accomodation - specular vs ocular Flashcards
what is the accommodative demand for emmtropes and cl wearers
emmtropes and contact lenses - 1/ distance= accommodative demand
normal distance = 33 cm = 1/33= +3.0d
myopes and spectacle wearers have reduced accomodation
how does refractive error effect accommodative demand
the wearing of spectacles , myopic or hypermnetopic changes the accommodative demand of an individual
- if we are thinking about the accommodative demand of someone wearing spectacles then that is spectacle accomodation
what is ocular accomodatiion
an emmetrope or a hyperope/myope with contact lenses then we are measuring ocular accommodation
what is the ocular accomodation of emmetropes and contact Lens wearers
1/ distance
distance = 33cm = 0.33m
1/0.33 = 3 diopters
distance = 25cm
1/ 0.25= 4 diopters
how is myopia and accommodative demand related
myopes in spectacles have reduced accommodative demand compared to the emmetrope/ hyperope in contact lenses
with myopes we are putting the retinal conjugate and putting it at optical infinity
when myopes are corrected we have 0d of accomodation for distance viewing
we use the correct amount of - so that the retinal conjugate is at optical infinity
when myope is looking in the distance with spectacles - what is happening to accomodatio
- when myopes looks through spectacles at distance - no accomodation is being done
myopes have less accommodative demand
when you look at a distance object and then a near object your visual system perceives blur -(that you don’t notice) this drives accomodation but to get from blurry to clear myopes have to do less work
how do we correct hypermetropes
in hypermetropes we add + so that the retinal conjugate is back at optical infinity - in hypermetropes the optical infinity has passed and therefore we use + Lenses to bring them back to optical infintity
how to work out ocular accomodarion
accommodative demand =
L dist - Lnear
i.e. vergence of cornea when looking at a distance object -vergence of cornea when looking at a near object
what is the curvture of a light wavefront. for a an emmetrope looking at optical infinity
l corn is the distance from the eye to the object
l dist (curvture of light from a distance) for an emmetrope looking at optical infinity = 0d because 1/ optical infinity = 0d
and for near l corn will be 1/ -0.33 which will be -3
0- –3 = + 3 = accommodative demand
because accomodative demand = list- l near
a myope wearing contact lenses r and left are looking at an object 20cm away calculate the accomodative demand compared to when they Are wearing glasses vs contact lenses
step 1
for contact lenses 1/ distance = accommodative demand
and that will be 0 because distance = optical infinity and 1/ optical infinity = 0
and then for contact lenses part it will be
1/-0.2 because It is 20cm away which will be -5
and then you do 0 - - 5 which will be -5
how are myopes corrected
a myopes eye has too much + for distance therefore we have to correct it with a negative lens
the correcting lens has take the retinal conjugate and put it at optical infinity (shown by the parallel lines)
when we are wearing spectacles and we are looking at a distance object - the light hitting the cornea appears to be coming from somewhere that isn’t optical infinity
the light hitting the cornea is diverged
has divergence added to it
for a myope ion spectacles corn dosnt come from optical infinity and therefore the ver gence dosnt equal 0 because the lens has added divergence
this is why specular and ocular accommodation = offset
1
because the lens is altering the mergence
where is vergence at 0
vergence = 0 at the spectacle lens
what is image Vergence equal to
if you have got the object vergence you want to know what the image veregcne is
image vergence = L’SP= Lsp + Fsp
FSP= strength of the spectacle lens
LSP= objecte vergence
how to work out image distance when object vergence =0
l’sp =1/L’SP
which is basically the second focal length f’sp
1 divided by image mergence when object vergence =0
what is f’sp
f’sp is a specila case of image distance when the object is at optical infinity so the object mergence = 0
but it is an image distance
what are the priniciples of correction
principles of correction - take light that comes from optical infinity and make it look like it is coming from the far point because light that comes from the far point images on the retina
what is distance to the far point
distance to the cornea = image distance - back vortex distance
and cornea vergence is equal to 1/ distance to the cornea
how to work out the effect on the accomodative demand if ametropes are corrected by spectacles
when they are looking at a far point object then object veregence is equal to 0
so object vergence = LSP
LSP=0
L’SP = IMAGE VERGENCE
IMAGE VERGENCE = OBJECT VERGENCE + SRENGTH OF LENS
IMAGE DISTANCE = l’sp - WHICH IS EQUAL TO 1 DIVIDED BY IMAGE VERGENCE L’SP
distance to the cornea lcorn - which is equal to l’sp (image distance) - d which is back vortex distance
and vergence of cornea- L corn- is equal to 1 divided by the distance to the cornea which is corn
how to work out l corn
distance to the cornea =
l’SP-D
where l’sp = image distance and d = back vortex distance
what does light from real object do to light
light from real objects add divergence so diverging lens will diverge even more
how to work out object distance lsp and object verging when it isn’t coming from optical infintity
object distance = distance from lens to object L + back vortex distance
Bvd is always negative
object vergegence = 1 over object distance
a myope is wearing glasses right and left , they are looking at an object 20cm away , calculate the accommodative demand compared to when they are wearing contact lenses- spectacle lens power = -11.50
sign convention = -
near lcorn = 1/ l corn
= 1/ -0.2 = (20 cm = 0.2mm)
which = -0.5
distance l corn is equal to 0 because light is coming from the cornea
accommodative demand for contact lenses = l dist - l near
= 0 – 5 = 5 (ocular accomodation = +5 diopters)
i.e. mergence of cornea when looking at a distance object - veergence of cornea when looking at a near object
image vergence L’SP when object Vergence LSP = 0
is equal to 1/ L’SP
image vergence = L’SP= LSP + FSP (object vergence + strength of the spectacle Lens power)
0 + (-11.5)= -11.5 d
and then special case of image distance (l’sp) = 1/L’SP
L’SP= image vergence
1/ -11.5 = -0.0870m
what is the object distance for the cornea
the object distance for the cornea when looking at an object at optical infinity is the distance to the far point - bevcaude light that comes from the far point images on the retina
so the far point is equal to lower case k
which is equal to image distance - back vortex distance (which he says in the question weather it is positive or negative)
so if the image distance is equal to -.0870m then you subtract that from the back vortex distance which is 13mm
-0.0870 - ( +0.013) = -0.1
so L distance = 1/ -0.1 = -10.0
(the object is 20cm away from eye not spectacle lens we are trying to work out how far it is from the glasses to work out how much divergence or convergence it has added thus working out the far point)
so the far point is the distance that the object appears to be from the spectacles because the spectacle is adding vergence to the light not where the actual object is
how to work out object distance from the lens
object distance from the Lense is lower case lsp
l + d
(l= distance from lens to the object)
d is back vortex distance
therefore object vergence is equal to 1 /lsp
1/ lsp(object distance)
