Optics

Front 1

K-readings: +44.00 DK
Base Curve: 7.6 mm
Subj Ref: -1.00 D (measured at cornea)
What power would you choose for rigid contact lenses?

Back 1

There are a number of (nearly equivalent) ways to do this problem. First, let's put the base curve and the K-readings in the same units. We can either choose mm or DK. Here, we put them both in DK units. The K readings are already in DK units. To convert the base curve, we have:

BC = 337.5 / 7.6 = 44.41 DK

Using the exploded system, we can easily calculate the power of the tear lens.

F (tear lens) = BC - (K reading)
F (tear lens) = +44.41 - 44.00 = +0.41 D.

Since the tear lens is positive, we must account for this by adding a negative amount to the contact lens prescription. The prescription is therefore:

CL = -1.00 - 0.41 = -1.41 D

Of course we could round to the nearest 0.25 D, if needed.

We can confirm our answer using SAM-FAP: in this case, 'steeper add minus'. In fact, the base curve is steeper than the cornea, so we have done the problem correctly (we added minus to the prescription).

Front 2

K-readings: +42.00 DK (both principal meridians)
Subj Ref: +2.00 D (measured at cornea)
What power RGB would you choose for an on-K fit? What would the base curve be (in mm)? Instead, what if you choose the base curve to be BC = 41.50 DK?

Back 2

For an on-K fit, one chooses the base curve to align with the corneal curvature. So in this case, the base curve would be

BC = +42.00 DK
or
BC = 337.5/42 = 8.04 mm.

For an on-K fit, the power of the tear lens (in the exploded system) is zero. So we choose the RGP to match the subjective refraction:

CL = +2.00 D.

On the other hand, if we chose the base curve to be BC=41.50, the power of the tear lens (in the exploded system) is:

F (tear lens) = 41.50 - 42.00 = -0.50 D

We must therefore add positive power to the prescription to account for the negative tear lens power. We have

CL = +2.00 + 0.50 = +2.50 D.

Again, we can check using SAM-FAP (in this case, 'flatter add plus').

Front 3

Subjective Refraction: +5.00 D (measured at vertex of 12 mm)
K-readings: +44.00 DK (both meridians)
Base curve: 44.50 DK.
What power rigid contact lens would you prescribe?

Back 3

First, let's scale the subjective refraction to account for the vertex distance. A refraction of +5.00 at a distance of 12 mm corresponds to a refraction of Rc at the cornea. We have

Rc = + 5.00 / (1- (0.012 x 5)) = +5.32 D

So this is the refraction you would obtain at the cornea.

Now we must find the power of the tear lens (exploded system). This is given by:

F (tear lens) = 44.50 - 44.00 = + 0.50 D

Since the tear lens is positive, we must account for this by adding a negative amount to the prescription. We would have

CL = 5.32 - 0.50 = +4.82 D.

We can double check by using SAM-FAP (in this case, 'steeper add minus').

Front 4

You are fitting a toric contact lens. The prescription is given by +2.00 - 1.00 x 135. When the patient blinks while wearing the contact lens, the mark at the bottom of the lens rotates to the left (lens rotates clockwise) by 5 degrees. What new prescription would you choose?

Back 4

Since the lens rotates by 5 degrees, we must compensate by choosing the axis of the prescription to be offset by 5 degrees. In terms of the mark at the bottom of the lens, we can describe this using 'LARS' (left add, right subtract).

Since the lens rotates clockwise (the mark moves left), we choose the axis to be 135 + 5 = 140. So we have

+2.00 - 1.00 x 140.

Front 5

You are fitting a toric contact lens. The prescription is given by -2.00 - 1.00 x 165. When the patient blinks while wearing the contact lens, the mark at the bottom of the lens rotates to the right (lens rotates counterclockwise) by 5 degrees. What new prescription would you choose?

Back 5

Using 'LARS', we would choose the new axis to be 165 - 5 = 160 degrees. So we have

-2.00 - 1.00 x 160.

Front 6

Fluorescein staining (in contact lens fitting) show vertical touch and horizontal pooling. What type of astigmatism does this eye have?

Back 6

Technically, we wouldn't be able to tell what kind of astigmatism the eye, as a whole, has. However, we can say something about the CORNEAL astigmatism.

ATR astigmatism is characterized by pooling in the horizontal meridian and touch in the vertical meridian. WTR would show the opposite characteristics.

Front 7

Subjective Refraction (at cornea): +3.00 - 1.00 x 180.
K-readings: +43.00 DK (at 180), +44.50 DK (at 90).
Hard Contact lens is used for correction.
CL power is +3.00 D.
Base curve is 43.00 DK.
What is the over-refraction that would be measured?

Back 7

The subjective refraction (which the problem states has already been corrected for vertex distance and is therefore valid at the cornea) has -1.00 D of astigmatism (axis 180). The corneal astigmatism can be found from the K readings. The K readings themselves tell us that the astigmatism of the cornea itself is: -1.50 D, axis 90.

We want to compare the refraction (which is the 'correction', not the error of the eye itself) to the K readings (which are for the eye itself, not the 'correction'). Let's convert the K readings to 'correction', simply by inserting a minus sign or switching the axis by 90 degrees. In terms of a correction, the corneal astigmatism would be +1.50, axis 90 or, equivalently, -1.50 D, axis 180.

The residual (non-corneal) astigmatism is:

Residual Astig = -1.00 - (-1.50) = +0.50 D, axis 180.

Or, we could write this as -0.50 D, axis 90.

If we use a spherical hard contact lens, we will correct all of the corneal astigmatism. However, the residual astigmatism will remain. So we expect -0.50 D, axis 90, to remain even with the spherical hard contact lens.

So the over-refraction would be plano - 0.50 x 90.

Notice that the spherical portion of the correction is taken care of by the +3.00 D CL powers.

Front 8

Subj Refraction (at 12 mm): +2.00 - 3.00 x 90.
K-readings: +45.00 DK (at 180), +42.50 DK (at 90).
A spherical soft contact lens is used for correction. How much astigmatism remains? How much astigmatism remains if a spherical hard contact lens were chosen instead?

Back 8

First, let's find out what the refraction would be at the cornea. The power cross at 12 mm is +2.00 (vertical) and -1.00 (horizontal). First, the vertical power needed at the cornea would be:

F vertical = +2.00 / (1-0.012*2) = 2.05 D

The horizontal power needed at the cornea would be:

F horizontal = -1.00 / (1+0.012*1) = - 0.99 D

So the refraction at the cornea would be

+2.05 - 3.04 x 90.

Note that since the powers are small (+2.00 and -1.00 D in each meridian), accounting for the vertex distance makes very little difference!

A spherical soft contact lens will not correct any astigmatism, so there would be -3.04 D, axis 90 of astigmatism remaining.

A hard contact lens will correct only the corneal part of the astigmatism. Let's find the residual (non-corneal) astigmatism.

Res Astig = Refraction Astigmatism - Corneal Astigmatism.

We can get the corneal astigmatism from the K-readings. It would be -2.50 D, axis 90 (remember that K-readings are for the eye itself, and we want the corneal astigmatism in terms of the correction, not the eye, so we throw in a minus sign). We have

Res Astig = -3.04 - (-2.50) = -0.54 D, axis 90.

This is the amount of astigmatism that would remain if a spherical hard contact lens were used for correction. This is pretty large, so this patient might be a good candidate for toric lenses.

Front 9

A patient with no visual pathologies looks through a pinhole occluder of radius 1 mm. Estimate the person's acuity under these conditions.

Back 9

Since there are no pathologies, we will assume diffraction limited vision. Using Rayleigh's criteria, we have

d sin(t) = 1.22 lambda
0.002 sin(t) = 1.22 (555 x 10^(-9))
t = 0.019 degrees

Note that we used d = 2 mm (the diameter, not the radius).

To estimate acuity, we need t to be measured in arcminutes. Then it corresponds to the MAR. We have

t = 0.019 * 60 = 1.16 arcminutes.

So the Snellen acuity is

20/x = 1/1.16
x = 23

The acuity is approximately 20/23.

Front 10

What is the ideal index of refraction to be used for anti-reflective coating on a crown glass (n=1.52) lens?

Back 10

Assuming this coating will separate the lens from air, the ideal index of refraction is simply the square root of the index of the lens material.

n = (1.52)^(1/2) = 1.23.

Front 11

What is the minimum thickness for an anti-reflective film made of magnesium fluoride (n=1.38)?

Back 11

The minimum thickness is given by

t = lambda / (4 n);

We will take lambda = 555 nm. So we have

t = 555 / (4*1.38) = 101 nm.

