TPR Optics

[SoulinNeed]

If a certain eye can only focus on objects at least 50 cm away, which of the following lenses in front of the eye, would allow it to focus on an object 25 cm away?

A. A converging lens with focal length 17 cm
B.A diverging lens with focal length 17 cm.
C. A diverging lens with focal length 25 cm
D A diverging lens with focal length 50 cm

The answer is D, and I get the math behind it, but why a diverging lens? Isn't this situation a case of farsightedness? Shouldn't a converging lens be used? Also, the math indicates that the focal length is a positive 50 cm, and doesn't that mean a converging lens?

the math is 1/f= 1/25 + 1/-50

which leads to 2/50 - 1/50

and thus focal length is 50 cm.

Thanks for any help.

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[MedPR]

If a certain eye can only focus on objects at least 50 cm away, which of the following lenses in front of the eye, would allow it to focus on an object 25 cm away?

A. A converging lens with focal length 17 cm
B.A diverging lens with focal length 17 cm.
C. A diverging lens with focal length 25 cm
D A diverging lens with focal length 50 cm

The answer is D, and I get the math behind it, but why a diverging lens? Isn't this situation a case of farsightedness? Shouldn't a converging lens be used? Also, the math indicates that the focal length is a positive 50 cm, and doesn't that mean a converging lens?

the math is 1/f= 1/25 + 1/-50

which leads to 2/50 - 1/50

and thus focal length is 50 cm.

Thanks for any help.

I think you should do 1/f=1/50-1/25 = 1/50-2/50=-1/50 f=-50?

 

[MrNeuro]

yeah i got converging lens...the closer an object gets to your eye the further the image and the image will be produced behind the macula (focus/nerve of eye?)....so you'd need a converging lens to bend the light more or have your ciliary bodies contract more to bend this light....

and in terms of the math its

1/f=1/i+1/o

i=image
o=object

you don't know the distance to the image..... how do you solve for focal length w/o an image distance??? what am i missing here..

what question # and book is this from?

 

[ljc]

yeah i got converging lens...the closer an object gets to your eye the further the image and the image will be produced behind the macula (focus/nerve of eye?)....so you'd need a converging lens to bend the light more or have your ciliary bodies contract more to bend this light....

and in terms of the math its

1/f=1/i+1/o

i=image
o=object

you don't know the distance to the image..... how do you solve for focal length w/o an image distance??? what am i missing here..

what question # and book is this from?

You do know the distance of the image: it must be 50cm.

Math:

1/f=1/p+1/i
You know p=25cm and i=50cm, but since the object is on the opposite side of the eye, it must be a negative object, so you get:
1/f=1/(-25)+1/50
...=-0.04+0.02
...=-0.02
1/f=-0.02
f= -50cm

Since it's a negative focal length, you know it's a diverging lens. Choice D.

 

[MrNeuro]

You do know the distance of the image: it must be 50cm.

Math:

1/f=1/p+1/i
You know p=25cm and i=50cm, but since the object is on the opposite side of the eye, it must be a negative object, so you get:
1/f=1/(-25)+1/50
...=-0.04+0.02
...=-0.02
1/f=-0.02
f= -50cm

Since it's a negative focal length, you know it's a diverging lens. Choice D.

edit: i was wrong if you want to see my incorrect reasoning look below...

 

[SoulinNeed]

I'm giving you guys the answer explanation that TPR gives. The focal length is 50 cm, not -50 cm.

It says that the image would be on the same side of the lens as the object, so it's virtual, so it's -50 cm.

So,

1/f= 1/25 + 1/-50

1/f= 2/50 - 1/50

1/50, so f= 50 cm.

I think it's just a terrible question. For hyperopia, which this is clearly the case (AT LEAST 50 cm away), it should be converging, which is what the math suggests. But yet, the answer says diverging.

