ametropia

Question 1

differentiate spherical vs astigmatism (2v2)

Answer 1

astigmatism:

• irregular shape of the cornea
• power is not same in all directions

spherical:

• light pencil not focusing on retina
• has same power across all meridians

Question 2

outline refractive ametropia (3)

Answer 2

refractive power ≠ 60D

1. refractive myopia:
   - refractive surface too strong for the axial length
   - >60D
   - focus in FRONT of retina
2. refractive hyperopia:
   - refracting surfaces too weak
   - <60D
   - focus BEHIND retina

Question 3

outline axial ametropia (3)

Answer 3

eye radius ≠ 5.6mm

1. axial myopia:
   - axial length is LONGER than eye’s second focal length (F’e)
   - >5.6mm
   - focus in FRONT of retina
2. axial hyperopia
   - axial length is SHORTER than eye’s second focal length (F’e)
   - <5.6mm
   - focus BEHIND retina

Question 4

define near point (3)

Answer 4

• punctum proximum (Mp)
• nearest point an object can be focused with accomodation fully exerted

nb: emmetrope Mp in front of eye

Question 5

define far point (2+3)

Answer 5

• punctum remotum (Mr)
• furthest point an object can be seen clearly with the unaccomodated eye (relaxed)

nb:
- emmetropes Mr at infinity
- myopic Mr at finite distance in front of eye (real negative)
- hyperopic Mr behind eye (positive virtual)

Question 6

outline secondary focal point and its importance in correcting ametropia (2+1)

Answer 6

• secondary focal point is the location of the image of a distant object formed by the correcting lens
• if object at far point -> image will be formed on retina -> clear vision

nb: myopia + hyperopia corrected by ensuring secondary focal point of correcting lens is placed at eye’s far point

Question 7

define vertex distance and identify its importance in correcting ametropia (2)

Answer 7

• positive distance from back vertex of lens to the corneal apex (10-14mm)
• power of spectacle lens required to correct ametropia is dependant on vertex distance

Question 8

identify change in power between Fs and Fcl (m v h)

Answer 8

• myopic requires less minus power in Fcl than Fs

- hyperopic requires more plus power in Fcl than Fs

Question 9

outline scheiner principle (1+4)

Answer 9

the state of ametropia can be determined by using Scheiner disc by occluding one pinhole and finding out which of the two images has seen by the subject

• > two narrow pencils pass through pinhole apperture and focus INFRONT or BEHIND retina (subject sees both pencil due to doubling effect)
• > upper pinhole occlusion results in only one image viewed:
• myopic will see bottom image
• hyperopic will see upper image