So the minimum thickness would be about 0.1 micron

Front 12

Consider a simple eye model consisting of a +43.00 D thin lens (the cornea) located 3.5 mm in front of a +20.00 D thin lens (the lens). The refractive index everywhere inside this reduced eye can be taken to be 1.33. If the axial length of the eye is 24 mm and this eye looks -- without accommodation -- at a real object 40 cm in front of the cornea, where does the image fall relative to the retina?

Back 12

This is a simple geometric optics problem involving successive imaging through two lenses. We have

L = -2.50 D
L'= F + L = +43.00 - 2.50 = 40.50 D.
l' = 1.33/40.50 = 3.28 cm.

The image from the cornea is formed 3.28 cm behind the cornea, which is 2.93 cm behind the lens. This image is a virtual object for the second lens.

L = 1.33 / 0.0293 = 45.4 D.
L'= F + L = +20.00 + 45.4 = 65.4 D.

That is, the image through both lenses falls 65.4 D, or 20.3 mm behind the second lens. From the axial length of the eye, we know that the retina falls 24-3.5 = 20.5 mm behind the second lens.

Therefore, the image falls 0.2 mm in front of the retina.

Front 13

A spectacle lens has a front surface power of +11.00 D, a refractive index of 1.5, a center thickness of 2 mm, and a back vertex power of +3.00 D. What is the power of the back surface of the lens?

Back 13

We know that the back vertex power, in terms of reduced thickness, t = 2/1.5 = 1.33 mm, is given by:

Fv = F2 + F1/(1-t F1);

We can solve this for F2 to get

F2 = Fv - F1/(1-t F1)
F2 = +3.00 - 11.00 / (1 - (0.00133)(11.00) )
F2 = - 8.16 D.

Front 14

The front and back sags of a bi-concave lens are measured to be 3.0 mm and 1.8 mm. If the center thickness of the lens is 4 mm, what is the edge thickness?

Back 14

For a bi-concave lens, it is easy to see that (in terms of absolute lengths) we have

ET = CT + s1 + s2
ET = 4.0 + 3.0 + 1.8 = 8.8 mm.

This lens would obviously have F < 0.

Front 15

The sag of the front surface of a 50 mm round glass lens is measured to be 3.6 mm. If this front surface is convex relative to air, what is its surface power?

Back 15

This problem relates sag to surface power. First we find radius of curvature. We have

r = h^2 / (2 s) = 25^2 / (7.2) = 86.8 mm.

Now surface power is simply

F = (1.5 - 1) / 0.0868 = 5.76 D.

Since the surface is convex relative to the lowest index of refraction medium, it is positive, so F = + 5.76 D.

Front 16

What prism is induced if we decenter a -5.00 D lens upward by 2 mm?

Back 16

Using Prentice's rule, we have

Z = 0.2 * 5.00 = 1.00 prism diopter.

Since the lens is negative, decentering upward corresponds to base down prism.

Front 17

What add would be prescribed for a patient whose near point was measured to be 25 cm when a +2.00 D assisting lens was used?

Back 17

The near point was measured to be 25 cm, or 4.00 D. However, one must remember that 2.00 D came from the assisting lens. Therefore, the amplitude of accommodation is +2.00 D.

If we assume the patient will be working at 40 cm, we must prescribe an add that requires the use of half the AoA (+1.00 D). The add would have power

F = 2.50 - 1.00 = 1.50 D.

Front 18

What is the accommodative demand when an uncorrected 2.00 D hyperope views a near object at 40 cm? Around what age will comfortably viewing such an object become difficult?

Back 18

The accommodative demand is 2.00 + 2.50 = 4.50 D. A hyperope must accommodate to see a distant object and then further accommodate to see a near object.

Since accommodation ability goes roughly as

AoA = 18.00 D - 0.3 * y

where y is age in years, and we can only use half of AoA comfortably, we have

9.00 = 18.00 - 0.3 y
y = 30 years old

So such near vision will become difficult after 30.

Front 19

Which Purkinje image is real and inverted?

Back 19

Image IV is real and inverted. The remaining three are virtual and upright.

Front 20

Which Purkinje image is formed ONLY by reflection?

Back 20

Purkinje image I is due to reflection only. The remaining images are due to reflection and refraction.

Front 21

What is the main clinical use for the Haidinger's brush phenomenon?

Back 21

Haidinger's brush can be used to diagnose eccentric fixation. Remember, Haidinger's brush is due to the birefringence of the macula.

Front 22

Which visual phenomenon is associated with the peripheral retina?

Haidinger's brush
Maxwell's Spot
Moore's lightning streaks

Back 22

Moore's lightning streaks is the only listed phenomenon associated with the peripheral retina.

Front 23

A patient requires the following prescription. What type of astigmatism does this patient have?

-3.00 + 2.00 x 125

Back 23

The principal meridians of the presciption have the following powers:

-3.00 D at 125
-1.00 D at 35

So this eye is 3.00 D myopic at 125 and 1.00 D myopic at 35. The eye meridian with the most plus power is the 125, so this is oblique.

So this is compound, myopic, oblique astigmatism.

Front 24

Cross cylinders (+-0.50 D) are placed before a patient's eyes while the patient views a near grid with horizontal and vertical lines. The plus meridian of the cross cylinder is vertical. Assuming that no add power is needed for this patient, which set of lines (horizontal or vertical) appear the most clear?

Back 24

The alignment of the JCC means that the horizontal line focus of the Conoid of Sturm will fall before the vertical line focus. If the vertical line appears brightest, the patient has been able to accommodate sufficiently and does not require an add.

On the other hand, if the horizontal line appears brightest, one should add +0.25 D increments until the images become identical or the vertical lines appear brightest.

Front 25

Explain why the duochrome test for refining the sphere in a refraction can still be used for a protanope.

Back 25

The duochrome test is based on chromatic aberration. Different wavelengths have (effectively) different indices of refraction, meaning that images of different colors will be focused to slightly different locations. The effect is exploited by showing the patient stimuli of two different colors.

The duochrome test is based on a physical property of light and is completely independent of the perception of color, which is altered in protanopes (who are missing the L cone photopigment).

Front 26

When a stenopaic slit is aligned horizontally, subjective refraction determines the best sphere to be -1.50 D. The slit is then rotated and the best sphere is measured to be +0.5 D. What prescription should be prescribed?

Back 26

The power cross of the lens would have -1.50 D (horizontally) and +0.50 D (vertically). This corresponds to +0.50 D - 2.00 D x 90

Front 27

A patient has a binocular PD of 64. The prescription is

OD: -4.00 D
OS: -9.00 D

If the optical centers of the lenses are incorrectly placed at a PD of 60, how much prism is erroneously induced?

Back 27

Assuming the eyes are symmetrically located about the midline, the PD is incorrect by 4/2 = 2 mm per eye. This leads to prism of

OD: Z = -4.00 x 0.2 = 0.8 prism diopters base out

OS: Z = -9.00 x 0.2 = 1.8 prism diopters base out

The erroneously short PD is like having both lenses decentered inward. As a result, the base induced by each is base out, since both lenses are negative lenses.

The total prism is therefore 2.6 prism diopters base out.

Front 28

What is the size of a letter on the 20/100 line of a typical Snellen chart?

Back 28

The 20/100 line corresponds to an MAR of 5 arcminutes. To be resolved, a letter must be at least 5 MAR, so this letter is 25 arcminutes. This corresponds to 25/60 = 0.41 degrees. Since the Snellen chart is used at 20 ft, the size of the letter is

x/20 = tan (0.41)
x = 0.14 feet

This is about 1.75 inches.

Front 29

Using a lens clock, you measure the front and back surface powers of a lens to be +4.00 D and -6.50 D, respectively. If the lens clock is calibrated for n=1.53 and the lens is made of material with n=1.45, what is the front vertex (neutralizing) power of the lens? The lens has a center thickness of 2 mm.

Back 29

First we find the actual surface powers.

F1 = +4.00 (0.45/0.53) = + 3.40 D

F2 = -6.50 (0.45/0.53) = -5.52 D

The front vertex power is then given by

Fn = F1 + F2/(1-t F2)

where t = (0.002/1.45) = 0.0014 m.

So

Fn = +3.40 - 5.52 / (1+0.0014*5.52) = -2.08 D.

One could similarly find the back vertex power as well, using

Fv = F2 + F1/(1-t F1).

Front 30

A patient wears the following prescription:

OD +4.00 - 2.00 x 180
OS +3.00 - 3.50 x 90

If the patient looks down 10 mm to read, what vertical imbalance is induced?

