loveoforganic

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What kind of lens would you use for each and why? This isn't registering with me.

Thanks
 

sleepy425

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What kind of lens would you use for each and why? This isn't registering with me.

Thanks
k, for nearsightedness (myopia), you use a diverging lens. for farsightedness (hyperopia), you use a converging lens.

here's why:

in a normal eye, the lens of the eye focuses the incoming light so that the image is formed exactly on the retina (since the retina is the part of your eye that detects light). if the image forms in front of the retina, or behind the retina, then you have a problem, and the image will be blurry since it is not focused.

in myopia, the image forms in front of the retina (so basically, the lens focuses the light to some point before it gets to the retina). To correct this, we need to make the image form a little later. Since diverging lenses "spread out" the light (as opposed to focusing the light), they can be used to correct myopia. In other words, say the retina is 2 cm away from the lens of your eye, but the lens of your eye makes the image form at 1.8 cm away. The diverging lens will diverge the image so that it doesn't focus until it has traveled 2 cm from your eye's lens.

in hyperopia, the image forms behind the retina. So we need to have the image form a bit earlier. For example, if we again say that the retina is 2 cm away from your eye's lens, and you're hyperopic so the image forms at 2.2 cm, basically your lens isn't focusing well enough. So you add a converging lens, which focuses the light so that it will form the image at 2 cm instead of 2.2 cm.
 
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loveoforganic

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Awesome, got it, thanks a bunch.

Congrats on the MD/PhD acceptance :thumbup:
 

BerkReviewTeach

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Sleepy's explanation is excellent.

I want to add a different approach to the problem that may also help. If you are one who visualizes ray diagrams well, then this method should help.

With a diverging lens, some of the virtual rays initiate from the virtual focal point, and thus intersect other rays between the f and lens. This means that the image is ALWAYS found between the lens and the virtual focal point. Such an image will always be near to the lens, relative to the position of the object. For someone who is nearsighted, a lens that can take a faraway object and generate a near image will be the ideal corrective lens.

With a converging lens when the object is inside the focal length, a virtual image is generated farther from the lens than the object. In other word, a converging lens can take a near object and generate a faraway image. Such an image will always be far from the lens, relative to the position of the object. For someone who is farsighted, a lens that can take a near object and generate a faraway image will be the ideal corrective lens.

While Sleepy's explanation is better and based on tangible fundamental information, this method can prove useful for people who are good at visualizing ray diagrams. It's quite fast if you can picture it.
 
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loveoforganic

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Sleepy's explanation is excellent.

I want to add a different approach to the problem that may also help. If you are one who visualizes ray diagrams well, then this method should help.

With a diverging lens, some of the virtual rays initiate from the virtual focal point, and thus intersect other rays between the f and lens. This means that the image is ALWAYS found between the lens and the virtual focal point. Such an image will always be near to the lens, relative to the position of the object. For someone who is nearsighted, a lens that can take a faraway object and generate a near image will be the ideal corrective lens.

With a converging lens when the object is inside the focal length, a virtual image is generated farther from the lens than the object. In other word, a converging lens can take a near object and generate a faraway image. Such an image will always be far from the lens, relative to the position of the object. For someone who is farsighted, a lens that can take a near object and generate a faraway image will be the ideal corrective lens.

While Sleepy's explanation is better and based on tangible fundamental information, this method can prove useful for people who are good at visualizing ray diagrams. It's quite fast if you can picture it.
I like the intuitive explanations :) And in both cases, you'd end up with an upright image, which would make sense, seeing as how that's necessary to process it correctly! Thanks
 

sleepy425

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Sleepy's explanation is excellent.

I want to add a different approach to the problem that may also help. If you are one who visualizes ray diagrams well, then this method should help.

With a diverging lens, some of the virtual rays initiate from the virtual focal point, and thus intersect other rays between the f and lens. This means that the image is ALWAYS found between the lens and the virtual focal point. Such an image will always be near to the lens, relative to the position of the object. For someone who is nearsighted, a lens that can take a faraway object and generate a near image will be the ideal corrective lens.

With a converging lens when the object is inside the focal length, a virtual image is generated farther from the lens than the object. In other word, a converging lens can take a near object and generate a faraway image. Such an image will always be far from the lens, relative to the position of the object. For someone who is farsighted, a lens that can take a near object and generate a faraway image will be the ideal corrective lens.

While Sleepy's explanation is better and based on tangible fundamental information, this method can prove useful for people who are good at visualizing ray diagrams. It's quite fast if you can picture it.
oh cool I'd never really thought of it that way. neat!
 

SuperSaiyan3

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another easy way to memorize the correcting lenses is:

FC (far : converging)

ND (near: diverging)



I don't know why, but I find it really easy to remember FC.

put a K in front of it and maybe that's why?

SS3 :luck:
 

stockraider

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myopic-diverging------>MD
hyperopic- converging

since most of us future MD's were nerds who needed glasses to see far distances, this works for me haha
 

wanderer

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k, for nearsightedness (myopia), you use a diverging lens. for farsightedness (hyperopia), you use a converging lens.

here's why:

in a normal eye, the lens of the eye focuses the incoming light so that the image is formed exactly on the retina (since the retina is the part of your eye that detects light). if the image forms in front of the retina, or behind the retina, then you have a problem, and the image will be blurry since it is not focused.

in myopia, the image forms in front of the retina (so basically, the lens focuses the light to some point before it gets to the retina). To correct this, we need to make the image form a little later. Since diverging lenses "spread out" the light (as opposed to focusing the light), they can be used to correct myopia. In other words, say the retina is 2 cm away from the lens of your eye, but the lens of your eye makes the image form at 1.8 cm away. The diverging lens will diverge the image so that it doesn't focus until it has traveled 2 cm from your eye's lens.

in hyperopia, the image forms behind the retina. So we need to have the image form a bit earlier. For example, if we again say that the retina is 2 cm away from your eye's lens, and you're hyperopic so the image forms at 2.2 cm, basically your lens isn't focusing well enough. So you add a converging lens, which focuses the light so that it will form the image at 2 cm instead of 2.2 cm.
This.

Also the change in focal length when focusing at far or near objects occurs due to flattening or rounding of the lens. The lens is a converging lens, and when viewing distant objects, the lens is pretty flat and has a longer focal length. When viewing closer objects the lens becomes more round and focuses light onto the retina. In one type of nearsightedness, the lens doesn't flatten enough for distant objects and so focuses light in front of the retina. To correct this you'd need a diverging lens to spread the light out more so that the object forms on the retina. This doesn't describe every form of nearsightedness, but helps me remember when to use converging and when to use diverging.