Question

In a LASIK vision correction, the power of a patient’s eye is increased by 3.00 D. Assuming this produces normal close vision, what was the patient’s near point before the procedure?

a) 33.3 cm
b) 25.0 cm
c) 50.0 cm
d) 66.7 cm

Answer 1

The patient's near point before the LASIK procedure was 25.0 cm. Thus option B is correct.

Step-by-step explanation:

The near point is the closest distance at which an eye can see an object clearly. In this scenario, the power of the patient's eye was increased by 3.00 D through LASIK surgery to achieve normal close vision. To find the patient's near point before the procedure, we can use the lens formula:

Lens formula:
`(1)/(f)`=
`(1)/(d_(0) )` +
`\frac{1}{d{i} }`

Here, f represents the focal length of the eye's lens
`d_(o)`, is the object distance, and
`d_(i)` is the image distance.

The formula for the lens power (P) in terms of focal length (f) is: (P =
`(1)/(f)` )

Given that the power change after LASIK surgery is 3.00 D and assuming the eye now has normal close vision, we can use the formula for lens power to find the original power
`P_(before)` of the eye:

`P_(before)` =
`P_(after)` +
`P_(added)`

`P_(before)` = 0 D + 3.00 , D = 3.00 D

Now, using the formula for lens power to focal length conversion P=
`(1)/(f)`, we can find the original focal length (
`f_(before)`) of the eye's lens:

`f_(before)` =
`(1)/(P_(before) )` =
`(1)/(3.00 D)` = 0.333m= 33.3 cm

Therefore, the patient's near point before the LASIK procedure, assuming normal close vision after the power increase of 3.00 D, was at a distance of 25.0 cm, which is the reciprocal of the original focal length
`f_{before`. Thus option B is correct.