power of an intrauouclar lens Flashcards
what is the typical amount of accomodative lag
humans don’t see blur - the typical amount of accomodative lag = 0.5 to 0.75d in a normal human
why is there a lack of perceived blur in the presence of optical blur at near
because of pupil miosis - the pupil constricts
pinhole - increases light scatter due to refraction
peripheral rays don’t contribute towards images formation - corneal rays go through pinhole - this reduces retinal illumination
pinhole reduces the size of the blur circle
pupil misos is what allows you to see clear at near and explains the accomodative lag in a normal human
what are the three main ways to correct aphakia
spectacle lenses- (lenticular lenses) - cosmesis is terrible , heavy , optical quality is poor due to prismatic effects - because they are looking away from the optical centre
contact lenses are optically Better- no prismatic effects because they remain on the eye and are centred
abberations = better
drawbacks of cl wear
dry eye , infections (microbial keratitis)
corneal hypoxia
cornea = avascular (prevents light scatter) gets 02 from diffusion , cornea can handle hypoxic stress as long as you get a chance to recover , if you keep depriving it of o2 corneal oedema
corneal sensitivity gros foin
as hypoxia increase a substance emitted a growth factor due to a lack of 02 - vaso endelthieliel Growth factor
consequences of vessels
cornea = a transparent structure shouldn’t be there , vessels are therefore fragiler and will leak in the cornea and start inflammation
aphasic cl - thick centrally - risk of hypoxia is increased
abberations associated with corrections are the least with iol correction
lense inserted into the capsule- capsule can tear or thicken afterwards
if you remove the lens out of an emmtrope what state are you leaving them in
- if you take out the lens you leave it in a hypermetropic state (crystalline lens) +20d
if you remove it from the emmetropic eye the eye is under +’d which = hyperopia for distance viewing
if they are emmetropic you are leaving them significantly hypermetropic
if a patient is myopic and you are removing the cyrystaline lens what refractive state are you leaving them in
if a patient is highly myopec then removing the cyrtaline lens could actually be correcting them for distance viewing if the axial length is 31mm
most people don’t have 31mm axial length
removing the lens will bring the image to the retina
an eye is made aphasic radius of the cornea = 8.750 mm
axial length is equal to 20.87mm
n eye = 1.333
the intraocular lens is placed 4mm behind the cornea , calculate fIOL
step 1 - calculate the power of the cornea f corn
n eye -1 / r corn 1.333-1/ 0.00875 = 38.075d
(calculate the image vergence) L’corn = Lcorn(object vergence) + Fcorn
object vergence = 0
0 + 38.057d = +38.057d
then calculate image distance l’corn = n eye / image vergence L’corn
then calculate the object distance of the intraocular lens - Liol = image distance - acd
0.035-0.04= 0.031
then calculate object vergence = n eye / object distance = 43 diopters
then calculate the image distance to the intraocular lens
= axial length - anterior chamber
then calculate the image vergence of the intraocular lens
neye/ image distance
Then calculate the power of the intraocular Lens
L’IOL- LIOL
iol of the image vergence - object vergence of the intrauouclar Lens
= 79.02- 43.73
= 35.29d
35.25ds
36.0ds
how to work out the power
work out the power of cornea - fcorn = neye -1 / rcorn
work out image vergence of cornea = L’corn = Lcorn + Fcorn (Object vergence for the cornea= usually 0 )
work out the image distance to cornea = n eye / image vergence
work out object distance to the intraocular Lens
image distance - anterior chamber depth
workout object vergence of the iol - neye/ object distance
work out the image distance of the iol = l’iol
k - acd (k = axial length) - anterior chamber depth
work out image vergence of iol - n eye / image distance
power of iol = fiol - L’IOL- LIOL
(image vergence of the intraocular lens - object vergence of iol)
if an aphakik eye is corrected with contact lenses , dcl +8.00ds
if the axial length is +21.25mm , calculate the iol power if placed 3mm behind the cornea n eye = 1.333
ocular refraction is equal to contact lenses in this question so k would = +18.00ds
f corn = k’ - k ( power of the cornea is equal to axial length in diopters - ocular refraction )
K’ (axial length in diopters is equal to n eye / axial length)
1.333/ 0.02125
= 62.73 diopters
f corn = k’-k
62.73- 18.00= 44.73 (fcorn)
L corn = (object vergence is equal to 0d )
l’corn = lcorn+ f corn
therefore 0 + 44.73 = 44.73(image vergence)
image distance to the cornea = 1.333/ L’corn (image vergence)
1.333/ +44.73 = +0.0298
object distance to the intraocular lens
liol
l’corn - acd (image distance - the anterior chamber depth
0.0298 - (+0.03) = 0.0268m
object vergence = Liol = neye / liol (neye / object distance)
power of introcular lens is equal to image vergence of intraocular lens - object vergence of intrauouclar lens
image vergence of intraouclar lens= n eye / axial length (given in question) - anterior chamber depth
= 1.333/ 0.02125- 0.03= 0.01825= +73.04
Fiol= L’IOL- Lill
73.04 - + 49.74
= +23.3d
what is the power of the intraocular lens equal to
power of iol = image vergence of iol - object vergence of iol
An aphakik eye is corrected with spectacles fsp + 14.00ds
dvd = 12mm
if axial length = +23.75mm . calculate the iol power if placed 4mm behind the cornea
n eye = 1.333
step 1 - working out ocular refraction
K= 1/k
fsp = k + d
rearrange for axial length
k = fsp - d
fsp = 1/fsp
1/ 14.00 = 0.071428
axial length = 0.074- 0.012= +0.0594m
K = 1/ +0.0594 = 16.84d (ocular refraction)
now work out axial length expressed in diopters = n eye / axial length given in question
1.333/ 0.02375= 56.13
fcorn = K’- K (axial length expressed in diopters - ocular refraction)= f corn = 56.13- 16.84 = 39.29
= 39.29
l corn = 0
threfore image vergence of cornea will be
+39.29 because
L’corn = l corn - fcorn
0- 39.29= 39.29
image distance to the cornea
l’corn = n eye / L’corn = 1.333/ 39.29 = +0.0339m
how to work out image distance for the iol - image distance for the iol = image vergence - anterior chamber depth
= +0.0299m
power of the intraocular lens is
image vergence of the iol L’iol - Liol( object vergence of the intrauouclar lens)
object vergence of intraocular lens = neye/ l’iol (image distance for the iol )
image veregnce for the iol= n eye / axial length - anterior chamber
1.333/ 0.01975 = 67.49
67.49 - +44.58= 22.9 = 23d
how to work out the power of an implant that will replace a contact Lense - what are the steps
ocular refraction is equal to contact Lens strength
fcorn = K’ - K ( K’ = axial length expressed in diopters)
K’ = n eye / k’ (axial length)
then using these values work out power of cornea
fcorn = K’-K
object vergence in a cornea = 0d because light is coming from optical infinity therefore image vergence will be equal to f corn
L’CORN= L CORN - F CORN
then work out image distance l’corn = neye/ L’corn
then work out object distance to the iol
image distance = l’corn = neye / L’corn
object distance to the cornea = image distance to the cornea l’corn - anterior chamber depth
then calculate object vergence of the iol = n eye / object distance to the iol
then the power of the intraocular lens = fiol= L’IOL- LIOL
image vergence of the lens - object vergence of the iol
L’IOL (image vergence of a iol) = neye/ k’- acd
