Lens Question

[Dr Gerrard]

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This is from the EK Physics study guide, page 157, problem 189.

A lens is manufactured in such a way as to allow the object and the image to be at the same distance from the lens. If the lens is not flat, the only way this could be true is if the lens were:

A. a diverging lens with the object at the focal distance.
B. a diverging lens with the object at twice the focal distance.
C. a converging lens with the object at the focal distance.
D. a converging lens with the object at twice the focal distance.


I thought the answer was B. The answer is D.

However, I do not understand why, in terms of the equation 1/f = 1/di + 1/do ---- where di is distance to image and do is distance to object.

If it were a converging lens, that would mean that the object and the image were on opposite sides. If they both were the same distance from the image, than one would be positive while the other would be negative, cancelling each other out.

 

[Charles_Carmichael]

This is from the EK Physics study guide, page 157, problem 189.

A lens is manufactured in such a way as to allow the object and the image to be at the same distance from the lens. If the lens is not flat, the only way this could be true is if the lens were:

A. a diverging lens with the object at the focal distance.
B. a diverging lens with the object at twice the focal distance.
C. a converging lens with the object at the focal distance.
D. a converging lens with the object at twice the focal distance.


I thought the answer was B. The answer is D.

However, I do not understand why, in terms of the equation 1/f = 1/di + 1/do ---- where di is distance to image and do is distance to object.

If it were a converging lens, that would mean that the object and the image were on opposite sides. If they both were the same distance from the image, than one would be positive while the other would be negative, cancelling each other out.

Positive and negative image distance depend on which side of the lens the image appears. For a converging lens, if the image appears on the right side of the lens (which it will always, unless the object is between the lens and the focal point), it's a real image and the distance is positive. If the image is virtual (in the case of converging lenses, the image will appear on the left side), then the image distance is negative. Also, remember that the focal length is negative for a diverging lens. For diverging lenses, the image is always virtual.

For the question, I'm thinking they're asking about real images, so it can't be the virtual image then. When solving for the answer using the lens equation, substitute do for di since they're the same distances. So your equation will look like this:

1/do + 1/di = 1/f
1/d0 + 1/do = 1/f
2/do = 1/f
do = 2f

So, you now know that the object has to be at a distance of 2f away for the image to have the same distance. Once again, I think they're asking for real images and that's why the answer is D (since diverging lenses can never have a real image). Hope this helps.

 

[Dr Gerrard]

So can I always just assume that no matter what happens, unless the object is within the focal length for a converging lens or a concave mirror, the image and the object distance will ALWAYS be the same sign?

While a diverging lens and a convex mirror will always have the image and object distance to have the opposite signs?

Because if the above were true, that problem would make sense.

I guess I just didn't realize that sign is not determined by what side the object is on, but rather is something that is just assigned to fit the magnification equation.

 

[Charles_Carmichael]

So can I always just assume that no matter what happens, unless the object is within the focal length for a converging lens or a concave mirror, the image and the object distance will ALWAYS be the same sign?

While a diverging lens and a convex mirror will always have the image and object distance to have the opposite signs?

Because if the above were true, that problem would make sense.

I guess I just didn't realize that sign is not determined by what side the object is on, but rather is something that is just assigned to fit the magnification equation.

Yup! Same side of mirror = real image, while behind mirror = virtual image. Behind lens = real image, while in front of lens = virtual image. And diverging lens/mirrors always produce virtual images. If the image is virtual, the sign is negative. That's why the diverging answer choice doesn't work for this problem. Hope this helps! 🙂

 

[LaTuaCantante]

Bump.

I'm confused about this question too, and I don't really understand the explanations given above, so I'm hoping someone might be able to shed some more light? It would be much appreciated! 🙂

Here's how I approached the problem:

- I eliminated A and C immediately - It's pretty clear that the distance is going to be 2f

For a converging lens
i) The focal point is on the left side (i.e. the opposite side from the object, the positive side), and the object is on the negative side.
ii) If the object is between the pseudo focal point and the lens, the image will be on the same side as the object (i.e. the negative side)
iii) If the object is farther from the lens than the pseudo focal point, the image will be on the opposite side (i.e. the positive side)

In this case, d=2f, so we know that iii) is correct, and not ii). Therefore, d(object) is negative and d(image) is positive .
This would mean that (1/f) = 0, which is not possible.

For a diverging lens
i) The focal point is on the right side (i.e. the same side from the object, the negative side)
ii) Regardless of where the object is in relation to the focal point, the image will be on the same side as the object (i.e. the negative side)

In this case, both d(object) and d(image) are negative.
This would mean that -|1/d(object)|-|1/d(image)| = -|1/f|, which is consistent with the focal point being on the right side of the lens.


Am I misunderstanding the material, or is my logic sound?

Thanks, everyone!