Lenses work by refracting, or bending, light. In order for an object to be in focus, the diverging rays that come off of any point on that object must be brought to a single point of focus on your retina. The amount of refraction for any shaped spherical lens is summarized as Lenz's Law. In other words, any time you are asked to understand image formation by a lens, it is likely to be a geometric optics problem.
When diverging rays from a single point on an object reach your eye, they refract. The shape of your eye and the lens in your eye all act as converging lenses, bringing those rays together so that ideally they cross at your retina. If the rays from a point on the object do not cross to a point at your retina, you see a blurred image because multiple nerves are sending information about that part of the object to your brain. In that case, you either need a lens to help converge the rays (if your eye is unable to converge them enough) or to diverge the rays (if your eye converges them too much) before they reach your eye.
Light typically reflects or is emitted off of objects in all directions. In other words, as you can see by the picture below, diverging rays of light come from every point.
When light enters a new medium, it changes speed. If part of the light wave enters the new material first, it will slow down or speed up before the rest of the wave. As you can see in the picture below, the net effect is that the wave changes direction. The amount by which they change direction depends on two things: the medium itself (the higher the index of refraction, the greater the speed change) and the shape of the interface.
In this problem, you are not told the shape of the lens and so you need to draw a picture to visualize the situation. Begin with a picture of nearby object that you are able to focus.
If you now draw a more distant object you can see that the rays that reach your eye from that object are less divergent (they have a smaller angular spread.) Therefore, your eye has a problem of converging rays too much.
To correct your vision, then, you need a diverging, or concave, lens to diverge light rays further before they reach your eye.
With this picture and understanding, you have already answered the first question--you need a diverging, or concave, lens to correct your vision.
Now that you know the shape of the lens, you can draw a ray diagram to help you further understand the relationship between the distant object and image formed by the lens.
You can't draw the diagram to scale without knowing focal length, but you can still use the ray diagram to get a better visualization of the situation. Just draw two focal points equi-distant from the lens, one on each side, to be able to visualize what happens to the light rays as they reach the lens.
It is true that in terms of equations this is just a quick "plug and chug" problem once you know the shape of the lens. But it is well worth your time to draw the principle rays for two reasons. The first is pragmatic--your instructor will likely require you to draw principle rays on your exam. The second relates to learning. By drawing the principle rays you have a strong visual to help you understand the physics of how the image is formed. That visual will also help you catch common errors in your math--for example, using an incorrect sign for image distance or focal length.
Your brain is a pretty remarkable organ and quickly learns to interpret images. If you ever got a new pair of glasses with a very different prescription from your previous pair, you probably noticed that you felt disoriented at first. That is because your brain was relearning depth in response to the new image locations. You can see for yourself that you are actually looking at an image in two ways. First, you can hold your glasses away from your face so that you are looking at an object partially through your glasses and partially without. If your correction is bad enough that this experiment is too blurry, look in the mirror. The part of your face behind your glasses likely appears smaller than the rest of your face. You are seeing the image of your eyes behind your glasses.
Rays spread out off the object in all directions. However, your pupil is of a fixed size and so only intercepts a small set of those rays. As you can see in the picture, if you are very close to an object that fixed size of your pupil intercepts a greater angular spread of rays than it does if you are far from the object.
Your eyeball itself is a convex shape and filled with fluid with an index of refraction greater than that of air. As a result, your eyeball is a lens that causes rays to converge. The lens in your eye can change shape and acts to fine tune the amount of convergence. If you are nearsighted, even with the lens stretched as flat as it can go, if the incoming rays aren't sufficiently divergent your eyeball will converge them too strongly for the shape of your eye and they will cross before they reach the retina.
As always, the image location is the place where the rays appear to originate. In other words, where rays from one point on the object appear to have come from a single point. To find the image location, trace the rays that get to your eye straight back to their point of apparent convergence (as shown by the dotted lines.)
If you wear glasses, you may not feel like you look at images all the time. But note that your brain is a pretty remarkable organ and quickly learns to interpret images. If you ever got a new pair of glasses with a very different prescription from your previous pair, you probably noticed that you felt disoriented at first. That is because your brain was relearning depth in response to the new image locations. You can see for yourself that you are actually looking at an image in two ways. First, you can hold your glasses away from your face so that you are looking at an object partially through your glasses and partially without. If your correction is bad enough that this experiment is too blurry, look in the mirror. The part of your face behind your glasses likely appears smaller than the rest of your face. You are seeing the image of your eyes behind your glasses.
