Help Understanding Focal Lengths

[aspiringdoc09]

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Sorry if this is posted elsewhere, this link was the only thing I found: http://forums.studentdoctor.net/showthread.php?t=629547

I am having a hard time understanding conceptually how the focal lengths of converging/diverging lenses and concave/convex mirrors work?

I don't understand how to quickly answer questions about them if asked.

I understand that concave (+f), convex (-f), converging (-f) and diverging (+f) but that is it.

I am working on EK 1001 #976 and it made me recognize that I don't understand this topic very well.

Please help.

 

[SaintJude]

Don't have EK 1001 Physics, so can't help you with that problem unless you post it.

One thing I will say is that focal lengths really depend upon understanding how ray diagrams work. I do know that EK says it's a waste of time to draw them and they are right. But in order to understand where the focal point will converge, one still has to be able to generally predict how the rays will look like. Which is why still some test preps, like Kaplan, recommend understanding them.

Concave lenses and convex mirrors both have diverging properties (-f)
While convex lenses and concave mirrors have converging properties (+f). Is that obvious to you?

 

[aspiringdoc09]

Don't have EK 1001 Physics, so can't help you with that problem unless you post it.

One thing I will say is that focal lengths really depend upon understanding how ray diagrams work. I do know that EK says it's a waste of time to draw them and they are right. But in order to understand where the focal point will converge, one still has to be able to generally predict how the rays will look like. Which is why still some test preps, like Kaplan, recommend understanding them.

Concave lenses and convex mirrors both have diverging properties (-f)
While convex lenses and concave mirrors have converging properties (+f). Is that obvious to you?

Yes, I understand that much.

Questions 976-982 refer to the figure. It gives a figure with two converging lenses (combinations). Object is x distance in front of lens 1 and both lenses are l distance apart.

Q976. If x is less than f1 and l is greater than f2, then the final image must be:
a. virtual and upright.
b. virtual and inverted.
c. real and upright.
d. real and inverted.

If I would answer based on what I understand with general rules than I will choose A because convex lenses and convex mirrors can only make virtual and upright images. But I don't understand all the rules with combination lenses either. The image of the 1st would be virtual and upright but does that mean the other lens does the opposite to make D? which is the answer in the back. Should I always expect the 2nd/last mirror in the combination to make the opposite image of the one prior to it? Is the information about the object being less than f1 and l greater than f2 extraneous info?

 

[MedPR]

Sorry if this is posted elsewhere, this link was the only thing I found: http://forums.studentdoctor.net/showthread.php?t=629547

I am having a hard time understanding conceptually how the focal lengths of converging/diverging lenses and concave/convex mirrors work?

I don't understand how to quickly answer questions about them if asked.

I understand that concave (+f), convex (-f), converging (-f) and diverging (+f) but that is it.

I am working on EK 1001 #976 and it made me recognize that I don't understand this topic very well.

Please help.

Converging lenses have positive focal lengths, diverging have negative focal lengths. Likewise, a concave lens (which is a diverging lens) has a negative focal length while a convex lens has a positive focal length.

 

[MedPR]

Yes, I understand that much.

Questions 976-982 refer to the figure. It gives a figure with two converging lenses (combinations). Object is x distance in front of lens 1 and both lenses are l distance apart.

Q976. If x is less than f1 and l is greater than f2, then the final image must be:
a. virtual and upright.
b. virtual and inverted.
c. real and upright.
d. real and inverted.

If I would answer based on what I understand with general rules than I will choose A because convex lenses and convex mirrors can only make virtual and upright images. But I don't understand all the rules with combination lenses either. The image of the 1st would be virtual and upright but does that mean the other lens does the opposite to make D? which is the answer in the back. Should I always expect the 2nd/last mirror in the combination to make the opposite image of the one prior to it? Is the information about the object being less than f1 and l greater than f2 extraneous info?

The image of the first lens becomes the object of the second lens. So you are correct about the first image being virtual and upright (to the left of lens1). Now you have to make this image the object for lens 2. Since l is greater than f2, object2 is outside of the focal length of f2. Then you use the thin lens equation again and you find that the image distance is a positive number, which makes it a real image. Then you use the magnification equation m=-di/do and you get a negative magnification number, which means that the image is inverted.

 

[aspiringdoc09]

Converging lenses have positive focal lengths, diverging have negative focal lengths. Likewise, a concave lens (which is a diverging lens) has a negative focal length while a convex lens has a positive focal length.

Are you sure about this? TPRH says that all things remain the same except for converging lenses (convex) have negative focal lengths and diverging lenses (concave) have positive focal lengths. Also, where the image forms (i.e. same side/opposite to object) differs when you look at lenses. I don't have my book on me to check.

 

[MedPR]

Are you sure about this? TPRH says that all things remain the same except for converging lenses (convex) have negative focal lengths and diverging lenses (concave) have positive focal lengths. Also, where the image forms (i.e. same side/opposite to object) differs when you look at lenses. I don't have my book on me to check.

