lecture 16 - visual acuity and functional contrast sensitivity

Question 1

how is visual acuity measured ?

Answer 1

. visual acuity is measured with high contrast letters

the highest one can produce about 90% contrast

Question 2

what do we measure in VA ?

Answer 2

. the smallest optotype size which the subject needs to be able to read or carry out some task

Question 3

what kind of task is involved in VA ?

Answer 3

it is not strictly spacial resolution task , its more a task of letter recognition affected by other parameters

Question 4

which is typical letter size one needs under photopic daylight levels of illumination ?

Answer 4

5 min of arc with 1 min of arc spatial resolution

1min = amount of smallest detail that can be resolved

Question 5

what are the principal parameters that affect VA and functional contrast sensitivity ( FCS ) - what we can see ?

Answer 5

. stimulus size - its angular subtense ( min arc ) at the eye
. stimulus contrast - on chart and on the retina
. retinal sensitivity to contrast ( function of retinal illuminance )

Question 6

how do we measure VA and FCS ?

Answer 6

1 - we can use single object ( unequal space - averaged luminance ) - we need to be able to control the contrast - use weber contrast
C= Lo/Lb -1

2. we can use spatially periodic pattern either sine waves or square wave
   - use Michelson contrast - which represent luminance difference either an increment or decrement with respect to average luminance , either bright or dark bars
   C=δL/ Lb

Question 7

what is contrast ?

Answer 7

. the luminance difference between the optotype Lo and the adjacent background Lb
. this luminance difference with respect to the background luminance provides measure of weber contrast

Question 8

what is difference between single object measurement and spatially periodic stimuli ?

Answer 8

. single object ( unequal space - averaged luminance )

. spatially periodic stimuli ( equal space-averaged luminance )

Question 9

what does the spatially periodic stimuli do ?

Answer 9

. spatially periodic stimuli arrows interest from a different point of view - they provide the means of measuring what is known as the modulation transfer function ( MTF )
. in spatially periodic stimuli the average light level remains unchanged but contrast of stimulus can change from contrast of 0 with nothing to contrast of 1

Question 10

what is MTF ( modulation transfer function ) ?

Answer 10

. ratio of contrast in image formed by lens to contrast of the grating in object
. MTF = c’/c

Question 11

what is modulation transfer curve ?

Answer 11

. plots the amount of contrast transferred by the lens at each spatial frequency

Question 12

what is MTF limit caused by ?

Answer 12

diffraction

Question 13

how do we measure MTF in human eye ?

Answer 13

. ideally we would like to know contrast in the retinal image but we can’t do this
. we measure threshold contrast for gratings of different spatial frequencies
. measure the smallest object contrast (Ct ) the patients needs to just see anything different to a uniform field - until they fail to see bright and dark bars

Question 14

how to measure threshold contrast ( Ct) ?

Answer 14

the patient needs to see the bars in the grating for every object spatial frequency of interest

Question 15

what is spatial frequency ( cycles / deg )

Answer 15

equals the reciprocal of spatial periodicity ( number of deg/cycles )

Question 16

what is the equation for CS contrast sensitivity ?

Answer 16

CS= 1/ Ct

when Ct is small CS is high and when Ct is high CS is low

Question 17

what does high spatial frequency involve ?

Answer 17

. high number of cycles per degree of visual angle

. width of bright and dark bars becomes very small

Question 18

what does low spatial frequency involve ?

Answer 18

. involves only a few cycles over visual fields

. width of bright and dark bars increases

Question 19

what are the advantage of measuring CS?

Answer 19

. sensitive to residual higher order aberrations , diffraction and increases scattered light
. can reveal presence of subclinical retinal disease

Question 20

how do we combine measure of VA and CS?

Answer 20

. using modern display techniques it is possible to measure VA and CS for both increment luminance as well as decrements luminance

Question 21

describe acuity plus test ?

Answer 21

. PX is presented with landlot C , the gap of the optotype can point diagonally toward the bottom left or bottom right , top right or top left
. px is simply to press one of four buttons to indicate positions of gap

Question 22

what are the acuity plus test properties ?

Answer 22

. four - alternative , forced choice procedure
. gap orientation task
. use positive and negative contrast
. measure both VA and FCS
. test low light level ( mesopic performance )
. with or without visual crowding

Question 23

what happens when we add distraction in acuity plus test ?

Answer 23

VA worsens

Question 24

why do we need to measure FCS ?

Answer 24

. VA is not sensitive to early changes in the optics of the eye ( i.e.residual refractive errors , higher order aberrations , increased scattered light )
. VA is not very sensitive to early structural changes in the retina caused by disease

Question 25

what does FCS test involve ?

Answer 25

FCS is efficient test for CS
. involves a fixed optotype usually 15 min or arc , 3 times the average VA
. can measure smallest contrast needed to locate the position of the gap using both negative and positive otptotype

Question 26

What imposes the ultimate limit on the resolving power of any optical system?

