Answer5

Chapter 5.

Multiple Choice Questions

I. Answers A, B, C and D are true statements, whereas E is false.

II. Answers B, C, and D are true statements, whereas A is false.

Chapter 5. Case History A

1. The patient has facultative hypermetropic eyes. In order to foveate diverging light she is forced to accommodate relatively more than a normal eye during near vision. Thus, the ciliary muscles are fatigued, and she gets eyestrain.

2. The accommodative power is calculated as follows: 1/F – 1/N = 4 – (-3) = 7 D. This result is exactly the same as that of a normotropic person of 30 years.

3. A fine text can be read further away and until the near point, 1/3 m in front of the eyes. Within the range of 0.2 m, she cannot obtain a sharp picture on the retina (foveation).

4. Hypermetropic persons usually have too short eyeballs – the saggital diameter is too short.

5. The fluid outflow at the iridocorneal junction is reduced, because such flat eyes implies a small chamber angle. This leads to increased intraocular pressure or glaucoma, a frequent disorder in hypermetropic patients

5. Case History B

1. The threshold pressure for a healthy person is 20 mPa or 2×10^-5 Pa (Po). Thus, the relation is: dB, SPL = 20 log (P/Po).

dB, SPL = 20 log (P/(2 10^-5)). Accordingly the sound pressure in the power plant is 2 Pa, and the ratio is 10⁵.

2. 26 dB, SPL = 20 log (P/(2 10^-5)).

The threshold pressure for the patient is (3.99 10^-4) Pa. This pressure is 20 times as high as that of the normally hearing person.

5. Case History C

1. 1/N = - 4 D or N = -0.25 m. The nearpoint (N) is located 0.25 m in front of her eye (- 0.25 m). The far point (F) is calculated as follows: 1/F = 5 D.

F is located 0.20 m behind her eye.

2. The accommodative power is: 1/F - 1/N = 1/0.20 - (1/-0.25) = 9 D. This is almost as good as normal although the girl has facultative hypermetropia.

3. The patient needs correction with the strongest convex lens for optimal visual acuity related to her far point: + 5 D. Lenses with slightly lesser refractive power would allow her to see sharply as well, but she would have to use her accommodation all the time and probably develop eyestrain.

5. Case History D

1.The accommodative power is 10 D equal to 1/F - 1/N. Accordingly, 1/F = 10 D - 1/0.05m or 10 D - 20 D = - 10 D. This is a concave lens with a refractive power of 10 D. This lens is the weakest concave lens that would refract parallel light beams; so they are directed as if they came from a point 0.1 m in front of the eye (F).

2. A refractive disorder with both N and F located in a finite distance in front of the eye must be myopia.

The patient is not allowed to drive a car without spectacles. The disorder is severe. The patient with - 10 D must be regarded as functionally blind, with a far point 10 cm in front of his eyes

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