Question

a patient undergoes a procedure of laser vision correction that reduces the power of her relaxed eye by 7.00 d, producing normal distant vision for her (that is, her new far point is at infinity.). assuming her lens-to-retina separation in her eye is 2.00 cm, calculate what the far point of her eye was prior to the procedure.

Answer 1

The original far point of a patient who underwent a laser vision correction reducing their eye's power by 7.00 D was 17.5 cm. This was calculated using the lens power formula, considering the normal distance vision power of 50.0 D plus the 7.00 D reduction.

Step-by-step explanation:

The previous far point of a patient who had laser vision correction can be calculated using the formula for lens power, which is P = 1/f, where P is the power in diopters and f is the focal length in meters. The procedure reduced the power of the patient's eye by 7.00 D, which implies that the eye's ability to focus on distant objects improved, and the far point became infinitely far away (normal distant vision).

To find the original far point, we can use the reduced power to calculate the original focal length.

The power of the eye for normal distance vision is considered to be around 50.0 D.

Since the power was reduced by 7.00 D, the original power was 50.0 D + 7.00 D = 57.0 D. Using the formula, the original focal length f would be f = 1/P.

Therefore, f = 1/57.0 D = 0.0175 meters or 17.5 cm.

The patient's original far point was 17.5 cm before the laser vision correction procedure.