Dark Adaption

Question 1

What is the Purkinje shift?

Answer 1

The purkinje shift is the shift in wavelengths that result in the maximum sensitive. The purkinje shift for vision is from 507 nm to 555 nm when transitioning from scotopic to photopic conditions.

Question 2

What is a photochromatic interval?

Answer 2

It’s the difference of sensitivities of a certain wavelength when comparing scotopic and photopic conditions and their luminous efficiencies.

Question 3

Where is the rod-free zone?

Answer 3

It is in the foveola that spans 350 microns or 0.8 arc degrees.

Question 4

Describe spatial summation and other factors involved.

Answer 4

Spatial summation is the eye’s ability to sum, or add, quanta over a given area. That area is called critical area. As the critical area (diameter) increases, the sensitivity to the stimulus increases because it can take in and add more photons together. It has to do with the neural convergence of photoreceptors, bipolar, and ganglion cells. We know rods have more convergence so the critical area is larger than that of cones. That is why we have a greater sensitivity to light in scotopic conditions. However, since the critical area is so broad, it can not detect exactly where each photon is coming from so that would lead to poorer resolution.

Question 5

At 4-7 degrees retinal eccentricity, what is the critical area?

Answer 5

About 30 arc minutes

Question 6

At 35 degrees retinal eccentricity, what is the critical area?

Answer 6

about 2 arc degrees.

Question 7

As you increase distance from the fovea, the critical area ______________. This will allow an _________ in light sensitivity for a fixed number of photons.

Answer 7

Increases; increase

Question 8

What is Ricco’s Law?

Answer 8

Ricco’s Law says that for a test spot stimuli under 10 arc minutes, the same amount of light energy will have the same effect no matter how it is distributed as long as its within the critical area. In other words, two stimuli will be equally detectable if the product of their luminance and area is the same.

L x A = C

L = stimulus luminance
A = stimulus area
C = constant

Question 9

Pros and cons of a low critical area

Answer 9

Pro: Greater resolution
Con: Less light sensitivity

Question 10

Pros and Cons of large critical area

Answer 10

Pros: Increased greater light sensitivity
Cons: Decreased resolution

Question 11

What happens when light falls outside the critical area in photopic conditions?

Answer 11

The critical area actively decreases its perceived brightness. This is considered spatial antagonism.

Question 12

Describe temporal summation.

Answer 12

Temporal summation is when the eye adds the stimulus together that are present during a certain time. Critical duration is the amount of time that goes by before sending a signal. When stimuli are present within the critical duration, it will add all the stimuli together as one stimuli.

Question 13

What is Bloch’s Law?

Answer 13

Bloch’s Law says that the stimulus will have the same effect, no matter how it is distributed across time as long as it is received within the critical duration for complete temporal summation.

C = LT

L = stimulus luminance (I = radiance)
T = stimulus duration
C = constant

Question 14

Dark adaption testing procedure

Answer 14

1) Light-adapt for 5 minutes (Bright, even illumination of the entire visual field. Head in bowl, light on)
2) Turn off the background light in bowl, leaving the patient in total darkness. (All room lights also off.)
3) Over the next 40-45 minutes (longer, if needed), repeatedly measure the patient’s light detection threshold using the ascending method of limits (know why!)
4) Plot the measured thresholds (logarithmic values, y-axis) as a function of the time (in minutes, x-axis) at which they were measured after turning off the adaptation light.

Question 15

Describe the Dark Adaption function graph.

Answer 15

It will plot the relative/increment threshold on the y axis and time in darkness on x axis. At first, you should see the response of the cones on the photopic limb because they recover quicker than cones. The value will decrease as the eyes are getting more adapted to dark and pick up on the stimulus easier. Then it shifts at the rod-cone break and the rods complete the graph and is called the scotopic limb. The line dips lower considering they are more sensitive to light in scotopic conditions and have smaller increment threshold.

Question 16

Photopic threshold (cones fully dark-adapted) is about ___ log units higher than scotopic threshold (rods fully dark-adapted).

Answer 16

4

Question 17

Reducing the _______ of the adapting light results in shifting the dark adaptation function down and to the left.

Answer 17

intensity

Question 18

Reducing the duration of the adapting light results in a more rapid decrease in the
 visual threshold during dark adaptation (i.e., steeper slope).

Question 19

What influences would be necessary to have a dark adaption graph have the smallest/latest rod-cone break?

Answer 19

For the smallest rod-cone break, you would need the shortest degree eccentric without being in the center of the fovea (just over 2 degrees), the smallest spot size stimulus, and a red stimulus present.

Question 20

What influences would be necessary to have a dark adaption graph have the greatest/soonest rod-cone break?

Answer 20

In order to have the greatest rod-cone break, you would need the stimulus to be present at the eccentricity to be greatest that has most rods, the largest spot size and also present a violet light.

Question 21

Under 2 degrees eccentricity, what does the dark adaption graph look like?

Answer 21

Only contains cones because rods are not found within 2 degrees eccentricity.

Question 22

What is fundus reflection densitometry? What does the graph show?

Answer 22

it is a technique that measures the levels of visual pigment in the living eye. The graph shows the recovery of rhodopsin density (fraction) after bleaching over time. The results typically show a linear graph until the 70% mark then curves off to 1.0 (100%).