Could someone explain this optics problem?

[thechairman]

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When a nearsighted person uses prescription glasses to view an object on the horizon, the image formed by the glasses is:

Answer:

Virtual, and the image is enlarged

 

[saidel273]

To my understanding, a nearsighted individual would be assigned a diverging lens. The image formed by a diverging lens is virtual, upright, and enlarged.

 

[changarang]

i don't know how to explain it in terms of formulas, but in a single-lens system, upright images are always virtual and real images are always inverted (this isn't always true in multiple-lens systems though). the glasses wouldn't be of much use if the person saw things upside down, so it's assumed that he's seeing an upright image, which would be virtual. and since this is a nearsighted person trying to see something far away, the object had to be "brought closer," which would magnify the image.

 

[detvar]

where'd u get that?

i'm fairly sure that images produced by diverging mirrors are smaller, not larger

 

[thechairman]

i don't know how to explain it in terms of formulas, but in a single-lens system, upright images are always virtual and real images are always inverted (this isn't always true in multiple-lens systems though). the glasses wouldn't be of much use if the person saw things upside down, so it's assumed that he's seeing an upright image, which would be virtual. and since this is a nearsighted person trying to see something far away, the object had to be "brought closer," which would magnify the image.

then, how would a person see the image? wouldn't a virtual image be formed on the same side as the object? doesn't the object need to form an image on the other side of the lens?

 

[BrokenGlass]

When a nearsighted person uses prescription glasses to view an object on the horizon, the image formed by the glasses is:

Answer:

Virtual, and the image is enlarged

Being nearsighted means the image forms in front of the retina. In other words, the rays converge too soon. This is corrected by a diverting lens. Diverging lenses always form virtual upright images.

Being farsighted means the image forms behind the retina. In other words, the image doesn't coverge soon enough. This is corrected by a converting lens. Converting lenses form real inverted images if the object is outside the focal length and virtual upright images if the object is inside the focal length.

In a person with normal eyesight, the image forms on the retina, so there is no need for corrective lenses.

But again, I am not an optometrist or an opthalmologist.

 

[changarang]

BrokenGlass is right about what exactly is it that prescription glasses do. at least, that's the same explanation i got. 😛 as for your question about how would a person see an image if it's virtual...we don't see only real images. we can see virtual images as well. "virtual" as applied here doesn't mean that the image doesn't exist, or that it can't be seen. i think "virtual" technically means that light seems to be converging at a place where the light doesn't actually converge, whereas "real" means light is converging at the same place where it seems to be converging.

i'm trying to think of another example of virtual images that we see. the closest thing i can think of is the "objects in mirror are closer than they appear" thing that you see on the side mirrors of cars. i'm not 100% sure if that's an example of a virtual image, but i think it is.

an example of a real image would be the image that forms on camera film. if the light didn't actually converge where the film was, exposing the chemicals there to light, then you wouldn't get an image that you could develop later on.

i hope someone can explain this better. 😳

 

[thechairman]

i think i got it. thanks for clarifying.

i got confused on what a virtual image entails, but i remember that an image in a regular mirror is a virtual image (located on the opposite side from the eye) as well, same as the situation in the glasses.