50 x100, OS: -3.25 DS with 3 base out (BO) prism. From a cosmetic standpoint, on which lens should the prism be prescribed?

1 BO left eye, 2 BO right eye

Correct answer 3 BO left eye

The prism is very mild and should not be prescribed

3 BO right eye

Answer 21

Because the lens of the right eye is going to be quite thick temporally, it is a good idea for cosmetic purposes to place the prism over the left eye in an attempt to equalize temporal edge thickness

Question 22

An object located in air (n=1.00) measures 2.5 cm in height and is 12.0 cm in front of a -12.00 polycarbonate (n=1.586) surface. Which of the following will correctly calculate the location of the secondary focal point?

f’=0.025/(1.586-1.0)

f’=1.586/-12.00

f’=(1.586-1.0)/-12.00

f’=1.0/-12.00

Answer 22

f’=1.586/-12.00

To correctly solve for the location of the secondary focal point, one must use the equation P=n’/f’. Inputting the values from the above question yields: -12.00= 1.586/f’. Solving for f’ yields= -0.132 or 13.2 cm. Because the answer is negative, the secondary focal point will be located to the left of the diverging surface.

Question 23

The right optical center (OC) of your patient’s glasses is not aligned with his pupil. The right OC is 4 mm higher than his right pupil. Calculate the induced prism and base direction base from the following Rx: OD +2.00-2.00 x 030.

0. 6 prism diopters BU
1. 2 prism diopters BD
2. 6 prism diopters BD
3. 2 prism diopters BU

Answer 23

0.2 prism diopters BU

The power in the vertical meridian of the above lens is +0.50D. This is calculated by using the formula: F = sphere + cylinder(sin2 theta). Theta is the difference between the meridian in question (vertical or 90 degrees) and the lens axis (030 degrees). Using Prentice’s rule: prism diopters = Fd. F is calculated as +2.00 + (-2.00)(sin2060)= +0.50D. d is how far away the OC is from the patient’s pupil or 4mm. Remember to express d in cm, not mm. +0.50 x 0.4 = 0.2 prism diopters. The patient is viewing through the lower half (base up) of the plus lens power since the OC is above the pupil.

Question 24

What separation distance will make the combination of a +3.00 and a +10.00 thin lens afocal?

2.3 cm

43 cm

1. 7 cm
2. 43 cm

23 cm

17 cm

Answer 24

43 cm

For this question, the equation for equivalent power of a thick lens system should be used, solving for thickness (t).

De = D1 + D2 - (t/n) x D1D2 
De = equivalent power, D1 = front surface power, D2 = back surface power 
t = thickness of lens system, n = index between the 2 surfaces 

An afocal system has its focal points (F and F’) located at infinity. Therefore, an incident parallel pencil of light rays will emerge into image space as a parallel pencil as well. Another way to characterize an afocal system is that the equivalent power (De) is 0.

In the above question, De = 0, D1 = +3.00, D2 = +10.00, n= 1 
0 = 3 + 10 - ((t/1) x (3) x (10)) 
0 = 13 - (t x 3 x 10) 
0 = 13 -30t 
30t = 13 
t = 0.43 m (or 43 cm) 

If the two lenses are separated by 43 cm, the lens system can be considered afocal. This type of combination of two plus lenses is also an example of a simple astronomical (Keplerian) telescope. Keep in mind that the image in this type of optical system is inverted.

Question 25

Which of the following formulas can be used to directly determine the lateral magnification produced by an optical system?

Image size/radius of curvature of refracting lens

Index of refraction/primary focal point

Object size/index of refracting medium

Object vergence/image vergence

Answer 25

Object vergence/image vergence
The lateral magnification of an image produced by an optical system can be calculated by dividing the image size by the object size or, alternatively, by dividing the object vergence by the image vergence.

Question 26

our -10.00 D patient wishes to have the thinnest edges possible for his glasses. Which of the following actions will result in a reduced edge thickness?

Decreasing the refractive index of the lens

Increasing the minimum blank size of the lens

Increasing the center thickness of the lens

Correct answer Choosing a smaller eyesize
Explanation - For a myope, the goal is to reduce the overall edge thickness. Decreased edge thickness can be achieved by decreasing the eyesize, increasing the refractive index of the lens, or by minimizing the center thickness.

Answer 26

Choosing a smaller eyesize
Explanation - For a myope, the goal is to reduce the overall edge thickness. Decreased edge thickness can be achieved by decreasing the eyesize, increasing the refractive index of the lens, or by minimizing the center thickness.

Question 27

A patient who recently picked up her new glasses returns to your office complaining that the new glasses make her eyes feel tired. The prescription in the right eye is +2.00 -2.00 x090 and the left eye is plano -2.50 x180. Her pupillary distance (PD) is 64 mm and the distance between the optical centers is 74 mm. How much prism was induced and in which direction for the right eye?

