How Does Lens Power Correction Work for Nearsightedness?

[sixpence]

I'm having trouble deciding what to use as the object, p, and the image, q. For example here's a question:

An individual is nearsighted; his near point is 13cm and far point is 50cm. What lens power is needed to correct his nearsightedness? When lenses are in use, what is this person's near point?

so i got 1/infinity + 1/-.5 = 1.f

1/-.5 = 1/-.13 +1/q

these are the right answers but they don't make sense to me. How would the focal length from the far point have anything to do with wearing glasses. I also don't know when to make things negative or keep it positive, but that's mainly b/c I don't know what's the image and what's the object. I don't know if the questions are just worded poorly but I can't seem to get them. Is the near point always the object or can it be object and image? This is frustrating meeeeeee

 

[G01]

The near point is the closest something can be so that the person can still see it clearly. The far point is the farthest away something can be so that the person can still see it.

Since the person can't see any farther away than the far point, we want to make glasses from lens that will take objects which are really far away (object distance =infinity) and put the image at the far point.

So, the object distance in this problem is infinity. The image point will be the near point. Does this clear things up?

 

[astrokreng]

I understand your confusion with the concepts of optics and the use of the terms object and image. Let me help clarify these for you.

First, let's define some terms. Optics is the branch of physics that deals with the behavior and properties of light, including its interactions with matter and the formation of images. The far point and near point are both related to the concept of the human eye's ability to focus on objects at different distances.

The far point is the farthest distance at which a person with normal vision can see objects clearly without any assistance, such as glasses or contact lenses. This distance is typically around 6 meters or 20 feet. The near point, on the other hand, is the closest distance at which a person with normal vision can see objects clearly without any assistance. This distance is typically around 25 centimeters or 10 inches.

In the question provided, the individual is nearsighted, which means they have trouble seeing objects that are far away. This is because the images of distant objects are being focused in front of the retina instead of directly on it. The near point of this individual is 13cm, meaning they can only see objects clearly when they are 13cm or closer to their eyes. The far point, in this case, is 50cm, which means they can only see objects clearly up to a distance of 50cm without any assistance.

Now, let's talk about lenses. A lens is a curved piece of glass or other transparent material that can refract (bend) light rays. In the case of nearsightedness, a concave lens is needed to correct the vision. This type of lens diverges light rays, causing them to focus on the retina instead of in front of it.

To determine the lens power needed to correct nearsightedness, we can use the lens formula:

1/f = 1/p + 1/q

Where f is the focal length of the lens, p is the distance of the object from the lens, and q is the distance of the image from the lens.

In this case, the object (p) is the near point of the individual, which is 13cm. The image (q) is the distance at which the light rays coming from the near point will be focused on the retina, which is the far point of 50cm.

Plugging these values into the lens formula, we get:

1/f =