The easiest principle ray to draw is the ray that goes straight from the object to the center of the lens and continues in the same direction. It is nice to draw this ray first because it can help you catch mistakes. For example, if you used the wrong focal point for the other rays, you would see them converge rather than diverge and would catch your mistake.
The definition of focal point is the point at which a ray initially traveling parallel to the axis crosses that axis. There are two focal points for lenses. For a concave lens, rays go out through the focal point on the near side of the lens (from the object.) To draw this principle ray, first draw a line from the object to the lens parallel to the optical axis. Then draw a guide line for the outgoing ray--if you aim from the near focal point to the place that the ray hits the lens, the outgoing ray will continue along that line of sight.
It can be confusing to remember that this ray goes out through the near focal point. You can always check yourself if you don't remember for sure. In air, a concave lens acts to diverge the rays--and so the angular separation between the outgoing rays should be greater than the angular separation between the incident rays.
For a concave lens, a ray that goes out parallel to the optical axis comes in to the lens aiming towards the focal point on the far side of the object from the lens. The blue dotted line shows how you should aim the incoming ray.
One common mistake is to use the wrong focal point. Remember that for a concave lens, the "in" focal point is on the oppose side of the lens as the object, and the "out" focal point is on the same side of the lens as the object.
There are two equations that summarize our understanding of geometric optics:
The first equation locates the image for a given lens some distance o from the object, and the second relates the size of the object to the size of the image. In this case, you are asked for the focal length given information about image and object locations and so will use the first relationship.
In this equation, o refers to the distance between the object and the cener of the lens. i refers to the distance between the imge and the center of the lens. Some books use s and s' respectively, and others use p and q. All refer to the same understanding. I like o and i because they name the concept more directly.
o is the distance between an object and the center of the lens. When light passes from the object to the lens to your eye, o is positive. Note that this is almost always the case. Negative values for o may arise, however, in situations with multiple lenses.
i is the distance between the image (where the object appears to be) and the center of the lens. When the lens is between the image and your eye, i is positive and we refer to this as a real image. When the image and object are on the other side of the lens, as is the case here, i is negative and we refer to this as a virtual image.
To correct your vision, you need a lens with a focal point of -1.3 m.
In this case, yes. If you recognize that for an object at infinity the focal distance is the same as the image distance (definition of focal length) that is great. Make sure to fully understand that definition and why you can take the object distance to be infinity. You may also want to use the steps in this problem to understand other single lens problems where o is less than infinity.
Be careful not to read the statement about focusing on an object as being the object distance. The problem tells you that you cannot focus any further away than 1.3 m. This means that all images must be located within 1.3 m.
The image is a virtual image (the rays do not actually cross at the image location and it is on the same side of the lens as the object.) i is negative for virtual images.
Your glasses need to allow you to see objects that are very far away. If you take the distance to the object to be infinity, the lenses will correct for any distant object.
1 divided by infinity equals 0.
Concave lenses have negative focal lengths. The math and your picture (and conceptual understanding) are all in agreement.
Any time you are asked to understand what you see through a spherical lens or mirror, it is likely to be a geometric optics problem. Ray diagrams help you to visualize how the lens bends the light and how your brain understands those light rays, and also to catch mathematical mistakes. In this case, you are able to focus on close objects (highly divergent rays) but not on distant ones (less divergent rays) and so you understand that your eye converges rays too much. A lens that causes rays to diverge (a concave lens) is therefore needed to correct your vision. Both your understanding and your math agree, because concave lenses have negative focal points.
The definition of focal point is the point at which rays coming from infinity converge and cross the optical axis. Therefore, the focal distance to correct nearsightedness should, indeed, be the far point of the eye.
The far point is the furthest distance at which your eye can focus.
Although numbers are not given specifically, with a little understanding of the situation you can figure them out. If your eyes can only focus out to 1.3 m, the image from all objects more distant than that must be located within your region of focus. The further away an object is, the further away its image will be. Therefore, if you want to be able to see as far as possible you can approximate the object distance to be infinity and its image at the farthest point of focus--1.3 m.
Draw a picture! In this case, the first step is to figure out the lens shape. A picture can help you visualize the rays and what needs to happen in order for your eyes to focus.
Yes. If you recognize that for an object at infinity the focal distance is the same as the image distance (definition of focal length) that is great. Make sure to fully understand that definition and why you can take the object distance to be infinity. You may also want to use the steps in this problem to understand other single lens problems where o is less than infinity.