1 prism diopter base out

None

2 prism diopters base out

1 prism diopter base in

2 prism diopters base in

Answer 27

None

Placing the prescriptions for both eyes on optical crosses reveals that both eyes are plano in the 180 degree meridian, which means that no matter how much your PD is off, prism will not be induced.

Question 28

A myopic patient wears a contact lens with a power of -7.00 D. She wishes to purchase a pair of glasses. How must her prescription be altered so that her spectacle lenses provide a clear retinal image?

The minus power of the spectacle lenses must be increased

The optical center of the lens must be decreased so that it coincides with the lower edge of her pupil

The minus power of the spectacle lenses must be decreased

The center thickness of the lens must be increased to compensate for a change in vertex distance

Answer 28

The minus power of the spectacle lenses must be increased

remember mina my brother is -9 and his CL is -7

When a minus powered lens (greater than 4.00 D) is moved from the corneal to the spectacle plane, the power needs to be adjusted by adding more minus to ensure a clear retinal image. Logically, the far point of the eye must correspond with the secondary focal point of the lens to achieve a clear retinal image. The far point of a myopic eye is an unchanging fixed distance from the eye. Spectacles are further from the eye but closer to the far point of the eye (by the vertex distance of the spectacle) and therefore a higher power will achieve a reduced focal length, which makes sense, as the lens is physically closer to the far point of the eye. Contact lenses sit closer to the eye and further from the actual far point of the eye and therefore require less minus power at the corneal plane. Therefore, to ensure that the far point of the eye coincides with the secondary focal point of the lens, the power of the contact lens must be decreased for a myopic eye when going from the spectacle to the corneal plane.

Question 29

Which of the following alterations of a plus powered lens will INCREASE the spectacle magnification experienced by a patient?

Decrease the center thickness

Decrease the base curve

Decrease the vertex distance

Decrease the index of refraction of the lens

Answer 29

Decrease the index of refraction of the lens

Spectacle magnification is a difference in the size of the retinal image caused by the use of corrective lenses. The higher the prescription, the greater the magnification (or minification) experienced by a patient. There are several ways to decrease spectacle magnification. For a plus powered lens, increasing the index of refraction, decreasing the base curve, vertex distance or center thickness will reduce magnification. For a minus powered lens, spectacle minification can be lessened by decreasing the vertex distance or the index of refraction, and increasing the center thickness or base curve. Although this sounds backwards, remember that the goal for minus lenses is to reduce minification, or to increase magnification; and increasing the center thickness of the lens or decreasing the index will achieve this goal.

Question 30

Which of the following is TRUE for an image created by a plane mirror?

The image is real

The image is the same size as the original object

The image is located closer to the mirror than the object

The image is upside-down

Answer 30

The image is the same size as the original object
Images created by a plane mirror are virtual, reversed, laterally inverted and upright, appear the same size as the object and appear to be located as far from the mirror as the object is situated in front of the mirror

Question 31

One prism diopter is roughly equivalent to how many degrees?

1 degree

0. 57 degrees
1. 12 degrees
2. 20 degrees

Answer 31

0.57 degrees

One prism diopter is roughly equivalent to how many degrees?

1 degree

Correct answer 0.57 degrees

0.12 degrees

0.20 degrees
Explanation - A prism diopter is equivalent to 0.57 degrees, or conversely, 1 degree is equal to 1.75 prism diopters.

Question 32

If a thick lens with a positive front and back surface increases in thickness, what effect will this have on the net equivalent power of the optical system?

The net equivalent power will become less positive

The change in net equivalent power cannot be determined without knowing the index of the material

The net equivalent power will become more positive

The net equivalent power will remain unchanged

Answer 32

The net equivalent power will become less positive

If both of the surface powers of a thick less are either negative or positive, increasing the separation between the surfaces will cause the net equivalent power of the optical system to become more negative (or less positive).

The effective power equation is below:
De = D1 + D2 - (t/n) x D1D2

Question 33

Your patient wishes to know how much magnification he experiences while wearing his glasses. The power magnification factor is 1.12 and the front surface power of his lenses is +1.25 D. The center thickness of his lenses is 1.8 mm and the index of refraction is 1.50. What is the total magnification (in percentage) experienced by this patient?

11. 2%
12. 3%

Correct answer 12.2%

7.2%

Answer 33

In order to determine the total magnification (Mspec) experienced by this patient we require two pieces of data: the shape magnification (Ms) and the power magnification (Mp). The formula for Ms=1/[1-F1(tc/n)] where F1= the power of the front surface of the lens, tc= the center thickness of the lens (in meters), and n= the index of the lens. Mp=1/[1-h(Fv)] where h= the vertex distance and Fv= the back vertex power of the lens. In the above question the Mp is given. Solve for the Ms by imputing the appropriate values: Ms=1/[1-1.25(0.0018/1.50], Ms= 1.0015. To determine the total magnification multiply Ms and Mp. Mspec=1.0015x1.12= 1.122. Expressed as a percentage, Mspec%= (Mspec-1)x100.

Solve for Mspec%=(1.122-1) x 100= 12.17%.