Back 30

In the vertical meridian, we have powers of:

OD: +2.00 D
OS: +3.00 D

So the prism effect from looking down 1.0 cm is

OD: 2.00 prism diopters base up
OS: 3.00 prism diopters base up

So the prism would be 1.00 prism diopter BUOS or, equivalently, 1.00 prism diopter BDOD.

Note that we could also do this problem if the axes were oblique (not 90 or 180) by using the sine squared law to find the approximate power in the vertical meridian.

Front 31

Which of the following aberrations involve(s) more than one wavelength of light?

Spherical aberration
Radial Astigmatism
Chromatic aberration
Coma
Distortion
Curvature of field

Back 31

Chromatic aberration arises from the difference between refractive indices of different wavelengths of light. The other aberrations are monochromatic and therefore involve only a single wavelength.

Front 32

Which aberration can be controlled by choosing the correct base curve for a given back vertex power?

Back 32

Radial astigmatism (also called marginal or oblique astigmatism) can be controlled by choosing the base curve according to the flatter curve from Tscherning's ellipse.

Front 33

Which aberration is typically associated with a teacup and saucer image?

Back 33

Radial/oblique/marginal astigmatism is associated with a teacup and saucer image. Specifically, the Teacup is formed by the Tangential rays and the Saucer by the Sagittal rays. You will recall that radial astigmatism arises from an asymmetry between the tangential and sagittal merdians, and that is exactly the origin of the teacup and saucer.

Front 34

A patient has a distance prescription of +2.50 - 3.00 x 135 and a +2.50 D add. The seg height is 17 mm, the seg drop is 4 mm, and the add is a flat-top 28. What is the image jump in this lens?

Back 34

Image jump involves only the add power. Using Prentice's rule and the fact that the distance from the optical center of the seg to the edge of the seg is 5 mm for a flat-top 28, we have

Z = 0.5 x 2.50 = 1.25 prism diopters base down.

It is base down because we are looking above the optical center of a plus lens (because image jumps happen at the seg line, which is above the seg optical center).

Front 35

What is the image jump for an executive bifocal of power +2.00 D?

Back 35

The image jump is 0, because the optical center of the seg is at the seg line, so using Prentice's rule, we have

Z = 0 x 2.00 = 0 prism diopters.

Front 36

A patient has a -3.00 D distance prescription and a +2.00 D add. When the patient reads, he looks 10 mm below the optical center of the distance lens. If the bifocal is a straight-top 25, what is the total prism induced when the patient is reading? The seg drop is 4 mm.

Back 36

We must calculate the prism from the near and distance portions individually and add them together. For the distance portion, we have

Z = 3.00 x 1 = 3 prism diopters base down.

It is base down because he is looking below OC in a divergent lens. For the near portion, we first calculate how far away from the optical center we are looking. The optical center of a straight-top 25 seg is 5 mm below the seg line. Therefore, it is 5 + 4 = 9 mm below the distance optical center. When the patient looks down 10 mm, he is looking 1 mm below the optical center of the seg. So the induced prism from the seg is

Z = 0.1 x 2.00 = 0.2 prism diopters base up.

It is base up because he is looking below the OC of a plus lens. The total prism is therefore

Z = 3 - 0.2 = 2.8 prism diopters base down.

Remember the big picture here: first calculate the prism from looking off-center in the distance lens, then calculate the prism from looking off-center in the near lens. Add these two prisms together to get the total prism.

Front 37

Find the prism induced in a +5.00 D lens decentered by 2 mm down and 3 mm out.

Back 37

The prism in the horizontal direction is

Z = 0.3 x 5 = 1.5 prism diopters base out.

In the vertical direction, we have

Z = 0.2 x 5 = 1 prism diopter base down.

To add the two prisms, we use vector addition:

Z^2 = 1^2 + 1.5^2 = 3.25
Z = 1.8 prism diopters down and out.

We can find the angle at which the base points as well. Using geometry (tangent equals opposite over adjacent), we have

tan x = 2/3
x = 33.7 degrees.

So the prism is pointing base out at an angle of 33.7 degrees below the horizontal axis.

Front 38

An aphakic eye has a far point located 10 cm in front of the eye. Where is the near point? What is the refractive error in this eye?

Back 38

The refractive error is simply

F = 1/0.10 = 10.00 D, myopic.

Since there is no lens, accommodation is not possible, so the near point and the far point are the same (10 cm in front of the eye).

Front 39

The keratometry readings for a patient's central corneal curvature are found to be +45.00 D in both principal meridians. If you prescribe hard contacts with a base curve of 8.00 mm, what is the power of the tear lens (in the exploded system)? How would you account for this?

Back 39

we can estimate the answer if we know the corneal curvature in mm. Given the keratometry reading, we know the radius of curvature is

45.00 = (1.3375 - 1)/r
r = 7.5 mm.

We therefore know that the cornea is steeper (smaller r) than the contact lens, so the tear lens must be negative.

The difference in radii of curvatures is 0.5 mm. This corresponds to a tear lens of power

F = 0.5 * 0.25 / 0.05 = - 2.50 D.

Again, since the contact lens is flatter (higher r), we know the tear lens is negative, so we would need to compensate by adding +2.50 D to the prescription.

The exact answer is found by treating the tear lens as two SSRI's. The relevant indexes of refraction are 1.49 for the contacts, 1.336 for the tear layer, and 1.3375 for the cornea. However, it is convenient to assume a thin layer of air (n=1) separates all materials. This optics assumption is known as the 'exploded system' method.

As a result, we only need the tear index of refraction. We have

F1 = (1.33-1)/0.008 = 41.25 D

This power is positive, because the surface is convex.

F2 = (1.33-1)/0.0075 = -44.00 D

The power is negative because the surface is concave.

The total tear lens power is therefore

F = -2.75 D

which is pretty close to our approximation. Note that because the tears and the 'effective' cornea have similar indices of refraction, one could use the K-readings directly.

Front 40

A 6X Galilean telescope has two lenses which are separated 15 cm. What are the powers of the ocular and objective lenses?

Back 40

The telescope is 6x, so the ratio of powers is

Foc / Fobj = -6.
Foc = -6 Fobj

Also, the separation distance is 15 cm, which means

|1 / Fobj| - |1 / Foc| = 0.15,

where I have used absolute values to avoid sign confusions. Combining our two equations, we have

1/ Fobj - 1/(6 Fobj) = 0.15
Fobj = 5.56 D

So Foc = -33.3 D.

Front 41

A lens has a front surface of +6.00 D and a back surface of -2.00 D. The lens is 48 mm round, made of glass (n=1.5), and it has an edge thickness of 2.2 mm. What is the center thickness of the lens?

Back 41

We first must draw a picture of the lens. The front surface is positive, so it must be convex relative to air. The back surface must be concave relative to air. Judging from the picture, we know that

CT-ET = s1 - s2

where all variables are absolute values (no sign convention is used). So we need to find the sags of the front and back surface. But that is easy. First find radii of curvature.

r1 = (1.5 - 1)/6.00 = 8.33 cm = 83.3 mm
r2 = 0.5 / 2.00 = 25 cm = 250 mm.

Again, I am not worried about a sign convention, just absolute values. From these, we can get the absolute values of the sags:

s1 = h^2/(2r) = 24^2 / (2*83.3) = 3.46 mm
s2 = h^2/(2r) = 24^2 / (2*250) = 1.15 mm

So

CT - ET = 3.46 - 1.15 = 2.31 mm.

Therefore, we have

CT = 2.31 + 2.20 = 4.51 mm.

As a check, note that the total power of the lens would be approximately +4.00 D (neglecting thickness), and in fact, the lens is thicker in the center than on the edge, as expected for a plus lens.

Front 42

A telescope is constructed from two lenses of power +60.00 (ocular) and +10.00 (objective). The telescope has an adjustable tube length, so it can be used to view near objects. By how much must the tube length be changed to see an object (using no accommodation or add power) that is 40 cm away?

Back 42

First, let's find the tube length of the telescope under normal (distance viewing) conditions. This is just the sum of the focal lengths of the lenses. So we have

l = 1/60 + 1/10 = 11.67 cm

To see a near object at 40 cm, we must adjust the tube length accordingly so that plane waves leave the ocular lens when viewing the near object. We can use a trick to find this. Specifically, we can add the vergence of the near object to the power of the objective lens and then find the new tube length for an afocal telescope with these powers. Can you see why this trick works?