Yes. Diverging lenses and mirrors always have a negative focal length.

 

[aspiringdoc09]

The image of the first lens becomes the object of the second lens. So you are correct about the first image being virtual and upright (to the left of lens1). Now you have to make this image the object for lens 2. Since l is greater than f2, object2 is outside of the focal length of f2. Then you use the thin lens equation again and you find that the image distance is a positive number, which makes it a real image. Then you use the magnification equation m=-di/do and you get a negative magnification number, which means that the image is inverted.

Honestly, that doesn't seem like an MCAT way to solve. I'm not saying you're wrong because you're not. However, if I had to answer a question like this on the MCAT, it seems time-consuming, especially when you throw in the equations. Another thing I don't understand is how do they make an image formed to the left on lens 1 the object of lens 2? It would make more sense if the image was formed to the right of lens 1.

 

[MedPR]

Honestly, that doesn't seem like an MCAT way to solve. I'm not saying you're wrong because you're not. However, if I had to answer a question like this on the MCAT, it seems time-consuming, especially when you throw in the equations. Another thing I don't understand is how do they make an image formed to the left on lens 1 the object of lens 2? It would make more sense if the image was formed to the right of lens 1.

You're right. The quicker way to solve is to know the patterns, but if you don't know the patterns you have to do it the long way (as above). If you know the patterns, you know that for a converging lens, if the object is inside the focal length, then the image it makes is virtual and upright. You also know that, for lenses, virtual means to the left whereas for mirrors virtual means to the right. Moving right along, you know that for a multilens system, i1=o2. And since they tell you that l is greater than f2, and you just found out that i1 is to the left of lens1, then you know that o2 is outside the focal length of lens2. If you know the patterns, you know that for a converging lens when the object is outside the focal length, a real inverted image is formed.

In not so many words:

2 lens system
o1 < f1, thus i1=virtual, upright, larger
i1=o2
o2 > f2, thus i2=real, inverted, larger
done

Edit: The point here is that if you don't know the shortcuts, you are stuck with doing it the long way. Alternatively, the long way for lens problems really doesn't take that long and, even though I know the patterns, I sometimes prefer to do it the long way just to be completely sure. If I don't have time, I'll refer to the shortcuts.

 

[aspiringdoc09]

Yes. Diverging lenses and mirrors always have a negative focal length.

I think you are right. I am confusing myself. I think when they said that diverging lenses act like concave mirrors I automatically decided to remember positive focal length for both. You are right! I'm stupid. They gave me that little piece of information and I ran in the wrong direction with it. LOL! Thanks.

 

[aspiringdoc09]

You're right. The quicker way to solve is to know the patterns, but if you don't know the patterns you have to do it the long way (as above). If you know the patterns, you know that for a converging lens, if the object is inside the focal length, then the image it makes is virtual and upright. You also know that, for lenses, virtual means to the left whereas for mirrors virtual means to the right. Moving right along, you know that for a multilens system, i1=o2. And since they tell you that l is greater than f2, and you just found out that i1 is to the left of lens1, then you know that o2 is outside the focal length of lens2. If you know the patterns, you know that for a converging lens when the object is outside the focal length, a real inverted image is formed.

In not so many words:

2 lens system
o1 < f1, thus i1=virtual, upright, larger
i1=o2
o2 > f2, thus i2=real, inverted, larger
done

Edit: The point here is that if you don't know the shortcuts, you are stuck with doing it the long way. Alternatively, the long way for lens problems really doesn't take that long and, even though I know the patterns, I sometimes prefer to do it the long way just to be completely sure. If I don't have time, I'll refer to the shortcuts.

Thanks MedPR. I will have to learn the shortcuts. How did you go about learning them or is physics your strong suit. It's my worst subject. Also, concave mirrors can make real or virtual images. Does that mean that the focal lengths, whether the object is within or not affect whether it is real of virtual?

 

[MedPR]

I think you are right. I am confusing myself. I think when they said that diverging lenses act like concave mirrors I automatically decided to remember positive focal length for both. You are right! I'm stupid. They gave me that little piece of information and I ran in the wrong direction with it. LOL! Thanks.

Diverging lenses act like convex mirrors, not concave mirrors.

 

[aspiringdoc09]

Diverging lenses act like convex mirrors, not concave mirrors.

All my reading was a LIE! They said that the diverging lens may be referred to as concave while the converging lens may be referred to as convex. Now, I am completely confused. I thought I understood but I guess not. I HATE PHYSICS!

 

[MedPR]

All my reading was a LIE! They said that the diverging lens may be referred to as concave while the converging lens may be referred to as convex. Now, I am completely confused.

Diverging lens = concave
Converging lens = convex

Diverging mirror = convex
Converging mirror = concave

Mirrors: image and object on the same side = real image
Lenses: image and object on same side = virtual image