Question 34

Transpose the following Rx: +2.00 - 2.00 x 045 to plus cylinder format

Plano +2.00 x 135

Plano - 2.00 x 135

Plano + 2.00 x 045

Plano - 2.00 x 045

Answer 34

Plano +2.00 x 135

he rules for transposition are as follows:

1. Add the sphere and cylinder powers together, keeping in mind the signs of each power.
2. Change the original cylinder power to the opposite sign.
3. If the original axis is 90 degrees or less, add 090 degrees to determine the new axis. If the original axis is greater than 090 degrees, subtract 090 degrees to determine the new axis.

Question 35

Your patient complains that she sees better out of the left eye of her glasses when she rotates the round lens counterclockwise (from her perspective) 45 degrees. The present axis of her OS Rx is 090 degrees but she obviously prefers which axis?

067 degrees

045 degrees

112 degrees

135 degrees

Answer 35

045 degrees

The orientation of the astigmatism axis is designated from zero degrees to 180 degrees. As the patient is wearing the glasses, zero degrees is to the patient’s left while 180 degrees is to the patient’s right. Think of a number line with the larger numbers increasing to the right. This patient is rotating her lens counterclockwise or towards the left towards zero. Thus, she prefers an axis of 045 degrees (90 degrees minus 45 degrees) in order to see more effectively.

Question 36

You check your new glasses and find that the distance between the optical centers measures 66 mm vs. your PD of 62 mm. How much prism is induced in the right eye and in what direction? Your prescription is:

OD -2.00 - 1.00 x 060
OS +4.00 - 4.00 x 030

0. 45 prism diopters BI
1. 55 prism diopters BO
2. 45 prism diopters BO
3. 55 prism diopters BI

Answer 36

0.55 prism diopters BI

The distance between the optical centers of the glasses measures wider than the patient’s PD, thus the OD lens optical center is located temporal to the pupil. The induced base direction, or the point in the lens the patient is looking through, is BI. The prism power is calculated by the sum of the sphere and a portion of the cylinder power. The lens power is based on the following formula: Sphere + cylinder (sin2theta) where theta is the degrees away from the meridian in question. In this example theta is 60 degrees, which results in a -2.75D power. Now use Prentice’s rule (prism = F x d) where d is expressed in cm and is half the discrepancy between the distance between the optical centers (DBOC) and the patient’s PD, or 2 mm, to determine the prism amount. -2.75 X 0.2 = 0.55 BI.

Question 37

Which of the following frame materials requires HIGH heat for lens insertion and frame adjustments?

Carbon fiber

Polyamide

Polycarbonate

Cellulose propionate

Cellulose acetate

Correct answer Optyl

Answer 37

Optyl

The steps utilized in inserting a lens into a frame during fabrication and adjusting the frame are standard for most plastic frames; however, the major variant is whether or not heat is used during the process (and how much). Some frame materials do not use heat at all when inserting lenses (this is called “cold snapping”) and some materials require high amounts of heat so that the material easily bends with only mild pressure.
Optyl is a frame material that requires high amounts of heat. Heat is typically applied to the frame until it can bend under it’s own weight and then lenses can be snapped into the frame, or adjustments to the frame can be made as necessary. Optyl material can safely be heated to temperatures of up to 200 degrees Celsius.
Most other materials utilize only minimal amounts of heat if lenses cannot be “cold snapped” (cellulose acetate, cellulose proprionate, and carbon fiber). Polycarbonate and polyamide materials should not be heated at all for the lens insertion process as this may lead to lens warpage; therefore, edged lenses must be fabricated exactly to size.

Question 38

Determine the reverse slab-off amount for the following Rx. Assume the reading level is 10 mm below the distance optical center.

OD -4.00 -1.00 x 045
OS -1.00 -1.00 x 090
Add +2.00 OU

Correct answer 3.5 prism diopters

2. 5 prism diopters
3. 0 prism diopters
4. 0 prism diopters

Answer 38

3.5 prism diopters
Reverse slab-off is often prescribed to correct a vertical prism imbalance due to anisometropia. Reverse slab-off adds base down prism to the most plus lens or the least minus lens. To solve this problem, calculate the vertical power difference between the lenses. Unless specified, assume the reading level is 10 mm below the distance optical centers. Using Prentice’s rule, calculate the reverse slab-off amount. The total vertical power of the right lens is calculated using the F=sphere + cylinder (sine2 theta) equation. For the right lens, the the calculation is -4.00 + -1.00(sine2 45)= -4.50 D in the 90 degree meridian.
The left lens has -1.00 D in the vertical meridian. The difference between the two in the vertical meridian is 3.50 D.
Prism = 3.5 x 1 or 3.5 reverse slab-off.

Question 39

Which of the following clear lens materials inherently blocks the MOST amount of ultraviolet-A (UV-A) light?