 

[MrNeuro]

how is the image 50 cm away doesn't it need to be located inside the eye??

what am i missing here

this is how i interpreted the question

a normal eye can focus at 50 cm (so object min distance to eye is 50 cm)
what kind of lens is required to allow the eye to focus at 25 cm (so what lens is needed to focus on an object 20 cm from the eye)

is that right? if so how is the image at 50 cm???


ahhhhh i think i get it now when it says the whole 50 cm spiel its saying that the focal length of the lens is 50 cm hence the image needs to form at 50 cm within the eye (positive side) in order for you to see the object........aaaaaahhhhh i got it

so tell me if im working through this correctly

b/c the focal length of the eye is 50 cm the a real inverted image must form at 50 cm so we know the distance of the image must be 50 cm

the object is now < focal length (50 cm) so it will create a virtual upright image and because its virtual the value for the image will be (-25 cm)

1/f = 1/-25 + 1/50 = -1/50 focal length must be -50 cm and the only lenses that have negative focal lengths are diverging lenses

You do know the distance of the image: it must be 50cm.

Math:

1/f=1/p+1/i
You know p=25cm and i=50cm, but since the object is on the opposite side of the eye, it must be a negative object, so you get:
1/f=1/(-25)+1/50
...=-0.04+0.02
...=-0.02
1/f=-0.02
f= -50cm

Since it's a negative focal length, you know it's a diverging lens. Choice D.

I think you should do 1/f=1/50-1/25 = 1/50-2/50=-1/50 f=-50?

The above is wrong...i think i got it. It has to be converging w/ a focal length of 50 heres the math

object at 25
focal length at 50

1/50= 1/25 + 1/i
1/i = 1/50 - 1/25
1/i = - 1/50 (makes sense because the Object < Focal length would yield a virtual upright enlarged image)

image at -50
object at 25

1/f = 1/25 - 1/50
1/f= 1/50

focal length is 50 and it must be a converging lens as this is a common case of farsightedness where the image focus is conventionally behind the focus of the retina so in order to bring towards the retina must bend the light using a converging lens

I think its a typo where D was meant to be

A converging lens with a 50 cm focal length

I'm giving you guys the answer explanation that TPR gives. The focal length is 50 cm, not -50 cm.

It says that the image would be on the same side of the lens as the object, so it's virtual, so it's -50 cm.

So,

1/f= 1/25 + 1/-50

1/f= 2/50 - 1/50

1/50, so f= 50 cm.

I think it's just a terrible question. For hyperopia, which this is clearly the case (AT LEAST 50 cm away), it should be converging, which is what the math suggests. But yet, the answer says diverging.

i agree i think its a typo

 

[SaintJude]

Man, this terminology still trips me up.

So for a lens, if the image is in front of the eye (on the side of the object), it is negative and virtual? And that's why TPR's designation for 1/i = - 1/i?

And for a mirror, if the image is on the same side of the object, it is real and positive? Why is that? Just "cause"?

 

[SoulinNeed]

Man, this terminology still trips me up.

So for a lens, if the image is in front of the eye (on the side of the object), it is negative and virtual? And that's why TPR's designation for 1/i = - 1/i?

And for a mirror, if the image is on the same side of the object, it is real and positive? Why is that? Just "cause"?

It has to do with the fact that a lens is two mirrors (convex or concave) essentially put together, I believe. For the sake of the MCAT, however, just cause is probably good enough.

 

[milski]

That's annoying, I would have answered A since it's the only converging one and would not have bothered with the math. 👎

 

[Dasypus]

Hmmm... according to Wikipedia (http://en.wikipedia.org/wiki/Corrective_lens#Sphere_component) and this site here (http://physics.info/lenses/) hyperopia (farsightedness, the inability to see up close) is fixed with converging lenses. So I'd certainly have gone with milski's reasoning.

But as people have said, the numbers work out to |f|=50 cm.

I'm voting for misprint.

Edit: the way I always thought about real/virtual images with lenses versus mirrors was about what happens when you go stand where the image is. If it's a real image, looking back at your optical instrument will let you see the object. So images in a mirror that look like they're on the other side (you know, like anything at all in a plane mirror) are virtual, because when you go stand there, all you see is the back side of the mirror. Same with images that form on the same side of the lens as the object: look through the lens from where the image forms, and there's nothing there, just ... I don't know, whatever's off the page on a physics diagram. So it's a virtual image. Real = you can stand there and see the source.