EXAM 1 - Self Assessment Questions

Question 1

The eyes of a rod monochromat are missing most of their cones, and vision is
dominated by rods. A. Describe the symptoms you expect a patient who is a
rod monochromat to manifest. B. What color sunglasses would you recommend
for a rod monochromat?

Answer 1

A. Photophobia, squinting, poor daytime vision, reduced visual acuity
(_20/200), nystagmus, poor fixation.
B. Long-pass sunglasses (red) are helpful because long wavelength light is
least effective at bleaching rhodopsin

Question 2

Photopic stimuli of 507 and 555 nm are placed side by side. An observer is
asked to adjust the intensities of these stimuli so that they are equally bright.
Their intensities are then reduced by the same amount such that they are
detected by the scotopic system. Under these scotopic conditions, which stimulus
appears brightest?

Answer 2

Locate these wavelengths on the photopic and scotopic spectral sensitivity
functions. The 507- and 555-nm stimuli are equally bright under photopic
conditions only when the 507-nm stimulus has more energy. (This is because
the photopic system is less sensitive to 507 nm than to 555 nm.) If the intensities
of these two equally bright stimuli are decreased by equal amounts, the
507-nm stimulus continues to contain more energy. Under scotopic conditions,
the 507-nm stimulus appears brighter because it has more energy and
because the scotopic system is more sensitive to 507 nm than to 555 nm.

Question 3

Which wavelength is most effective at bleaching rhodopsin? Explain.

Answer 3

The scotopic visual system is most sensitive to 507 nm because this wavelength
is most effective at bleaching rhodopsin.

Question 4

After 20 minutes of dark adaptation, what is the color of
the 610-nm stimulus at threshold? B. Answer the same question for the 465-
nm stimulus.

Answer 4

A: A. Red, because threshold detection of 610 nm, after 20 minutes of dark
adaptation, is mediated by the cones.
B. The 465-nm stimulus has no color after

Question 5

After 1 hour of dark adaptation, what is the difference in
sensitivity between the rods and cones for a stimulus of 465 nm? B. Answer
the same question for the 610-nm stimulus.

Answer 5

A. About 4 log units (photochromatic interval).

B. About 0.5 log unit (photochromatic interval).

Question 6

A. The cones are exposed to a bright light source that bleaches much of their
photopigment. After 3 minutes in the dark, what percentage of the bleached
cone photopigment has recovered? B. What percentage of the rod photopigment
has recovered at the rod–cone break for 610 nm (see Fig. 3–12)?

Answer 6

6. A. After 1.5 minutes, 50% of the photopigment has regenerated and 50%
   remains bleached. Over the next 1.5 minutes, 50% of the remaining photopigment
   recovers. Consequently, the total amount of cone photopigment
   that recovers after 3 minutes is 50% plus 25%, or 75%.
   B. The rod–cone break occurs at about 35 minutes. The half-life for rhodopsin
   regeneration is 5 minutes. Therefore, after 5 minutes, 50% has regenerated;
   and after 10 minutes, 75% has regenerated. Carrying out this calculation to
   35 minutes shows that at this time 99.22% of the rhodopsin has regenerated.

Question 7

Refer to Fig. 3–10. A. What is the rod threshold after approximately 11 minutes
of dark adaptation for a large 420-nm stimulus? B. For the same 420-nm
stimulus, what is the cone threshold after 20 minutes of dark adaptation?
C. What is the photochromatic interval for 420 nm?

Answer 7

A. Early during dark adaptation (see Fig. 3–10B), the rod function has a vertical
slope; this indicates an infinitely high threshold. After 11 minutes of
dark adaptation, rods remain saturated. Only after at least 12 minutes can
rods detect the stimulus.
B. After 20 minutes of dark adaptation, the cone threshold is about 1.75 log
units.
C. The difference between the cone and rod plateaus is about 3 log units.

Question 8

A. Two patches of light are adjacent to each other. The conditions are scotopic.
One patch emits light of 507 nm and the other emits light of 620 nm.
Both patches produce 40 quanta of light. Which patch is brighter? B. A patch
of 507 nm and a patch of 620 nm each bleach 30 rhodopsin molecules. Which
patch is brighter?