Hi-index glass

CR-39 (plastic)

Glass

Correct answer Polycarbonate

Answer 39

Polycarbonate

Polycarbonate blocks 100% of UV-B and most of UV-A light (some studies state that it blocks all UV-A) and therefore does not require the addition of a UV protective coat. CR-39 blocks 100% of UV-B light but only some of UV-A light. Glass transmits both UV-A and UV-B light to a much greater degree than CR-39 or polycarbonate. Remember to prescribe a UV coating to patients who insist on tinting glass or CR-39 lenses because many patients associate a tint with adequate UV protection.

Question 40

A Keplerian telescope possesses an eyepiece with a corresponding power of +20.00 D and a +5.00 D objective lens. Which of the following will CORRECTLY solve for the resulting angular magnification of the telescope?

M= (+5.00)/(+20.00)

Correct answer M= -(+20.00)/(+5.00)

M= -(+5.00)/(+20.00)

M= (+20.00)/(+5.00)

Answer 40

M= -(+20.00)/(+5.00)

Using the formula M= -Doc/Dobj, where Doc= dioptric power of the oculars and Dobj= dioptric power of the objective lens one can solve for the angular magnification provided by the telescope. Solving for M= -(+20.00)/(+5.00)= -4.0 x.

Question 41

A patient brings in an executive bifocal. Using a lens clock, the top portion of the executive measures +5.00 and the reading portion measures +7.25. What is the approximate power of the reading add?

+1.12 D

+2.25 D

+5.00 D

+7.25 D

Answer 41

+2.25 D

A lens clock can be used to determine the add power of multifocal lenses that are constructed of the same material, such as an executive bifocal (a one-piece). In order to calculate the add power, the lens clock should be placed on the distance portion of the lens parallel to the segment line and as close to the line as possible. The value is then read from the lens clock. The lens clock is then placed in the near portion of the lens with the pins oriented parallel to the segment line, and the value on the lens clock is noted. Lastly, the difference between these two values is calculated, rendering the reading addition power of the lens.

Question 42

What is the base curve of the following Rx if the back (ocular) curve measures -4.00 in the vertical meridian and -8.00 in the horizontal meridian?

+2.00 -4.00 x 090

+4.00

Correct answer +6.00

+8.00

+2.00

Answer 42

+6.00

The first cross represents the base curve on the front of the lens. The second curve represents the back or ocular curve. The third cross represents the final Rx. Place the Rx on the third of the 3 optical crosses. The vertical cross is +2.00; the horizontal cross is -2.00. Now fill in the second cross with the values given in the question, namely -4.00 on the vertical cross and -8.00 on the horizontal cross. Adding the vertical meridians of cross number 1 added to the vertical meridians of cross number 2 should equal the vertical meridian value of cross number 3. Repeat for the horizontal meridians of each cross. Base curves for modern lenses should always have a plus value.

Question 43

A prism is used to deviate light by 5 degrees. What is the power of the prism (in diopters)?

5 prism diopters

10 prism diopters

8. 75 prism diopters
9. 85 prism diopters

Answer 43

8.75 prism diopters

A prism diopter is equivalent to 0.57 degrees, or conversely, 1 degree is equal to 1.75 prism diopters. To solve this problem, multiply 5 by 1.75 to determine the power of the prism in diopters: 5*1.75 = 8.75 prism diopters.

Question 44

Which of the following heating methods is BEST for adjusting frames fabricated from nylon?

Hot water

Hot air

Glass beads

Sand

Salt

Answer 44

Hot water

The preferred method utilized for heating nylon frames is hot water. Nylon frames cannot be heated uniformly enough through other methods as hot forced air and salt pans do not penetrate the material enough, leaving the outer layers hot and the deeper portions too cool to stretch or bend. Hot water tends to penetrate the nylon material best. When adjusting a nylon frame, it is best to hold the frame into its desired position as it cools. If the frame is released before it cools, it tends to resume its initial conformation. The addition of cold running water can help speed the process along.

Question 45

You decide to prescribe your patient round top bifocals due to them requiring unequal bifocal heights. What adjustment must be made to the segment height when ordering round top bifocals vs. flat top bifocals?

Add 0.50mm to the segment height

Add 1mm to the segment height

Subtract 1mm from the segment height

Subtract 0.5mm from the segment height

Answer 45

Add 1mm to the segment height

• Round segments are a great choice when cosmesis is a concern in patients where unequal bifocal segment heights are necessary because they are less conspicuous than a flat-top bifocal. However, round segments must be positioned about 1.0mm higher than flat-top segments to take the “roundness” at the top into consideration so that the patient may enter the near add power at the proper location of the lens

Question 46

Which of the following patients, each possessing an amplitude of accommodation of 1.50 D, would possess the BEST near visual acuity if left UNCORRECTED?

