ANSWERS
(The following are the answers to mock examination 2. To return to each station click on the number.).
1.
a. Absolute hypermetropia  = + 2.00D
   (absolute hypermetropia is defined as the least amount of plus lenses needed for clear vision
    without cycloplegia)

b. Manifest hypermetropia = + 4.00D
    (manifest hypermetropia is defined as without cylcoplegia, the most plus correction that can be 
     tolerated without blurring of vision)

c. Facultative hypermetropia = + 2.00D
    (facultative hypermetropia is defined as the difference between absolute and manifest hypermetropia
     + 4.00D - + 2.00D = + 2.00 D)

d. Latent hypermetropia = + 2.00 D
    (latent hypermetropia is defined as the difference between manifest hypermetropia and hypermetropia
     measured with cycloplegia + 6.00D - + 4.00D = + 2.00D)


2.

a. You can make a Galilean telescope whereas your friend can make an astronomical telescope.
    (You use +2.00D as the objective lens and -8.00D as the eyepiece lens. Your friend use +2.00D as the 
      objective lens and +5.00D as the eyepiece lens.)

b. Your telescope length is 37.5cm whereas your friend's  telescope is 70cm.
    (In Galilean telescope, the secondary focal point of the plus lens ie. objective lens coincides with the primary 
     focal point of the minus lens ie. eyepiece lens. Therefore 50cm - 12.5cm = 37.5cm.
     In astronomical telescope, the secondary focal point of the objective lens coincides with the primary focal 
     point of the eyepiece lens. Therefore 50cm + 20cm = 70cm.)

c. Your telescope will provide more magnification.
   (The magnification of a telescope is calculated as the power of the eyepiece divided by the power of the 
    objective.
    Therefore your telescope magnification = 8 / 2 = 4X
                    your friend's telescope         = 5 / 2 = 2.5X)

d. The image formed by Galilean telescope is erect whereas the image formed by astronomical 
     telescope is inverted and real.
 

3.
a. Hyperostosis of the left lateral portion of the sphenoid with left proptosis.

b. Sphenoid wing meningioma.

c. 

  • Better bony definition than MRI especially in detecting orbital fractures and bony metastasis
  • Detecting metallic foreign body within the orbit or globe (contraindicated in MRI)
  • Shorter running time than MRI
  • Less expensive than MRI


4. 

a. Humphrey field analyser is a static automated perimetry.
( In this test, the patient maintains fixation on a central target and the computer randomly presents a brief  (about 0.2 seconds) and non-moving ie. static light stimulus at different loci throughout the visual field. The intensity of the light stimulus that the patient can see is then recorded.)

b. The total deviation measures the difference (in db) between the patient's threshold values 
    and that of the age-corrected values.

c. The pattern deviation adjusts the total deviation for any shift in the patient's overall 
    sensitivity. This allows localised area of field loss to be clearly demonstrated.
   ( Many conditions other than glaucoma can cause poor vision for eg. cataract or corneal oedema. Therefore, to find out how much of a patientís relative insensitivity to light is due to glaucoma rather than to something else, it is important to "subtract out" these other factors. This can be done because these others conditions tend to produce a similar pattern of diffuse visual field loss, while glaucoma tends to produce localized areas of visual field loss.)

d. Superior arcuate scotoma.

e. 

  • open angle glaucoma with inferior loss of arcuate nerve fibre layer
  • optic disc pit
  • inferior branch retinal vein occlusion
    (Visual field should not be interpreted without reference to ocular examination. An arcuate scotoma 
      can occur  in other conditions other than open angle glaucoma as mentioned above)


5.
 

a. + 4.00 / - 1.50 x 70 = + 3.50 / + 1.50 x 160 ; compound hypermetropic astigmatism.

b. + 1.25 / - 3.00 x 90 = - 1.75 / + 3.00 x 180 ; mixed astigmatism.

c. PL / + 1.50 x 45 = -1.50 / - 1.50 x 135 ; simple hypermetropia astigmatism.

d. - 2.00 / + 2.00 x 50 = PL / - 2.00 x 140 ; simple myopic astigmatism.

e. - 1.75 / - 2.00 x 135 = - 3.75 / + 2.00 x 45 ; compound myopic astigmatism.

Compound astigmatism occurs when the two principal meridians of an eye are either both hypermetropic
ie. compound hypermetropic astigmatism or both myopic ie. compound myopic astigmatism.  Mixed 
astigmatism occurs when one principal meridian is hypermetropic and the other myopia. Simple astigmatism
occurs when one principal meridian of the eye is emmetropia and the other myopia ie. simple myopic
astigmatism or hypermetropic ie. simple hypermetropic astigmatism.
 

compound 
hypermetropic 
astigmastism
compound
myopic 
astigmatism 
mixed
astigmatism
Simple 
hypermetropic
astigmatism
simple 
myopic
astigmatism
6. 
a. OS and therefore left eye.
    (OS = right eye)

b. A = anterior lens echo
    B = posterior lens echo
    C =  retinal echo

c. Power of the intraocular lens needed for emmetropia 
    = 118 - 2.5 (22.3) - 0.9 (43)
    = 23.55D
   The power needed would be 23.50D as the implant comes in step of 0.5D

d. The resultant refraction would be hypermetropic if the axial length is overestimated.
 
 

7.
 
a. Infinity to 3.3cm.
    (The range of accommodation is the linear distance between the far point and the near 
     point. In the absence of accommodation, the boy is in focus at infinity. With maximal
     accommodation, he is in focus at 1/30 = 0.033m = 33cm)

b. 10cm
    (To be in focus in infinity, the man needs to use up +10.00D of his accommodation 
     amplitude. As a result, he has only 20 - 10 = 10 D for near vision. His near point is
     1/10 = 0.10 m = 10 cm)

c. The near point is 5 cm and the range of accommodation is infinity to 5 cm.
   (With hypermetropic correction he does not need to use up any of his accommodation
    and can therefore use all the 20 D for near vision. The near point is therefore
    1/20 = 0.05m = 5cm. The range of accommodation is therefore from infinity
    to 5 cm)
 


8.

a. The base curve is 8.6mm, the diameter of the lens is 13.5 mm and the power is +10.00D
    (The first number denotes the base curve, the second the diameter and the last the lens power.)

b.  + 9.09D
     (Using the formula for lens effectivity the new lens power 
         = 10 / (1- (-0.01)10)
         = 10 / 1.1
         = 9.09D )

c. Problems of aphakic glasses include:

  • ring scotoma. When the patient looks laterally the object seems to disappear and this 

  • occur due to the prismatic effect of the lenses. As a result, the patient has problem 
    with peripheral vision
  • pincushion effect because the periphery of the image is more magnified than the centre
  • excessive magnification
  • altered depth perception
  • weight of the glasses

 
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