In other words, since the vergence of the near object (at the objective lens) is -1/.40 = -2.50 D, we must add this to the object lens power, giving an objective lens of power +7.50 D. So we need the separation between lenses of powers +60.00 D and +7.50 D. To create an afocal telescope from these lenses, they must be separated, again, by the sum of their focal lengths:

l = 1/7.50 + 1/60 = 15 cm

So to see the near object through the telescope without accommodation or add power, the tube length must be extended by

d = 15 - 11.67 = 3.33 cm.

Front 43

Keratometry readings from a patient are given below. What is the radius of curvature of the patient's central cornea?

OD: +44.00
OS: +42.00

Back 43

We can convert between the reading of the keratometer and the radius (in units of length) by using

F = (1.3375-1)/r

This is just treating the cornea like an SSRI with n=1.3375.

So we have

OD:
r = 0.3375/44.00 = 7.7 mm.

OS:
r = 0.3375/42.00 = 8.0 mm.

Note that this is the reasoning behind the clinical rule that says

r = 337.5 / F

This is the same equation, but r is now given directly in mm.

Front 44

What is the minus cylinder form of the following?

+2.00 + 1.00 x 90

Back 44

Drawing the power cross, we see that this is simply +2.00 in the vertical meridian and +3.00 in the horizontal meridian. So we would have

+3.00 - 1.00 x 180

Front 45

What minimum blank size is needed for a lens with effective diameter 62 mm and a decentration 10 mm (5 mm per lens)?

Back 45

The minimum blank size would be

BS = 62 + 10 + 2 = 74 mm.

The extra 2 mm is a fudge factor that accounts for edging.

Front 46

Which curve on the Tscherning ellipse is used to pick a base curve?

Back 46

While in principle either curve will reduce radial astigmatism for a given back vertex power, the flatter Ostwalt curve is typically used in practice to determine the base curve.

Front 47

An object is 2 m from a 7 prism diopter (base down) prism. By how much is the image of this object displaced when viewed through the prism? If the prism is 10 cm in front of the eye's center of rotation, by what angle must the eye rotate to see the image?

Back 47

We could use the prism effectivity equation to do this problem (see below). However, instead we will try to reason it out.

First, 7 prism diopters means the displacement would be 7 cm at a distance of 1 m. So at 2 m, the displacement is 14 cm. The prism is base down, so the image is displaced upward (it appears higher than the object).

Assuming the prism has no dioptric power, the image is located 210 cm from the center of rotation of the eye (this is just 2 m plus 10 cm). So the angle x that the eye must rotate is given by

tan x = 14/210
x = 3.81 degrees.

Here's how we do the same problem with the prism effectivity equation (which is just the algebraic statement of the method above). We have

Ze = Z/(1-C L)

with Z = 7 prism diopters, C= 0.10 m, and L=-0.5 D. So we have

Ze = 6.66 prism diopters

To convert prism power to angle, we have

tan x = 6.66/(100)
x= 3.81 degrees.

Front 48

Unpolarized light is incident on a polarizer-analyser setup. The intensity of the transmitted light is measured to be 20percent of the intensity of the original light. What is the angle between the transmission axis of the polarizer and that of the analyzer?

Back 48

This is a simple application of Malus' Law. The polarizer-analyser setup is simply two polarizers rotated at some angle relative to one another. So this is very similar to the Malus law problem in the text.

Let the original light intensity be I0. Then the final transmitted light has intensity 0.20 I0; this is just 20percent of the original value. Let's consider the path the light takes. After striking the first polarizer, the light loses half its intensity. So the light leaving the first polarizer has intensity 0.5 I0.

This light then strikes the second polarizer, which is rotated at some angle. To find the angle, x, we simply use Malus:

0.20 I0 = 0.50 I0 cos^2 (x)
cos x = 0.16
x = 50.8 degrees.

So the two polarizers are rotated at an angle of 50.8 degrees relative to one another.

Front 49

A patient wears +10.00 D spectacles (n=1.5) at a vertex distance of 14 mm. The lens has a base curve of +16.00 D and is 11 mm thick. What is the magnification of the retinal image when the person puts on these glasses? How would this change if this lens were made thinner (all else being equal)?

Back 49

The spectacle magnification is a product of shape and power. We have t=14 mm, n=1.5, and F1=+16.00 D (the base curve). So the shape factor is:

Ms = 1/(1-(t/n) F1)
Ms = 1.13

We also know that h=14+3 = 17 mm and Fv=10.00, so the power factor is

Mp = 1/(1-hFb)
Mp = 1.20

So

M = Mp Ms = 1.35.

The image is 35percent larger with glasses than without.

If the lens is made thinner, Ms would approach 1, so the total magnification would approach M = 1.20. That is, the image would get smaller.

Front 50

Consider a patient who wears a -10.00 D lens at a short vertex distance of 10 mm. By what factor does the retinal image size change when this lens is moved to a vertex distance of 14 mm, assuming the power doesn't change? How would your answer change if you changed the prescription to properly account for the new vertex distance?

Back 50

There are two things changing here: the vertex distance and the power needed in the prescription. We consider two cases. In the first, we just assume that the lens is being moved, but it's power is not changing. The spectacle magnification changes as a result of h going from 13 mm to 17 mm. Specifically, the power factor Mp changes. At a vertex distance of 10 mm (h=13 mm), we have:

Mp = 1/(1-h Fv)
Mp = 0.885

For a vertex distance of 14 mm (h=17 mm), we have

Mp = 1/(1-h Fv)
Mp = 0.855

So the magnifcation changes by a factor of 855/885 = 0.97; that is, the image size at the new vertex distance is only about 97percent of its size at the old vertex distance. Note that we didn't need the shape factor, because it cancels out in the ratio.

Now let's consider the case where we change the prescription appropriately. If we need -10.00 D at 10 mm, then we would need a power of -10.41 D at a distance of 14 mm. See Geometric Optics (or earlier problems) for a discussion of this point. So now, our power factor becomes, at 10 mm,

Mp = 1/(1+0.013*10) = 0.885

and at 14 mm we have

Mp = 1/(1+0.017*10.41) = 0.849

So the results, to this level of precision, are practically the same. The new image is about 96percent as big as the original.

Front 51

What is Knapp's law? How is it useful clinically?

Back 51

Knapp's law says that when a correcting lens for an axial ametrope is placed at the primary focal point of the eye, the relative spectacle magnification is 1. That is, the retinal image size will be the same as for an emmetrope.

Knapp's lie underlies the idea that axial ametropes should be corrected with spectacles, because the vertex distance (approximately 14 mm) is relatively close to the eye's primary focal point (approximately 17 mm). Therefore, the RSM remains approximately 1.

Note that either spectacles or glasses could be used to correct for the refractive error of the eye. However, only glasses would produce a retinal image SIZE approximately equal to that formed in an emmetropic eye.

Front 52

Consider a spherical glass lens of power -5.00 D with 8 degrees of pantoscopic tilt. What (un-tilted) spherocylindrical lens is equivalent to this tilted spherical lens?

Back 52

Let's find the power in the sagittal meridian first. In the case of pantoscopic tilt, this is the 180. To do this, we use

Fs = -5.00 (1+sin^2(theta)/3)
Fs = -5.03 D

Fs is the power in the sagittal meridian. Notice that this is the equation for sphere power of a tilted lens given on the equation sheet and the textbook. However, read below to see how to use this properly.

We also know there is some radial astigmatism induced. That is just the difference in the powers between the tangential and sagittal meridians. We have

Fc = -5.00 tan^2(theta)
Fc = -0.10 D

This is just the equation for cyl power (or radial astigmatism) of a tilted lens that is given in the book. It is saying that there is 0.10 D more negative power in the tangential meridian. The tangential meridian is 90 for pantoscopic tilt. We therefore have

Ft = -5.03 - 0.10
Ft = -5.13 D

where Ft is the power in the tangential meridian.

So the power cross would have -5.03 (180 meridian) and -5.13 (vertical meridian). This is equivalent to

-5.03 - 0.10 x 180

This is the case covered in the textbook. Note that for a minus lens like this one, the radial astigmatism is the cyl power (when written in minus cyl) and the sphere power is the same as the power in the sagittal meridian.

THIS IS NOT THE CASE FOR A PLUS LENS!

For a plus lens, the equation given above for the cyl power gives the cyl of the prescription when written in PLUS CYLINDER FORM. Then the new sphere power is also given by the equation for Fs above. Again, though, this is written in plus cylinder form. You would have to transpose it get the answer in minus cyl form.

Front 53

Consider a spherical glass lens of power +5.00 D with 8 degrees of pantoscopic tilt. What (un-tilted) spherocylindrical lens is equivalent to this tilted spherical lens?

