Optics - near sighted or far sighted

[triplezero24]

A pair of relaxed eyes of a patient have a refractive power of 48.5 diopters.

a) is this person near or far sighted?

I think its far sighted because of the short focal length.

b) Find the near point of this person. (The retina is 2.4cm from the lens).

Ok, so the f=.02 and I'm assuming we supposed to use 1/do + 1/di = 1/f.

I'm just not sure what I'm supposed to find. Which variable represents the length of the retina to the lens? Which one am I finding?

 

[Andrew Mason]

A pair of relaxed eyes of a patient have a refractive power of 48.5 diopters.

a) is this person near or far sighted?
I think its far sighted because of the short focal length.

Are we to assume that this person has retina placement of 2.4 cm behind the lens? If so this person is nearsighted. A refractive power of 48.5 dioptres means that the eye has a focal length of 2.1 cm (f = 1/P in metres). The images from distant locations will focus at the focal length which is in front of the retina. Since the retina is located 2.4 cm behind the lens, the focal length is too short, which means the person is near sighted. For a normal person, the retina is about 1.7 mm from the lens of the eye, so if we assume a normal retina placement, the focus would be behind the retina, in which case the person would be farsighted.

b) Find the near point of this person. (The retina is 2.4cm from the lens).

Ok, so the f=.02 and I'm assuming we supposed to use 1/do + 1/di = 1/f.

I'm just not sure what I'm supposed to find. Which variable represents the length of the retina to the lens? Which one am I finding?

The near point is the distance of the closest object that is in focus (with no accommodation by the eye, so the focal length does not change). So:

\frac{1}{f} = \frac{1}{o} + \frac{1}{i} where f = 2.1 cm and i= 2.4 cm

So \frac{1}{o} = 48.5 - 41.7 = 6.8

Object distance = 15 cm.

AM

 

[thepatientmental]

Based on the given information, it is likely that this person is far sighted. This means that they have difficulty seeing objects up close, but can see objects far away more clearly. The short focal length mentioned in the question supports this assumption, as it indicates that the person's eyes are not able to focus on objects that are close to them.

To find the near point of this person, we can use the formula 1/do + 1/di = 1/f, where do represents the distance of the object from the lens, di represents the distance of the image from the lens, and f represents the focal length. In this case, we are trying to find the near point, which is the closest distance at which the person can see an object clearly. This would be represented by di in the formula.

Substituting the given values, we get 1/do + 1/2.4 = 1/48.5. Solving for di, we get di = 2.47 cm. This means that the person's near point is approximately 2.47 cm away from their eyes, which is quite far compared to the average near point distance of 25 cm. This further supports the assumption that the person is far sighted, as their near point is farther away than normal.