ANSWERS 
(The following are answers to mock examination 5. To return to the question click on the number.)
1.
(Remember that : i) the power of the cylinder is 90^{0} to the axis ii) 2/3 metre is equivalent to 1.50 which has to be taken off) a. b. OD 1.00 / 1.50 X 90
a.
ii. The following treatment is useful:
a. There is a 4D difference between the two meridians. Therefore,
the required cylinder
b. The tight suture is at 140^{0} and
need to be removed.
a. More myopic. b. Dislocation of the implant. Astigmatism can occur due to tilting
of the lens.
a.
b. The nodal point is 17 mm anterior to the retina and all refraction
occurs at the plane through
Object height / retinal image height = distance
from nodal point / 17 mm
a. The CT scan shows a right subdural haematoma. (The haematoma
appears as a lenticular
b. The angiography shows a left internal carotid artery aneurysm
in the region of the cavernous
c. The MRI scan shows a lesion arising from the dorsal midbrain.
(This can lead to raised
a. Optic disc drusen. (The ultrasound shows an area of high echo
density at the optic nerve head, this is
b. Vitreous haemrrhage. c. There is a raised lesion arising from the choroid. Malignant melanoma
should be top of the
8.
U + D = V U = object vergence in dioptresi) The image is 0.333 m to the left of the lens and is inverted. ( U = 1/50 cm = 1/0.5 m =  2D, the sign is negative because the vergence of the object is divergence U + D = 2 + 5 = +3 V = 3 and therefore the image vergence = 1/3 = 0.333m) ii) The image is at infinity and erect.

Click here to return to mock examination 5 Click here to return to OSE