Back 53

Now we will try the tilted lens problem with a plus lens. For a plus lens, the equation for sphere power gives the new sphere power of the lens when written in PLUS CYLINDER FORM.

So we would have

Fs = 5.00 (1 + sin^2 (8)/3)
Fs = 5.03 D

So this is the sphere power when written in plus cylinder form. IT IS NOT THE SPHERE POWER IF WRITTEN IN MINUS CYLINDER FORM.

Also, we can find the amount of radial astigmatism (which is the difference in powers between the two principal meridians). We have

Fc = 5.00 tan^2(8)
Fc = 0.10 D.

This would be the cylinder power if written in PLUS CYL form. So in plus cyl form, the equivalent lens is

+5.03 + 0.10 x 180

Again, the axis is 180. If we write this as a power cross, we have

F = 5.03 (horizontal)
F = 5.13 (vertical)

In minus cyl form, this corresponds to

5.13 - 0.10 x 90.

Front 54

We all know how to find the far point of an ametropic eye. As the eye rotates, the far point will also rotate, and the spherical arc the far point traces out is called the FAR POINT SPHERE. Assume a patient wears a -8.00 D lens at a vertex distance of 14 mm. Take the center of rotation of the eye to be 14 mm behind the cornea. What is the curvature of the far point sphere? What is the RADIUS of curvature of the far point sphere?

Back 54

The lens is -8.00 D, so the far point is located at 1/8 = 12.5 cm in front of the lens. So this is a distance of 13.9 cm from the cornea. Similarly, this is a total of

x = 12.5 + 1.4 + 1.4 = 15.3 cm

from the center of rotation of the eye. As the eye rotates, the far point traces out a sphere (well, not a whole sphere, since your eye can't rotate fully) with a radius of 15.3 cm.

As a result, one would say that the radius of curvature of the far point sphere is 15.3 cm. Similarly, the curvature of the far point sphere is 1/0.153 = 6.54 (units are inverse meters).

The radius of curvature of the far point sphere can be compared with the radius of curvature of the Petzval surface (for example, the review book by Bennett et al has a problem similar to this). Recall that the radius of curvature of the Petzval surface is just 1/K, with

K = F/n.

Front 55

A frame measures 54 x 18. The distance PD is 66 mm, and the near PD is 62 mm. What is the total seg inset? What is the seg inset?

Back 55

First, let's find the total seg inset. This is the distance between the center of the lens (geometrically) and the seg location. First, we note that the segs should be placed symmetrically, so they are each 62/2 = 31 mm from the 'nose' (excuse my lax terminology here; I will refer to the 'nose' as the point in the very center of the frames, where your nose would be. Hopefully you see what I am talking about).

But we also know that the center (geometrical) of each lens is a total of 54/2 + 18/2 = 36 mm from the 'nose'. Can you see this from your picture?

The total seg inset is just the difference between these distances. So the total seg inset is simply 36-31 = 5 mm inward in each eye.

Now what about the 'seg inset'? Well, we already found where the seg is located. Now we just want to know how far that point is from the optical center of the distance lens optical center. So we need to know where the distance lens optical center is placed. Again, to place them symmetrically before each eye, they must each be at 66/2 = 33 mm from the nose. This is therefore 36-33 = 3 mm from the center (geometric) of the lens.

So the 'seg inset' is simply the difference: 5-3 = 2 mm.

This will always correspond to (Distance PD - Near PD)/2, of course. You can probably see this algebraically from what we've done above.

Front 56

A patient wears a rigid contact lens, -6.00 D, which completely corrects his ametropia. The base curve of the contact is 0.5 mm steeper than the corneal curvature (you can assume there is no corneal astigmatism). What is the actual refractive state of the eye?

Back 56

The tear lens, together with the contact lens, completely corrects the ametropia. So we simply need the power of the tear lens and the correcting lens together. We already know the latter (it is -6.00 D).

The tear lens is positive, because the base curve is steeper than the cornea. This means that the front surface of the tear lens has a higher absolute power (in magnitude) than the back surface of the tear lens. Recall that the front of the tear lens is positive and the back is negative. We can use our approximate forumla to calculate the power of the tear lens. There are 0.25 D for every 0.05 mm difference. So the power of the tear lens, Ft, is approximately

Ft = 0.25/0.05 * 0.5 = +2.50 D.

So the overall correction power is

Fcor = +2.50 - 6.00 = -3.50 D.

So the refractive state of the eye is 3.50 D myopic.

Regardless of the clinical relevance of the specific numbers chosen for this problem, do you see the idea? Could you do the problem if keratometry readings were given, and there was some corneal astigmatism? It would be identical, mathematically, but you would have to do each meridian separately.

Front 57

A patient's distance visual acuity is 20/200. What add would you expect them to need to read standard size print?

Back 57

Recall Kestenbaum's rule: The reciprocal of the distance acuity gives you the expected add for standard size print.

For this example: 200/20= +10 add

Front 58

T/F: A person can have 20/20 vision and be considered legally blind.

Back 58

True. Legal blindness can be based on visual acuity or visual field. If a patient has less than 20 degrees of visual field diameter in the better eye, they are considered legally blind.

Front 59

A pair of +8 high powered readers would required how much BI prism to allow the patient to see single?

Back 59

10 prism diopters BI prism over each eye

Remember: The amount of prism over each eye= The amount of plus power + 2

Front 60

A telescope is marked with 5x20. What is the magnification and the diameter of the exit pupil in mm?

Back 60

Magnification: 5x

Diameter of the exit pupil= 20/5=4mm

The 20 represents the size of the objective lens.

Front 61

If a patients best corrected acuity is 20/100 in both eyes and the goal is to be able to watch television (say this requires approximately 20/50 acuity). What is the anticipated magnification this patient will need in a telescope to achieve his/her goal?

Back 61

2x

100/50- Divide the denominator of the actual acuity by the denominator of the goal acuity.

Front 62

A simple eye model has axial length 18 mm, a thin lens of power +62.00 D representing the lens/cornea, and a vitreous with index n=1.3. What is the far point for this reduced eye?

Back 62

We place an object at the retina and find the corresponding image. We have
L = -1.3/0.018 = -72.2 D.
L' = F + L = 62.00 - 72.2 = -10.2 D.
The far point is 10.2 D behind the cornea, which corresponds to 9.8 cm. Clearly, the eye is hyperopic.

Front 63

A real object is imaged through an optical system. The image formed is virtual image. Is it inverted or upright?

Back 63

Since magnification is

m = U/V = L/L'

and L (or U) and L' (or V) have the same signs (negative), the image is upright.

Front 64

When a telescope is used to view a distant object, what telescope component typically serves as the aperture stop? What is the entrance pupil? the exit pupil?

Back 64

The objective lens of a telescope is typically the aperture stop. The entrance pupil is also the objective lens, while the exit pupil is the image of the objective lens through the ocular lens.

Front 65

A patient requires a prescription of +5.00 D in contacts. What power is needed at a vertex distance of 14 mm?

Back 65

The far point of the eye must correspond with the secondary focal point of the correcting lens. The far point of the eye can be found with knowledge of the contact prescription. It is located at
l' = -1/5.00 = -20 cm
behind the cornea. A lens at a vertex distance of 14 mm would require a lens with focal length
f=20+1.4 = 21.4 cm.
This corresponds to a power of
F = 1/0.214 = 4.67 D.

Front 66

Under what conditions do the nodal points and principal points of a multiple lens system coincide?

Back 66

The nodal points are displaced from the principal planes by a distance equal to the sum of the focal lengths of the system. When the index of refraction on the object space side of the system is the same as that on the image space side, the focal lengths are equal but opposite and the total displacement is zero, meaning the principle and nodal points coincide.

Front 67

What is the magnification of a Keplerian telescope with objective lens +20.00 D and and ocular lens +50.00 D?

Back 67

M = -50.00 / 20.00 = -2.5.
The image would be inverted.

Front 68

What are the two most important aberrations to control for when designing ophthalmic lenses?

Back 68

Distortion and radial astigmatism are the most critical aberrations to consider in ophthalmic lens design. Spherical aberration and coma are typically not critical because they are effectively eliminated by small aperture sizes (e.g. the pupil).

Front 69

What happens to magnitude of spherical aberration when the diameter of the objective lens of an afocal telescope is doubled?

Back 69

Spherical aberration scales with the square of the entrance pupil diameter. If the objective lens doubles in diameter, so too will the entrance pupil for an afocal telescope, where the objective lens serves as the aperture stop. If the entrance pupil diameter doubles, the spherical aberration increases by a factor of 4.

