( The following are answers to mock examination 4. To return to each question click on the number)
1.
a.
i) Location of the first local line (formed by the meridian with +3.00D)
Using the formula 1/u +1/v = 1/f
u = object distance from the centre of the lens
v = image distance from the centre of the lens
f  =  focal length of the lens
1/v = 1/f - 1/u
1/v = 3 - 1
v   = 0.5 metre
The first focal length is therefore 50 cm behind the lens
ii) Location of the second local line (formed by the meridian with +2.00D)
Using the formula in i)
1/v = 2 - 1
v  =  1 metre
The second focal length is therefore 100 cm behind the lens
b.  Location of the circle of least confusion
=  (length of first focal length + length of second focal length) / 2
=  (50 + 100) / 2
=  75 cm behind the lens

c. Diameter of the circle of least confusion
= (D1 - D2) / (D1 + D2)      X  aperture of the lens
D1 = the lens power that produces the first focal length = 3.00 D
D2 = the lens power that produces the second focal length = 2.00 D
aperture of the lens = 60 mm

= 1/5 (60)
= 12 mm

a. The resultant prism is found by adding the powers of the two prisms
= 7 prism dioptres, base-out

b. The resultant prism is found by subtracting the weaker prism from the
stronger prism
= 1 prism dioptre, base-in

c. The resultant prism is found by adding the powers of the two prisms
= 7 prism dioptre, base-down

d. The resultant prism is found by subtracting the weaker prism from the
stronger prism
= 1 prism dioptre, base-up

( For horizontal prismatic effects:
a. when prisms are placed with their bases in the same direction for each eye
(both base-in or base-out) the resultant prismatic effect is found by
adding the powers of the prism
b. when prisms are placed with their bases in the opposite direction and of
different powers, the power of the weaker prism is subtracted from that
of the stronger prism

For vertical prismatic effects:
a. when prisms are placed with their bases in the same direction for each
eye but of different powers, the resultant prismatic effect is found by
subtracting the weaker prism from the stronger prism

b. when prisms are placed with their bases in the opposite direction, the
resultant prism power is obtained by adding the two prisms)

3.
a. The power crosses translate to spherocylindrical corrections (with minus cylinder) of
RE    +1.50 / - 4.00 X 135
LE    -0.75 /  -1.25  X 90

As the working distance is 2/3 metre, -1.50 D is subtracted from the above giving
RE       PL / -4.00 X 135
LE   -2.25 / -1.25 X 90

b. The power crosses translate to spherocylindrical corrections (with positive cylinder) of
RE     -2.50 / +4.00 X 45
LE     -2.00 / +1.25 X 180

As the working distance is now 1/2 metre, -2.00D is subtracted from the above
RE     -4.50 / +4.00 X 45
LE      -4.00 / +1.25 X 180

a.
i.  In direct ophthalmolscopy, the optic of patient's eye acts as a simple magnifier
60D / 4 = 15 X
Therefore, the retina of an emmetrope appears 15X larger than normal.

ii.  Bigger. In a myope, the eye has more plus power and using the formula of
Magnification = power / 4
the optic disc will appear bigger

b.
i. The magnification of the an emmetrope's retina is the ratio of the patient's power
to the power of the condensing lens.
Therefore, the magnification = 60 / 20 = 3X

ii. When a 15 D lens is used. The retina magnification = 60/15 = 4X

a. Sturge-Weber's syndrome ( The skull X-ray shows the typical curvilinear calcification
or  "tram" line appearance, it is caused by ipsilateral arteriorvenous malformation

b. Right retinoblastoma ( There is calcification in the right posterior pole. In a child with poor
vision, retinoblastoma is the most likely diagnosis)

c. Foreign body in the left orbital region ( There is a small opacity in the left orbit, this may
be located anywhere as the anteroposterior radiograph does not provide enough
information about its exact location. A lateral X-ray can help location but CT scan
provide the best mean of localization.)

a. Fluorescence has an absorption peak from 465 to 490 nm (blue region of the visible spectrum)
and an emission peak of 520 - 530 nm (in the yello-green region of the visible spectrum)

b. Venous phase.
The picture shows :

• multiple small areas of hyperfluorescence which represent the microaneurysms
• increased area of hypofluorescence in the fovea avascular zone ie. ishcaemic maculopathy
c. Late venous phase.
The picture shows an area of hyperfluorescence in the macula. The hyperfluorescence assume
a smoke stack appearance. This is a classical appearance of central serous retinopathy

a. Fully accommodative esotropia = refractive accommodative esotropia
( Presence of esotropia which is usually slightly larger for near than distance.
Refraction reveals significant hypermetropia. The esotropia is eliminated with
wearing glasses.)

b. Non-refractive accommodative esotropia = accommodative esotropia with convergence
excess
( Presence of slight esotropia or esophoria for distance but with near vision the esotropia
is significantly increased. Refraction shows no significant refractive errors for the patient's
age.)

c. Right lateral rectus palsy
(The esotropia is typically more for distance than near. There is incomitant deviation.)

8.

a. Hess chart relies on two laws:

Sherrington's law ( law of reciprocal innervation) = the contraction of a muscle is accompanied
by simultaneous and proportional relaxation of its antagonist.

Hering's law (law of equal innervation) = innervation to the extraocular muscle is equal in both
eyes.

b. Right third nerve palsy ( The field of the affected eye is the small one)

c. Right gaze ( The hess chart shows right lateral rectus underaction and the most likely cause if abducent
nerve palsy.)