Answer 8

A. The 507-nm patch is brighter because it results in more quantal absorptions
by rhodopsin.
B. Both patches have the same brightness because they bleach the same
number of rhodopsin molecules.

Question 9

A light source and filter combination produces 1000 lumens. The filter is
monochromatic with peak transmission at 600 nm. How many watts are
transmitted by the filter?

Answer 9

number of watts) (visual efficiency) (680 lumens/watt) _ number of lumens
Solve for number of watts:
(number of watts) (0.62) (680 lumens/W) _ 1000 lumens

Question 10

Q1: A source produces 10 W at 500 nm, 5 W at 550 nm, and 20 W at 650 nm.
How many lumens are produced?

Q2: give the number of scotopic lumens produced

Answer 10

A1: Use the formula from Problem 4–1 to calculate the number of lumens for
each wavelength. Abney’s law allows the addition of these amounts to determine
the total number of lumens.
500 nm: (10 W) (0.35) (680 lumens/W) _ 2380 lumens
550 nm: (5 W) (1.00) (680 lumens/W) _ 3400 lumens
650 nm: (20 W) (0.10) (680 lumens/W) _ 1360 lumens
Total _ 7140 lumens

A2: Observe V_(_). There are 680 scotopic lumens/W at 555 nm. The number
of scotopic lumens per watt at other wavelengths is determined by a
proportionality factor.
500 nm: (10 W) (1.0/0.4) (680 scotopic lumens/W) _ 17,000 scotopic lumens
550 nm: (5 W) (0.50/0.40) (680 scotopic lumens/W) _ 4250 scotopic lumens
650 nm: (20 W) (0) (680 scotopic lumens/W) _ 0 scotopic lumens
Total _ 21,250 scotopic lumens

Question 11

An illuminance probe is used to measure the lighting conditions in a
classroom. The reading obtained is 70 foot-candles. The probe measures
3 _ 3 cm. How many lumens are incident on the surface of the probe?

Answer 11

Seventy foot-candles fall on a probe that measures 3 _ 3 cm. The area of this
probe is 0.0097 ft2. Therefore,
(70 lumens/ft2) (0.0097 ft2) _ 0.67 lumens.

Question 12

A device measures irradiance. You would like to convert this device into an
illuminance probe. How do you do this?

Answer 12

Place a filter with the transmission characteristics of the V(_) function in
front of the probe. The device would then need to be calibrated.

Question 13

A point source has an intensity of 100 cd. What illumination does it produce
at a distance of 2 ft? Give your answer in both foot-candles and lux
(lumens/m2). (Assume there are 10 lux per foot-candle.)

Answer 13

E _ I/d2
E _ 100 lumens/(2 ft)2 _ 25 lumen/ft2 or 25 foot-candles
Convert to lux:
(25 foot-candles) (10 lux/foot-candle) _ 250 lux

Question 14

A point source of 50 cd is 1 ft from a sheet of paper. Another point source at
3 ft from this same piece of paper illuminates the paper equal to the first
source. What is the intensity of this unknown point source?

Answer 14

E1 _ E2
I1/d1
2 _ I2/d2
2
50 cd/(1 ft)2 _ X/(3 ft)2
X _ 450 cd

Question 15

An illuminance meter is tilted 60 degrees from the perpendicular position
with respect to a point source of light. The meter reads 100 lux and is 2 ft
from the source. What is the intensity of the point source?

Answer 15

E _ (I/d2) cos _
As lux is metric, convert distance to meters:
2 ft _ 0.6096 m
Substitute:
100 _ I/(0.6096)2 (cos 60)
I _ 74.32 cd

Question 16

A neutral density filter absorbs 75% of the light incident on it. What is the
optical density of the filter?

Answer 16

OD _ log (1/T)
OD _ log (1/0.25)
OD _ 0.60

Question 17

A 0.5 neutral density filter is combined with a 1.0 neutral density filter. A.
How much light (as a percentage) does the combination transmit? B. How
much is absorbed?