Front 70

Which monochromatic aberration occurs for both on and off axis point sources?

Back 70

Spherical aberration occurs for on and off axis point sources. Coma occurs only for off axis point sources. Distortion, curvature of field, and radial astigmatism also occur for sources displaced from the optical axis.

Front 71

Which of the following materials gives rise to the most chromatic aberration (all else being equal)?

Crown glass
Polycarbonate
High index plastic
Trivex

Back 71

The Abbe values are given approximately below:

Crown glass: 59
Polycarbonate: 30
High index plastic: 30-36
Trivex: 43

As a result, chromatic aberration in descending order is:

Polycarbonate (highest)
High index plastic
Trivex
Crown glass (lowest)

Front 72

An object immersed in water is located 15 cm below the surface of the water. To a viewer above the surface of the water (in air), how far does the object appear to be from the surface? To ask it another way, by what amount does the object appear displaced from its actual position?

Back 72

This is a simple apparent depth problem. We treat the immersed object as a real optical object and find the conjugate image.

L = -1.3/0.15 = -8.6
L' = F + L = 0 - 8.6 = - 8.6
l' = - 1/0.086 = -11. 5 cm

The image appears 11.5 cm to the water side of the interface. Equivalently, the object appears to be displaced by 15-11.5 = 3.5 cm from its actual location.

Front 73

What is the back vertex power of a biconvex lens made of glass (n=1.5) with a center thickness of 2 mm, a front radius of curvature of 18 cm and a back radius of curvature of 22 cm?

Back 73

First we will find the surface powers. The front surface power is
F1 = (1.5-1)/ 0.18 = 2.78 D.
F2 = 0.5 / 0.22 = 2.27 D.

Since it is biconvex, both surfaces are convex relative to the lowest index of refraction (air), so both surfaces are positive. The back vertex power is then given by:

First, note that t/n = 0.002/1.5 = 0.00133

Fv = 2.27 + 2.78 / (1- 0.00133* 2.78 ) = 5.06 D.

Note that because the lens is so thin and the surface powers relatively small, the back vertex power is approximately the sum of the surface powers.

Front 74

A thin glass prism has an apex angle of 8 degrees. What is the prism power of this prism? How far will a beam of light passing through this prism be displaced on a wall 2 m away from the prism?

Back 74

We first must find the deviation angle.

d = (n-1) A = 0.5 (8) = 4 degrees.

Now we convert this to a prism power, Z.
Specifically, we want to know how far (in cm)
this beam would be displaced on a wall 1 m away.

Z = 100 tan(4) = 7 prism diopters.

This corresponds to 14 cm on a wall 2 m away.

Front 75

The symmetry points are the object and image locations that correspond to a lateral magnification of m=-1. For a lens of power F separating two materials (n1 and n2), find the symmetry points.

Back 75

We know that

m = L/L' = -1

So we have L = - L'. We therefore have

L' = F + L = F - L'
L' = F/2.

So the image is located at

l' = 2 n1 / F = 2 f2

which is twice the secondary focal distance (f2). Similarly, the object is located at twice the primary focal distance from the lens.

**Note that if n1=n2, the symmetry points are 4f apart, where f is the focal length of the system.**

Front 76

What is the focal ratio (f-number) of a typical human eye?

Back 76

The f-number is the ratio of the focal length to the diameter of the entrance pupil. For a typical eye, the power is about +60.00, so the focal length is 16.7 mm. The entrance pupil diameter will depend on lighting conditions, but we can approximate it as somewhere around 3 mm. Hence, the f-number is given approximately by f/5.6.

In reality, it typically varies from about f/8.3 (bright conditions) to about f/2.1 (dark conditions).

Front 77

What prism is induced by decentering the following lens by 4 mm downward?

+5.00 - 2.00 x 30

Back 77

We would like to find the approximate power in the vertical meridian. To do so, we use

F90 = +5.00 - 2.00 sin(90-30) sin(90-30)
F90 = +5.00 - 1.50 = +3.50 D.

This corresponds to a prism power of

Z = (0.4) 3.50 = 1.4 prism diopters.

Since it is a plus lens in this meridian, decentering down corresponds to base down.

Front 78

A person views an object through a prism that is 7 prism diopters base down. The object is located 3 m from the prism, and the prism is 18 cm from the eye's center of rotation. By what angle must the eye rotate to see the object?

Back 78

This problem uses the prism effectivity equation.

Ze = Z / (1 - C L)

We have
Z = 7 prism diopters
C = 0.18 m

*** Note that the center of rotation is about 14 mm behind the cornea. Therefore the problem could have stated that the prism was located 16.6 cm in front of the cornea. This would be the exact same problem. ***

L = - 1/3 = - 0.33 D.

Plugging these numbers in, we get

Ze = 7/(1+0.18*(0.33))
Ze = 6.6 prism diopters.

But the problem asks for the answer in terms of an angle, not in terms of prism diopters. In terms of an angle, this is simply

tan(x) = 6.6 / 100;
x = 3.8 degrees.

Front 79

Keratometry is based on the fact that the cornea acts as a mirror. Assume you place a small object, 5.5 cm tall, about 8 cm from the cornea. If the reflected, upright image is measured to be 2.3 mm, what is the radius of curvature for the cornea? What would the power readout from the keratometer be?

Back 79

The cornea acts as a convex mirror of unknown power. We want to find that power and then estimate the radius of curvature. This is just a geometric optics mirror problem.

m = 0.23 / 5.5 = 0.042.

But we also know that

m = L/L';
L = 0.042 L';

Since L = - 1/0.08 = - 12.50 D, we know that

L' = -12.50 / 0.042 = -299 D.

So

F = L' - L = - 286.5 D.

This corresponds to a radius of curvature of

286.5 = 2/r
r = 7.0 mm

Keep in mind that the mirror is convex. Some people use a sign convention for mirrors where the positive direction is to the left, in which case we might call this number negative.

A keratometer would NOT read off this value. Instead, it would treat the cornea as an SSRI separating air and n=1.3375 with radius of curvature 7.0 mm. The power readout would be

F = 0.3375 / 0.007 = 48.2 D.

Front 80

A real object 3 cm in height is placed 25 cm from a thin lens. The image is real and is measured to be 6 cm in height. Is the image upright or inverted? What is the power of the lens? Where is the image located?

Back 80

First, we know that the image is inverted since the object is real (L negative) and the image is real (L' positive). We also know the magnitude of m, so

m = -6/3 = -2.

But also m = L/L', so

L = - 2 L'

Therefore we have

F = L' - L = -L/2 - L = - 3 L/2.

Since L = -1/.25 = -4.00 D, we have

F = +6.00 D.

The image is located at L' = 2.00, which corresponds to

l' = 50 cm.

Front 81

An object is embedded at the center of a glass sphere of radius 10 cm. How far from the edge of the sphere does the object appear to a viewer outside the sphere (in air)?

Back 81

This is an apparent depth problem, but now the interface will have a nonzero power. Using the power formula for an SSRI, the power of the interface is

F = 0.5 / 0.10 = 5.00 D

and it is positive because the interface is convex relative to the lowest index of refraction. Now we must find the image of the embedded object.

L = - 1.5/0.10 = - 15.00.
L'= +5.00 - 15.00 = -10.00.
l' = - 1/10 = - 10 cm

So the image is 10 cm to the glass side of the interface. Thus, the object appears to be embedded exactly at the center of the sphere, which is where it is physically located! Note, however, that it will appear larger by a factor of m = L/L' = 1.5.

Front 82

A 3 cm real object is located (in air) 35 cm from a - 2.00 D lens. Water (n=1.3) is on the other side of the lens. Where is the image located, and what is its height?

Back 82

This is a standard vergence problem. Note that this type of problem is essentially identical to finding the far point in a model eye (in which case the object is placed at the retina). One can imagine many other cases, as well.

L = -1/.35 = -2.86 D
L' = F + L = -2.00 - 2.86 = -4.86 D
l' = 1.3/L' = -26.8 cm
m = L/L' = 0.59

The image is virtual, upright, and located 26.8 cm to the object side of the lens. The height of the image is simply

h = m x 3 = 0.59 x 3 = 1.76 cm.

Front 83

A stand magnifier is used to see an object located 8 cm from the magnifier. The power of the magnifier is +10.00 D. If the patient is presbyopic, what add must you prescribe if he/she plans to sit 25 cm from the magnifier lens while reading?

Back 83

To do this problem, we first find the vergence of light leaving the magnifier. The object being magnified is real, so we have

L = -1/0.08 = -12.5 D
L' = 10.00 - 12.5 = -2.50 D.