A 55 year-old, emmetropic female

A 55 year-old, 0.50 hyperopic female

A 55 year-old, 1.50 D hyperopic female

Correct answer A 55 year-old, 1.50 D myopic female

Answer 46

A 1.50 D uncorrected myopic female with 1.50 D amplitude of accommodation with have a near point of 33.3 cm. 1.50 D of myopia will allow her somewhat good near vision, which then adds to her amplitude of accommodation (which is 1.50 D). Therefore, 1.50 D + 1.50 D = 3.00 D. 1/3.00 D= 0.333 m or 33.3 cm. Her near point of accommodation is 33.3 cm; objects that are located closer than this will be blurry to the patient. An uncorrected emmetrope with 1.50 D of accommodation will be able to see objects that lie at 66.7 cm clearly when the eye is maximally accommodated (1/1.50=0.667 m or 66.7 cm). A 1.50 D hyperope needs to accommodate 1.50 D in order to achieve a clear image of a distant object. Therefore, a 1.50 D hyperope with a 1.50 D amplitude of accommodation will not be able to clearly visualize an object that is located closer than optical infinity.

Question 47

Your myopic patient desires his lenses to be as thin as possible. Select the frame size that will minimize his temporal lens thickness due to decentration from the following list. His distance PD is 64 mm.

54-20-135

54-16-135

54-22-135

Correct answer 54-12-135

Answer 47

54-12-135
In most cases, the optical center of a spectacle lens should be coincident with the patient’s pupils unless prism is desired. Decentration is the process of moving the lens optical center so that it no longer aligns with the patient’s pupils. As a minus lens is moved or decentered nasally, the temporal aspect of the lens gets thicker, since more of the minus power lens blank is utilized. To minimize temporal lens thickness due to decentration, select a frame where the frame PD (A + DBL) is equal to or as close as possible to the patient’s PD. In this case, the best frame has a PD of 54+12=66.

Question 48

Which of the following statements regarding indirect ophthalmoscopy is TRUE?

Decreasing the dioptric power of the hand-held lens increases the linear field of view of the resultant retinal image

The linear field of view is smaller than that offered by direct ophthalmoscopy

Decreasing the magnitude of the dioptric power of the hand-held lens increases the magnification of the retinal image

The viewed retinal image is upright

Answer 48

Decreasing the magnitude of the dioptric power of the hand-held lens increases the magnification of the retinal image

There is an inverse relationship between magnification and the power of the hand-held lens when performing indirect ophthalmoscopy. Increasing the dioptric value of the hand-held lens will reduce the overall magnification but will increase the linear field of view. The image viewed through indirect ophthalmoscopy is inverted therefore when interpreting results be sure to know what area of the retina you are actually viewing.

Question 49

In order to ensure proper measurements for a pair of progressive spectacles, which of the following should be completed during the fitting process?

Taking binocular interpupillary distance measurements

Drawing a line for the fitting height at the lower eyelid margin

Pre-adjustment of the frame on the patient’s face

Use of the standard progressive lens cut-out chart to verify that frame depth is adequate

Answer 49

Pre-adjustment of the frame on the patient’s face

•Measure distance PDs monocularly (not binocularly)
°Recommended method is to use a pupillometer (most accurate)
•Fit and fully adjust the desired frame on the patient’s face
°Pantoscopic tilt, frame height, vertex distance, face-form wrap, and nose-pad alignment
°If the frame is not pre-adjusted, fitting height measurements may be incorrect
•Ensure that the clear demo lenses are in place in the desired frame
°If lenses are not present, or are darkly tinted, place clear (transparent) tape across the eyewire of the empty frame
•Dispenser should be positioned with his or her eyes at the level of the patient’s eyes
°The patient should look at the bridge of the fitter’s nose
°Draw a horizontal line on the lens or tape through the center of the pupil for each eye (not the lower eyelid margin as for bifocals)
•Place the frame on the specific manufacturer’s centration chart (there is no standard progressive lens cut-out chart)
°Line up the bridge so that it is centered on the diagonally converging central alignment pattern
°Line up the marked horizontal fitting lines so that they are aligned with the horizontal axis of the chart
°Mark the previously measured PD for each eye as a vertical line that crosses the horizontal fitting line
•The fitting height of the lenses should then be measured as the vertical distance between the fitting cross and the level of the inside bevel of the lower eyewire of the frame
•Move the frame over to the lens picture portion of the centration chart so that the cross on the lens matches with the cross on the chart
°Verify that the add power circle fits within the boundaries of the frame
°Verify that one of the large overall circles completely encloses the frame’s lens shape

Question 50

All else being equal, what pupil diameter is associated with the least amount of spherical aberration?

3. 0 mm
4. 0 mm
5. 0 mm
6. 0 mm

Answer 50

3.0 mm
Spherical aberration increases with increased pupil size. Multiple sources state that the ideal pupil size is between 2.0-3.0 mm in size. Pupils greater than this size experience a larger degree of spherical aberration, while a pupil smaller than this size is optically limited by diffraction.