Answer 17

Combined OD _ 0.50 	 1.0 _ 1.5
OD _ log (1/T)
1.5 _ log (1/T)
101.5 _ 1/T
31.62 _ 1/T
T _ 0.032
Percent transmitted _ (100)(0.032) _ 3.2%
Percent absorbed _ 100 _ 3.2 _ 96.8%

Question 18

A matte surface with a reflectance factor of 0.7 has a luminance of 50 footlamberts.
What is the illuminance falling on the surface?

Answer 18

L _ rE
50 _ 0.7E
E _ 71.43 foot-candles

Question 19

By use of the inverse square law and an illuminance device, the intensity of a
point source is determined to be 25 cd. Assume this uniform point source
produces the same amount of light in all directions. How many total lumens
does it produce?

Answer 19

(25 cd) (4_ lumens/cd) _ 314 lumens

Question 20

A light source is located 3 ft from a matte surface. The illumination falling
onto the surface is 100 lux. The luminance of this surface, at an angle of
30 degrees, is 5 foot-lamberts. What is the reflectance factor of this surface?

Answer 20

L _ rE
Convert 100 lumens/m2 to foot-candles:
L _ rE
5 _ r(10)
R _ 0.5

Question 21

Malingerers fabricate symptoms to obtain a positive medical diagnosis (i.e.,
they fake a visual disorder). Most often, these patients claim a reduction in
visual acuity. Which electrodiagnostic test would be most useful in making a
diagnosis of malingering? Explain.

Answer 21

Visually evoked potential (VEP). If the VEP is normal, the visual pathways
are intact up through at least primary visual cortex.

Question 22

Discuss a potential pitfall of using electrodiagnostic tests to make a diagnosis
of malingering. (Hint: Could a patient with severely reduced vision show normal
electrodiagnostic test results?)

Answer 22

A patient with a lesion in the higher visual centers could manifest a normal
electrooculogram, electroretinogram, and visually evoked potential because
these tests sample the visual system at lower levels.

Question 23

Why must the patient look at the stimulus to obtain a normal evoked potential?

Answer 23

The evoked potential is primarily a foveal response. Aversion of gaze will
result in a reduced potential.

Question 24

The pattern electroretinogram (PERG) and steady-state visually evoked
potential (VEP) are obtained with the same stimulus. (This stimulus typically
is a checkerboard pattern that undergoes a phase shift.) In amblyopia, the
PERG is normal, whereas the VEP is abnormal. Explain.

Answer 24

The pattern electroretinogram (PERG) test retinal function, whereas the
VEP tests cortical function. Amblyopia affects the cortex, not the retina.
Consequently, the VEP is abnormal, and the PERG is normal.

Question 25

In Chapter 7, we discussed the concept of the visual system as a Fourier analyzer.
Explain how the receptive field properties of ganglion cells could be
considered consistent with this hypothesis.

Answer 25

Gratings are optimal stimuli for ganglion cells, which have evolved to detect
contrast. The various spatial frequencies may be coded by ganglion cells
with various receptive field dimensions. That is, low spatial frequencies may
be coded by peripheral ganglion cells with large receptive fields and high
frequencies may be coded by foveal ganglion cells with small receptive fields.

Question 26

Foveal visual acuity is 20/20, whereas peripheral visual acuity is on the order
of 20/200. How can this be explained in terms of the receptive field properties
of ganglion cells?

Answer 26

The receptive fields of foveal ganglion cells are smaller than those of peripheral
ganglion cells. These smaller foveal receptive fields manifest less spatial summation,
but higher spatial resolution, than the larger peripheral receptive fields.

Question 27

Psychophysical experiments show that the human visual system manifests
greater spatial summation under dark-adapted conditions than under lightadapted
conditions. How could the receptive fields of ganglion cells change
to account for this psychophysical finding?

Answer 27

The antagonistic surround of ganglion cells is less apparent under dark adaptation.
This allows greater spatial summation with a resultant increase in sensitivity
(and reduction in resolution) (Barlow et al., 1957).