However, we are not interested in the light leaving the lens, but rather in the light striking the eye. We therefore need the vergence 25 cm downstream.

To find this, draw a picture of the -2.50 D wavefront traveling, for example, left to right. The center of curvature of the wavefront is located 1/2.5 = 40 cm to the left of the wavefront. If we want the vergence 25 cm downstream (to the right), we want the vergence at a distance of 25+40 = 65 cm from the center of curvature. Therefore, we have

V = -1/.65 = -1.53 D

Thus, the patient needs an add of about +1.5 D to use this magnifier.

Here is another way to think about the last step. As we saw, the stand magnifier forms an image at L'= - 2.50 D, which is 40 cm in front of the magnifier. This is then a total of 65 cm from the patient's eye. So the add needed would be 1/0.65 = 1.53 D.

Front 84

The base curve of a hard contact lens (n=1.5) is 10 mm. When this lens is placed on the cornea, a tear lens is formed. What is the power of the tear lens in the exploded system, given that the radius of curvature of the central cornea is measured to be 8 mm? By 'exploded system' I mean the system in which the tear lens is surrounded by air on both sides. (This would provide a very poor fit!)

Back 84

First, remember that when the cornea is steeper (smaller r) than the base curve, you will always have a minus lens. This would mean you should add plus power to the prescription to compensate (but the question doesn't ask for that).

Typically with tear lenses, you can assume that a thin layer of air separates the tear lens from both the cornea and the contact. This is an optics trick known as an exploded system. Therefore, we only need to know the index of the tears (n=1.3366) and treat the tear lens as two SSRI's.

The power of the front surface of the tear lens is

F1 = (1.336-1.00) / 0.01 = + 33.0 D

This surface is positive, because the surface is convex relative to the lowest index of refraction (air).

The back surface has a power of

F2 = (1.336-1.00) / 0.008 = - 41.25 D.

This lens is negative, because the surface is concave relative to the lowest index of refraction.

The total power of the tear lens is therefore

F = -8.25 D.

This is quite significant because of the numbers chosen. Note that we could also estimate the magnitude of this power using the box on page 525. It would give 40 * 0.25 = -10 D, which is not too far off.

Front 85

A contact lens has a base curve that is steeper than the central curvature of the cornea. Would you add plus or minus power to the prescription to compensate.

Back 85

If the base curve is steeper, a plus tear lens is formed (in the exploded system). As a result, one would need to compensate by adding minus power to the prescription.

You can estimate the power of the tear lens by multiplying the difference in radii of curvatures (measured in mm) by 0.25 / 0.05.

Alternatively, you can calculate the power by treating the cornea as two SSRI's (that is, the tear lens is just a thin lens in air).

Front 86

A tear lens is formed between rigid contacts and the cornea. Assuming that a thin layer of air separates the tear lens from both the cornea, on one side, and the contact, on the other, what are the signs of the surface powers of the tear lens?

Back 86

This is a good conceptual exercise. The tear lens is just two SSRI's placed back to back. The first SSRI, formed between air and the tears, has a radius determined by the base curve of the contact lens. It is a convex surface relative to the lowest n, so it is positive.

The second surface is formed between the tears and air and has a radius of curvature determined by the central corneal curvature. It is concave relative to the lowest n, so it is always negative.

So you are adding a positive number and a negative number. If the base curve is steeper (smaller r), the tear lens will always be positive. Then you must compensate by adding minus power to the prescription.

On the other hand, if the cornea is steeper (smaller r), the tear lens will always be negative. Then you must compensate by adding plus power to the prescription.

Front 87

What does an orthoscopic doublet help correct?

Back 87

An orthoscopic doublet is a combination of lenses that eliminates distortion. It is typically found in eyepieces of optical instruments such as high-powered telescopes.

Front 88

A distant point object is imaged through the lens given below. Where is the vertical line focus? Where would a screen be placed--relative to the lens--to achieve the highest quality image?

The lens is +4.00 - 5.00 x 180.

Back 88

The lens has a power cross that is +4.00 D (horizontal) and -1.00 D (vertical). The horizontal line focus comes from the power in the vertical meridian, and vice versa. So the line images of a distant object are

horizontal line: 1/1 = -100 cm
(in front of the lens).

vertical line: 1/4 = 25 cm behind the lens.

The circle of least confusion, which indicates the position of the best image, is found at +1.5 D (the average of the powers in the two meridians). This corresponds to 1/1.5 = 67 cm behind the lens.

So the screen should be placed 67 cm behind the lens.

Front 89

How much cyl power is induced when a +3.00 D lens is tilted 8 degrees?

Back 89

The induced cyl power is given by

Fc = F tan(x) tan(x)
Fc = 3.00 tan(8) tan(8) = 0.059 D

This is also the radial astigmatism induced by rays striking a lens at 8 degrees.

Front 90

A person stands 0.5 m from a concave mirror with radius of curvature 2 m. Where is this person's image? Is it magnified?

Back 90

First, the power of the mirror is

F = 2/2 = +1.00 D.

The mirror is concave, so we know it is positive.

The person is 0.5 m away, so we have

L = -1/0.5 = -2.00 D
L' = -2.00 + 1.00
L' = -1.00 D

So the image is virtual and located 1 D from the mirror. This means the image is located 1 m behind the mirror.

m = L/L' = 2

So the image is upright and magnified.

Front 91

A patient's cornea is measured to have a central curvature of 6 mm. If this cornea reflects light (acts as a mirror), what is its power? If this cornea refracts light (acts as an SSRI), what is its power?

Back 91

The power of a mirror is

F = -2n/r = -2/0.006 = -333.3 D

We know it is negative because it is a convex mirror. Note that keratometry is based on the fact that the cornea acts as a mirror.

The power of an SSRI is

F = (n2-n1)/r = (1.33-1.00)/0.006 = +55.0 D.

We have taken the n of the cornea to be about 1.33. Relative to the lowest index of refraction (air), this is a convex surface, so it would be positive.

Front 92

A thick glass lens (thickness 5 mm) has a front surface power of -5.00 D and a back surface power of -3.00 D. What is the equivalent power of the lens?

Back 92

Equivalent power is given by

Fe = F1 + F2 - t/n F1 F2
Fe = -5.00 - 3.00 - (0.005/1.5) 15.00
Fe = -8.05 D

One can see that this power is very close to the sum of the surface powers. This is because the thickness (5 mm) is relatively small.

Front 93

A lens 15 mm thick has front surface power +15.00 D and back surface power -2.00 D. If the lens is made of glass (n=1.5), find the equivalent focal length and the back focal length of the lens.

Back 93

The equivalent power is

Fe = +15.00 - 2.00 + (0.015/1.5) 30.00
Fe = +13.3 D.

So the equivalent focal length of the system is

fe = 1/Fe = 1/13.3 = 7.52 cm.

The back vertex power is

Fv = F2 + F1/(1-(t/n) F1)
Fv = -2.00 + 15/(1-0.15)
Fv = +15.65 D

As a result, the back focal length is

fb = 1/Fv = 1/15.65 = 6.40 cm.

As a result, note that the secondary principle plane must be located 7.52-6.40 = 1.12 cm to the left (front-side) of the back surface of the lens.

The matches what we would get from Eq. 13.39 in the textbook.

Note that this is one way you can always find the principle plane locations WITHOUT having to memorize this formula. First find the equivalent focal length, then the back focal length. The location of the secondary principle plane will be the difference in the two. A similar logic applies to the primary principle plane and the front focal length.

Front 94

A thick lens (n=1.5) has front surface power F=+20.00 and a back surface power of +5.00 D. The lens is 10 mm thick at the center. If an object is placed 40 cm from this lens, where is the image formed?

Back 94

The equivalent power of the system is

Fe = F1 + F2 - (t/n) F1 F2
Fe = 20.00 + 5.00 - 0.67 = +24.33 D.

The back vertex power is

Fv = F2 + F1/(1-(t/n) F1)
Fv = +28.08 D

The front neutralizing power is
Fn = F1 + F2/(1-(t/n) F2)
Fn = +25.17 D

Therefore, our focal lengths are
fe = 1/24.33 = 4.11 cm
fv = 1/28.08 = 3.56 cm
fn = 1/25.17 = 3.97 cm

This means that the primary principle plane is located 4.11-3.97 = 0.14 cm behind the front surface. The secondary principle plane is located 4.11 - 3.56 = 0.55 cm in front of the back surface. We could also have gotten these directly from Equations 13.39.