Question 51

A patient has a spectacle correction of +12.00 D. The vertex distance is 13 mm. What is the power at the corneal plane?
+10.37 D

+11.25 D

+14.25 D

+12.00 D

Answer 51

+10.37 D

When a plus lens is moved to the corneal plane, the lens is perceived as having less power than in the spectacle plane (the effective power of the lens decreases). In order to determine the exact power of the lens at the corneal plane, the following equation can be used:
De = Ds / (1 + (d x Ds))
De = effective power at corneal plane, Ds = power at spectacle plane, d = vertex distance (in meters)
De = +12.00 / (1 + (0.013 x +12.00)0
De = +12.00 / +1.156
De = +10.38D

Don’t get this question confused with the adjustment that would need to occur to maintain an effective power of +12.00 at the corneal plane. In this case on would use the following equation:
Dc = Ds/(1 - (d x Ds)
Dc = power at the corneal plane, Ds = power at the spectacle plane, d = vertex distance (in meters)
Dc = +12.00/(1- (0.013 x +12.00) = +14.22 D

Question 52

During case history, your patient stated that he spends a lot of time fishing. Which of the following products should be recommended to him?

Sunglasses with a rose tint

Sunglasses with a green tint and an anti-reflective coating

Sunglasses with polarized lenses

Clear lenses with an anti-reflective coating

Sunglasses with a blue tint

Answer 52

Sunglasses with polarized lenses

ecause the patient is an avid outdoor enthusiast and particularly spends a great deal of time on the water, he should be informed about the benefits of polarized lenses. There is a great deal of horizontally polarized light that reflects off of the surface of the water. Polarized lenses are generally constructed with a vertically-oriented filter, allowing only light parallel to this orientation to pass through the lens and blocking the transmission of horizontally-positioned light rays. Because the lenses filter the reflection off of the water surface, fish also become more visible to the observer.

Question 53

A patient with basic exophoria who experiences asthenopia with reading is seen at your office. You decide to prescribe glasses with prism. What type of prism is required and how will the prism help to alleviate the patient’s symptomology?

Base in; the image of the object is shifted towards the base of the prism

Base out; the image of the object is shifted towards the base of the prism

Base out; the image of the object is shifted towards the apex of the prism

Correct answer Base in; the image of the object is shifted towards the apex of the prism

Answer 53

The clinical application of prisms may be useful for patients with phorias or tropias. Prisms cause light to deviate to the base, but the image will be displaced towards its apex. A symptomatic patient with exophoria will benefit from base in prism. A person with exophoria exhibits a tendency for the eyes to turn outwards. Placing base in prism over this patient’s eyes will cause the image to be displaced towards the apex, shifting the image outward and thereby reducing the total demand of the patient on convergence. This reduces the symptoms of muscle stress described as asthenopia.

Question 54

A patient reports to you that he cannot see clearly past 27.0 cm with his left eye. Based on this information, what power of contact lens (rounding to the nearest quarter diopter) will MOST likely allow for a clear retinal image when viewing an object at optical infinity?

• 3.75 D
• 3.50 D
• 4.00 D
• 2.75 D
• 3.25 D

Answer 54

-3.75 D

In order to solve for the power of the contact lens that will most likely correct the patient’s ametropia, the reciprocal of the far point of the eye (in meters) must be determined. 1/0.27=3.704 D, rounding to the nearest quarter diopter yields -3.75 D.

Question 55

A patient returns to your office reporting that her eyes feel strange when she reads 6 mm below the optical centers of her new glasses. The prescription in her right eye is -4.00 DS and -7.00 DS in her left eye. How much vertical prism is induced when she reads?

1. 8 prism diopters base down
2. 2 prism diopters base down
3. 6 prism diopters base down
4. 4 prism diopters base down

Answer 55

1.8 prism diopters base down

Use the Prentice rule to solve this problem: prism diopters(pd) =d*F, where d= the distance from the optical center in centimeters and F= the power of the lens in the desired meridian in diopters. In this instance, the patient is looking through base down prism in both eyes, which will cancel some of the prismatic effect as the bases are aligned. Solving for the amount of prism on the right eye, pd=0.6(-4.00)= 2.4 base down prism. Solve for the left eye: pd=0.6(-7.00)=4.2 base down prism. Subtract the two to determine the total prismatic effect experienced by the patient: 4.2-2.4=1.8 base down prism. Alternatively, you can omit one of the steps by initially determining the total power difference in the vertical meridian between the two lenses, which is 3.00 D (7-4=3). Then you can multiply this power difference by the distance between the patient’s line of sight and the optical center, which is 6 mm in this question. Pd=0.6(3)= 1.8 prism diopters base down over the left eye. Generally, vertical imbalances of smaller magnitudes do not pose too much of a problem for single vision lenses as the patient can tilt her head to re-align the optical centers with her line of sight, thus eliminating any possible diplopia.

Question 56

The office staff wants to know if Mrs. Jones’ glasses are acceptable to dispense. The glasses are off axis by 3 degrees in the left lens. The right lens is within ANSI standards. The Rx is +1.50-1.75 x 087 OS. Does the left lens meet ANSI standards?

Correct answer No

Yes

Answer 56

ANSI standards state that for a cylinder power of 1.75D the axis tolerance is +/- 2 degrees. The left lens in this example does not pass since the axis is off by 3 degrees.