Answer 26

-Airy disk
-these imaging systems represented by a single lens, or complex imaging system- many lenses etc
-Whenever you pass light through an optical system which has finite aperture - likely to get diffraction of light around the edge of the circular aperture- which causes formation of Airy disk- smallest image size
-

Question 27

What can the size of the airy disk be represented by?

Answer 27

the radius of the first dark ring in the diffraction pattern (airy disk)

Question 28

What is the expression of the first dark ring?

Answer 28

p= 0.61(wavelength) / n’u’

• radius increases with increasing wavelength
• decreases with n’ and u’

Question 29

What criterion can be used to define the resolving power of an optical system?

Answer 29

• When we bring 2 objects point closer together, their images in their image plane get closer together
• so how much closer we can bring them until we can no longer resolve the 2 objects as separate image points in the image plane is RAYLEIGH CRITERION

Question 30

What is the Airy pattern and how can it be used to determine the best spatial resolution one can achieve with a lens?

Answer 30

The Airy pattern - diffraction pattern caused by a circular aperture

Question 31

How do we quantify the size of this pattern?

Answer 31

use equation
p= 0.61(wavelength) / n’sin u’
or p = 0.61 (wavelength)/ n’u’

Question 32

How do we use the Airy disc in the case of a lens with finite conjugates (as above) or in angular space as in the case of a telescope objective or the human eye?

Answer 32

-The lens needs to be optically extremely well designed- which means aberration free- needs a good lens to even consider using the Rayleigh criterion.

Question 33

What is the airy pattern?

Answer 33

• The radius of the first dark ring in the Airy pattern equals p

Question 34

What can we see when we take a cross section through the airy pattern?

Answer 34

-can see the distances between the centre of the pattern, first point of zero intensity in the Airy pattern (away from the peak) corresponds to a distance, p, (first dark ring)

Question 35

What did Raleigh propose (demonstrate) experimentally?

Answer 35

• we can bring the 2 object points closer until the two images are separated by the radius of the first dark ring in the diffraction pattern (airy disk) .
• The Rayleigh criterion requires the separation between two adjacent object points to produce an image separation equal to p (the radius of the first dark ring in the Airy pattern).

Question 36

How can one use the Rayleigh criterion to predict how pupil size affects spatial resolution in the human eye ?- in case of telescope objective

Answer 36

• telescope objective
• object field angle is small
• focal length of telescope is large
• work in paraxial region- at least the doublet lens - which is used to correct spherical aberration.

Question 37

So if 2 stars are at infinity , How close together the stars can be in terms of object field angle at which they subtend to be able to resolve them as separate images in the image plane?

Answer 37

• given the parameters of the telescope objective - can find out the radius of the first dark ring in diffraction pattern- in one of these images
  -can subsitute for u’ with the diameter of the telescope objective divided by 2 (to get radius of objective divided by the focal length which gives the angle u’ )
  B (resolution power)= 1.22 (wavelength) / n’D - to find angle
• the smaller the wavelength the higher the resolving power)
  -if to see with human eye- makes sense to consider wavelength in the middle of visual spectrum ( eye most sensitive )
  D = Diameter of lens
  n’= R.I - Usually 1
  -The resolving power of the telescope objective increases as diameter of aperture increases- hence angle which can resolve becomes smaller.

Question 38

What does p determine?

Answer 38

the angular separation of two adjacent and distant object points (or stars) that can be resolved

Question 39

How does the limit of spatial resolution of the human eye varies with pupil size? (in the absence of higher order aberrations)

Answer 39

• SAME principles apply
• the pupil size of human eye can change from 1.5mm to 8mm
  -diffraction pattern in absence of aberrations ( looks perfect)
• to apply the Rayleigh criterion to the human eye we do the same thing as telescope objective however human eye is more complex as n’ is different to n. - beavsue of that Nodal point locations and Principle planes become important.
  -Make use of the
  Gullstrand Schematic Eye parameters- good approximation to the average good eye

Question 40

What is the Gullstrand Shcematic eye ?

Answer 40

• good approx to average good human eye hence use
• Use n’ = 1.336, P’F’ = 22 mm.
• cane estimate the angle u as the function of pupil size because the angle u’ is equal to~ D/2P’F’, where D is the diameter of the pupil and P’F’ is the focal length (fo’).
• Then calculate angle beta 0 - which radius of the first dark ring or separation between the centres of thew 2 images subtends at the nodal point N’ - which is the same angle as these 2 points (image) would project through In object space and would represent the angle fo separation of 2 point object.
• Use wavelength = 555 nm. Calculate, p and then Bo= p / N’F’

Question 41

What is the graph for the human eye on how diffraction determined by size of pupil limits the best spatial resolution ?

Answer 41

• angle beta 0 - function of pupil size
• VA (on y axis) , pupil size ( on x axis) - the line fundamental limit imposed by diffraction ( check graph on slideshow )
• On the right side is the equivalent Snellen acuities which correspond to the VA in mins of arc- if start with 6/6 it corresponds with 1min of arc - 2.5mm pupil size - in terms of diffraction limited
• in order to achieve 6/3 - to be able to resolve 5mins or arc - need 4.5 mm pupil size.