The vergence of light striking the first plane is

L = -1/(0.40+0.0014) = -2.49 D

Therefore, the light leaving the 2nd plane is

L' = Fe + L = 24.33 - 2.49 = +21.84 D.

So the image is located at l' = 1/21.84 = 4.58 cm behind the second principle plane. This is a distance of

x = 4.58 - 0.55 = 4.03 cm behind the back surface of the lens.

Front 95

A real object is located 25 cm from the primary principal plane of a thick lens. The image of that object is located 10 cm in front of the secondary principal plane. What is the lateral magnification of the image? If the object were 10 cm tall, how tall is the image?

Back 95

The object corresponds to a vergence of L at the primary principal plane. The value of L is

L = -1/0.25 = -4.00 D.

The image corresponds to a vergence of

L' = -10.00 D

at the secondary principle plane. So the magnification is given by

m = L/L' = 0.40

It is upright and only 40percent of the object height. So the image of a 10 cm object would be 4 cm tall.

Front 96

How much prism is induced when the following lens is decentered 5 mm out?

-3.00 -1.00 x 135

Back 96

First we must find the approximate power in the 180 meridian. We have

F180 = -3.00 -1.00 sin(180-135) sin(180-135)
F180 = -3.50 D.

The decentering is 5 mm, so the total prism induced is

Z = 3.50 * 0.5 = 1.75 prism diopters base in.

For a minus lens, if the decentering is out, the base is IN.

Front 97

How much cyl power is induced by tilting a +15.00 D lens by 8 degrees?

Back 97

This will induce

Fc = 15.00 tan(8) tan(8)
Fc = 0.3 D of cyl power.

Front 98

Light is incident from the air side of an air-water interface. If the light strikes at an angle of incidence of 25 degrees, what is the angle of refraction? What is the critical angle for total internal reflection in the air?

Back 98

We use Snell's law directly.

1 sin(25) = 1.3 sin(r)
r = 19 degrees.

To find the critical angle, c, we always find the angle of incidence that creates an angle of refraction of 90 degrees. This gives

sin(c) = n2/n1 = 1.3/1 = 1.3

There is no solution in this case. This is a general result. Total internal reflection can only occur when light travels from a higher index medium to a lower index medium.

Front 99

Does light travel faster in water or glass?

Back 99

The speed of light is given by a constant (c) divided by the index of refraction of the material.

v = c/n

Since glass has a higher index than water, v is smaller for glass than for water. Thus, light travels faster in water.

Note that the frequency of light is the same in all materials! Only the velocity and wavelength of light depend on medium!

Front 100

Which of the following has the smallest frequency?

UV light
IR light
microwaves
Visible light

Back 100

In terms of frequency, the ordering is (from lowest to highest):

microwaves
IR light
visible light
UV light

In other words, UV light has the highest frequency (and therefore the highest energy), but the shortest wavelength.

Front 101

Which type of UV light is most damaging to the eye?

Back 101

UV-C is technically the most damaging to the eye. However, UV-C light from the sun is blocked by the Earth's atmosphere, so optometrists are typically most concerned with preventing damage from UV-B light.

Front 102

Incandescent light source A has a color temperature of 5000 K, while incandescent light source B has a color temperature of 2700 K. What is the ratio of the wavelengths of these sources?

Back 102

Color temperature is inversely related to wavelength. The idea is based on the Blackbody radiation spectrum.

We have

lambdaA = constant / TA
lambdaB = constant / TB

So

lambdaA/lambdaB = TB/TA = 0.54

So source A has has a shorter wavelength.

Front 103

A patient is asked to move his head back and forth while observing the speckle pattern formed on a wall from the reflections of a coherent light source. If the patient is myopic, how will the motion appear?

Back 103

The myopic patient will see 'against motion'. This is an example of the speckle effect.

Front 104

Two objects are imaged through a camera with aperture of diameter 2 mm. If the objects are 3 meters from the camera, estimate the minimum separation between objects when they are just resolvable.

Back 104

We use Rayleigh's criteria, and we take lambda = 555 nm, which is in the center of the visible spectrum. Rayleigh's criteria will give us the angular separation.

d sin(theta) = 1.22 lambda
0.002 sin(theta) = 1.22 * 555 x 10^(-9)
theta = 0.019 degrees

So the angular separation of the just resolvable objects is 0.019 degrees. This corresponds to a separation distance of

tan 0.019 = x/3
x = 1 mm.

Notice that you can (if you prefer) combine these two steps into one step for very small angles. Then you have the equation:

x = 1.22 lambda D / d
x = 1.22 * 555 x 10^(-9) * 3 / 0.002
x = 1 mm.

Front 105

Two linearly polarized waves of equal frequency are out of phase by 90 degrees and have equal amplitudes. When these waves are added together, what is the polarization of the resulting wave?

Back 105

The resulting wave is circularly polarized. If the amplitudes had been different, the wave would be elliptically polarized.

Front 106

A glass lens (n=1.5) has center thickness 10 mm, front surface power +6.00 D, and a back vertex power of -2.00 D. What is the equivalent focal length of this lens?

Back 106

To find equivalent focal length, we need the equivalent power. To get equivalent power, we need F1 and F2--the surface powers--but we are only given F1. But we can find F2 from the equation for back vertex power

Fv = F2 + F1/(1-(t/n) F1)
so
F2 = Fv - F1/(1-(t/n) F1)
F2 = - 8.25 D

Note that this part could be a question on its own.

Now the equivalent power is

Fe = F1 + F2 - t/n F1 F2
Fe = +6.00 - 8.25 + (0.01/1.5)*49.5
Fe = - 1.92 D

Again, this could be a question on its own.

Finally, the equivalent focal length is

fe = 1/Fe = 52 cm.

Front 107

What is the critical angle for total internal reflection for light traveling from water to air? What about from air to water?

Back 107

We can find the critical angle, which we will call x, using Snell's law with the angle of refraction set to 90 degrees. We have

n1 sin x = n2 sin 90
sin x = n2/n1

For a critical angle to exist, we must be traveling from a material with higher index of refraction to one with lower index of refraction. There is no critical angle for light striking from the air side of an air-water interface. However, if light strikes from the water side, we have

sin x = 1/1.3
x = 50.3 degrees.

Front 108

You are wearing polarized sunglasses while driving on a wet road. Assume sunlight is incident on the road at a broad range of incident angles. For which incident angle will the reflected light (glare) appear the lowest?

Back 108

Light that is incident at Brewster's angle will be plane polarized and therefore completely blocked by the polarized sunglasses. So for this question, we simply want to find Brewster's angle, which we will call x.

We have

tan x = n2/n1
tan x = 1.3 / 1
x = 52 degrees.

Front 109

Light from a green (532 nm) laser pointer strikes a thin strand of hair. The image on a screen 1 m away shows a central diffraction peak of width 1 mm. What is the width of the hair?

Back 109

This problem is essentially a single slit diffration problem, with the hair serving as the 'aperture'. If you find this confusing, you should quickly review the single slit diffraction section of physical optics in the text.

For single slit diffraction, we have

d sin theta = m lambda

This equation will give us the angular width of (half of) the central diffraction pattern if we take m=1. However, the problem gives the spatial width (1 mm on a wall 1 m away), not the angular width. So we must use some geometry. The angular width of the pattern is

tan theta = 0.0005/1
theta = 0.029 degrees

Notice that we used 0.5 mm rather than 1 mm, because the equation we will use (the first one, above) only gives the angular width of half the peak. In other words, the angle from the center of the peak to the edge of the peak is 0.029 degrees (or 0.5 mm on a wall 1 m away).

Now we simply plug in. We have

d sin theta = 1 lambda
d sin(0.029) = 532 x 10^(-9)
d = 1.1 mm.

Here is another approach. We can write an approximate equation to do this problem in one step. In particular, for small angles y, we can write an approximation:

tan y = sin y = y.

Using this approximation, we can write an equation for the full width of the diffraction pattern, x, as

x = 2 D lambda / d

where D is the distance from aperture to the screen, and d is the diameter of the aperture. If you like, you can memorize this equation and start from here. For this problem, we would have

0.001 = 2 (1) 532 x 10^(-9) / d
d = 1.1 mm.

Front 110

A patient requires -3.25 D spectacles at 14 mm. What prescription would provide equivalent dioptric power in contacts?

Back 110

The patient is a myope, and we know that the far point of the eye is located at 1/3.25 = 30.8 cm in front of the spectacle plane. This is therefore a total of 32.2 cm from the cornea. So the prescription that should sit at the cornea would have a power of

F = -1/0.322 = -3.11 D,

and of course it would be still be negative.