Question 57

In order to maintain the same power of a lens, how must the radius of curvature be adjusted if the index of refraction of the lens is increased?

It must decrease

It must increase

It remains the same

Answer 57

It must increase

In order to prove that the radius of curvature must increase, use the equation for a single surface power and input any numbers into it. For example, if a lens in air had a power of +4.00 D and an index of refraction of 1.60, it would have a radius of curvature of F = n’-n/r Solve for r: 4.00 = 1.6-1/r; 4.00 = 0.6/r; r = 0.6/4.00 r = 0.15 meters or 15 cm. Therefore, if the index of refraction was to increase to 1.7 and we desired the same power of +4.00 D, the new radius of curvature would have to increase to 17.5 cm. 4.00 = 1.7-1.0/r Solve for r; 4 = 0.7/r; r = 0.7/4; r = 0.175 meters or 17.5 cm.

Question 58

The intensity of the light transmitted when unpolarized light is incident on a polarizer is equal to which of the following percentages?

75%

33%

Correct answer 50%

100%

25%

Answer 58

When unpolarized light reaches a polarizer, the intensity of transmitted light with respect the intensity of the incident light is 1/2.

Question 59

Your patient requires 2 BO prism. You decide to decenter the left lens in his glasses to induce the desired prism. The prescription is OD: -4.00 DS, OS: -4.75-0.75x178. The patient’s pupillary distance is 68 mm. How much and in which direction do you have to decenter the left lens?

4. 2 mm, out
5. 2 mm, in
6. 5 mm, out
7. 5 mm, in

Answer 59

4.2 mm, in

n order to create base out prism, one must decenter the left lens inwards (towards the patient’s nose). To determine the degree of nasal decentration, use Prentice’s rule: prism diopters= d*F where d is equal to the distance from the optical center in centimeters and F= the power of the lens in the desired meridian in diopters. The power of the left lens in the horizontal meridian is equal to -4.75 D. Put the values into Prentice’s rule, yielding: 2BO= (d)(-4.75) solve for d: d=2/-4.75, d=0.42 cm or 4.2 mm.

Question 60

Which 3 of the following lenses possess a negative dioptric power? (Select 3)

A plano-convex lens

Correct answer A concave-meniscus lens

Correct answer A plano-concave lens

Correct answer A biconcave lens

A convex-meniscus lens

A biconvex lens

Answer 60

Convex lenses are positively powered. A convex-meniscus lens, although it has both a convex and a concave surface, is positive in power due to the fact that the power of its convex surface is greater in magnitude than the concave surface. Conversely, concave surfaces possess negative powers.

Question 61

Which of the statements regarding pupil size is TRUE?

As the size of the pupil increases, the amount of image defocus experienced by an object located at optical infinity will decrease proportionally

As the pupil size decreases, the diameter of the corresponding blur circles increases and the amount of light reaching the retina will increase

As the pupil size increases the diameter of the resulting Airy’s disk increases accordingly and is directionally proportional to the increase in pupil size

Correct answer As the pupil size decreases, the distance between two point sources must be increased in order for the two point sources to be resolved

Answer 61

As the pupil size decreases, the amount of object defocus will decrease, as will the amount of light reaching the retina. Elderly patients often experience a phenomenon called senile miosis in which the size of the pupil is physiologically decreased, affording increased visual acuity secondary to a pinhole effect, along with an increased depth of field. However, the amount of light reaching the retina is diminished, which may cause a paradoxical and offsetting decrease in visual acuity.

A decreased pupil size increases the amount of diffraction experienced. A resulting retinal diffraction pattern caused by a point source appears as alternating light and dark bands of concentric circles in a bull’s eye configuration. The measured distance between the center light circle (peak) to the center of the adjacent dark band (trough) is known as Airy’s disk. The radius of Airy’s disk increases with decreased pupil size. If viewed through a small pupil, two point sources of light will create two retinal diffraction patterns. The two objects must be sufficiently far away from each other to be resolved as two separate objects. The minimum required distance separating the two objects to allow for proper resolution is the radius of Airy’s disk (or when the peak of one Airy’s disk falls on the trough of the other Airy’s disk). Because the radius of Airy’s disk increases (as does the size of the diffraction pattern) with decreased pupil size, the distance between two objects must also increase in order for the perception of two point sources to occur.

Question 62

A patient is seen at your office complaining that her right eye is physiologically higher than her left eye. She would like to know if glasses would help improve the cosmesis of her predicament. You know that prism will shift the image of an object. How would you orient a prism to help her appearance?

Correct answer Prescribe base up prism over the right eye

Prescribe base in prism over the left eye

Prescribe base down prism over the left eye

Prescribe base out prism over the left eye

Answer 62

A prism will bend light towards its base, but the image will be shifted towards the apex of the prism. Therefore, by prescribing base up prism over her right eye, its image will be shifted down towards the apex of the prism. Another way of remembering this is to think of the prism as an arrow that will point in the direction of the deviation (i.e., exotropia is neutralized with base in prism, the eye points outwards, the apex of the prism also points out). Prescribing prism for cosmetic purposes may not always be an option as significant vertical prism may induce diplopia or visual discomfort.

