WEEK 7: LAB

Question 1

A person who has an accommodative facility result of 4 cycles per minute binocularly when looking through +/-2.00 DS lenses has: 
A. no accommodative facility 
B. average accommodative facility
C. superior accommodative facility 
D. poor accommodative facility

Answer 1

D. poor accommodative facility

The normal expected results is 8 cycles or more in 1 minute binocularly. This is significantly below that value so indicates poor accommodative facility.

Question 2

Binocular measures of accommodation are about 0.50 D greater than monocular findings mostly likely because:

A. The method of limits is more effect under binocular stimulation.

B. As they are conducted after monocular measurements so the patient is more familiar with the test procedure

C. There is an accommodation, convergence, pupil constriction near triad.

D. Patient’s try harder when using two eyes rather than one.

Answer 2

C. There is an accommodation, convergence, pupil constriction near triad.

Question 3

A patient can see clearly from inifinity to 50 cm. The patient’s near point of accommodation is ______
m. (Enter your answer to 2 decimal places)The patient’s amplitude of accommodation is _______
Dioptres.

Answer 3

0. 50m

2. 00D

Question 4

A myope with refractive error of -2.00 DS is looking at an object located 20 cm away.

The myope will accommodate by _______
Dioptres (insert your answer to two decimal places) when wearing spectacles to correct their refractive error to see the object clearly. The myope will accommodate by _______
Dioptres (insert your answer to two decimal places) when not wearing any spectacles to see the object clearly.

Answer 4

5D
3D

When wearing spectacles, the patient is effectively emmetropic so the far point vergence = 0. Ffp = L + PA where FFp is the far point vergence, and L is the stimulus to accommodation and Pa is the accommodation as measured in the plane of the cornea, required to focus the object.

So Pa = Ffp-L = 0-(1/-0.20)=5.00 D

When not wearing spectacles, far point vergence = -2.00. Ffp = L + PA where FFp is the far point vergence, and L is the stimulus to accommodation and Pa is the accommodation as measured in the plane of the cornea, required to focus the object.

So Pa = Ffp-L = -2-(1/-0.20)=3.00 D

Question 5

The far point of a hyperope with refractive error of +1.00 DS is located at [A] m. Insert your answer to two decimal places.

Answer 5

0.1m

The far point is where objects are located and clearly imaged on the retina without accommodation or wearing spectacles.

The far point vergence = +1.00 DS.

So the location of the far point - 1/far point vergence

This is 1/1 = 0.10 m.

Question 6

The far point of a myope with refractive error of -6.50 DS is located at [A] m. Insert your answer to two decimal places.

Answer 6

-0.15m

The far point is where objects may be located and imaged clearly on the retina without accommodating or wearing spectacles.

The far point is where objects are located and clearly imaged on the retina without accommodation or wearing spectacles.

The far point vergence = -6.50DS.

So the location of the far point - 1/far point vergence

This is 1/-6.50 = -0.15 m.

Question 7

A hyperope’s eyeball is typically too ________
(insert either short or long) for the refractive power of the eye.

A myope’s eyeball is typically too _________
(insert either short or long) for the refractive power of the eye.

To view an object located 20 cm away from the front of the eye, the crystalline lens must accommodate by _______
Dioptres (insert a number to 2 decimal places)

Answer 7

short

long

5.00D

Myopes have longer eye balls than normal.

Hyperopes have shorter eye balls than normal.

The stimulus to accommodation is 1/0.20 = 5.00 D