Question 63

An object is 15.0 cm in height and is located 25.0 cm from a plane mirror. Given the angular magnification of the mirror, what is the size of the image?

10. 0 cm
11. 0 cm
12. 5 cm
13. 0 cm

Correct answer 15.0 cm

Answer 63

Plane mirrors do not alter the vergence of light, as they do not possess any power. The resulting image will therefore have a magnification factor of 1.0 x and will be the same size as the object, which in this example is 15.0 cm.

Question 64

You perform a manifest refraction on your patient and obtain a value of -5.50 -3.00 x 180 for the right eye. If the phoropter was placed with a vertex distance of 13mm, what is the refractive error at the corneal plane? (round to the nearest 0.12)

Correct answer -5.12 -2.50 x 180

• 5.12 -3.00 x 180
• 5.12 -2.87 x 180
• 5.12 -3.50 x 180
• 5.12 -3.12 X 180

Answer 64

Vertexing is the conversion between the powers at the spectacle and corneal plane. It is important to vertex the power at the spectacle plane when fitting a patient with contact lenses, as the required contact lens power will be different than value obtained during refraction with the phoropter. Typically, powers over 4D (either plus or minus) should be vertexed, while spectacle powers below 4D are not significantly different at the spectacle and corneal plane and do not typically need to be vertexed. It is important to remember that if there is a cylindrical component, each meridian must be converted separately.
The equation for conversion of power at the spectacle plane to the corneal is below:
Fc = Fs / (1 ? dFs)
Fc = power at corneal plane, Fs = power at spectacle plane, d = vertex distance (meters)
For the above patient, the power in the 180 meridian is -5.50 and the power in the 90 meridian is -8.50
Fc = -5.50 / (1 ? (0.013)(-5.50) = -5.132 in 180 meridian (or -5.12D)
Fc = -8.50 / (1 ? (0.013)(-8.50) = -7.654 in 90 meridian (or -7.62D)
Answer = -5.12 -2.50 x 180

Question 65

Which of the following equations may be used to determine the location of the secondary focal point of a +5.00 polycarbonate (n=1.586) thin lens that is located in air?

f’= -1.586/+5.00

f’= -1.00/+5.00

f’= 1.586/+5.00

Correct answer f’= 1.00/+5.00

Answer 65

A thick lens has two refracting surfaces. The refractive power of the first surface is determined by using the equation P=(n-n’)/r. The power of the second refractive surface is calculated via use of the equation P=(n’-n)/r. For a thin lens, in most circumstances, it can be assumed that all refraction occurs at a single surface despite the fact that the lens possesses two surfaces. In the case of a thin lens, the thickness is presumed to be negligible the majority of the time.

Question 66

You are trying to obtain some frame measurements for a patient and can’t find your ruler. You see some markings on the inside of the frame that reads “‘55 □ 19 145.” What do these measurements represent, respectively?

Frame PD (pupillary distance), “B” measurement, and temple length

“A” measurement, “B” measurement, and temple length

“B” measurement, DBL (distance between lenses), and temple length

Frame PD (pupillary distance), DBL (distance between lenses), and temple length

Correct answer “A” measurement, DBL (distance between lenses), and temple length

Answer 66

If you are trying to obtain some quick measurements of a frame, most frames are marked with the eye size, DBL, and temple length. When a marking with □ is seen, it means that the measurements are determined using the boxing system. For example a marking on a frame that reads: “55 □ 19 145” indicates that the “A” measurement is 55mm, DBL is 19mm, and the temple length is 145mm. On most frames these markings are found imprinted on the inner side of the temples in plastic frames and inside of the bridge on metal frames (but they may be present at other locations).

Question 67

Your patient’s glasses are crooked, with the right lens higher on his face compared to the left lens. Select the most likely frame adjustment solution.

Bend the right temple down

Correct answer Bend the right temple up

Bend the right temple in, decreasing temple spread

Bend the right temple out, increasing temple spread

Answer 67

If the right lens is sitting higher than the left lens, the right lens needs to be lowered. This is accomplished by raising the right temple or decreasing the pantoscopic tilt. By raising the right temple, a space is created between the top of the right ear and the temple allowing the frame to drop down on the ear and thus leveling the frame on the face.

Question 68

A Galilean telescope has an ocular lens with a power of -32.00 D and an objective lens with a power of +8.00 D. What is the magnification provided by the telescope?

8x

Correct answer 4x

256x

0.25x

Answer 68

o calculate the magnification (M) of a telescope divide the power of the ocular lens (Doc) by the power of the objective lens (Dobj): M=-Doc/Dobj. In the example above, M=-(-32 D)/8 D= 4x. The magnification of a Galilean telescope is positive due to the fact that its ocular has a minus powered lens. The magnification of an astronomical telescope is negative and therefore its image will be upside down.