Focal Length/Focal Point

[Dr Gerrard]

I cannot conceptualize this.

I never learned any of this in AP physics or college physics.

EK is just not clicking for me.
When does focal point happen?

I look at the diagrams in EK (149 Physics Book, if someone could look at these, it would be extremely helpful).

In those diagrams, none of those rays converge to one point, ever. It looks as though the focal point is just an arbitrary point.

Also, if someone could explain the diagram on 153, that would also be awesome.
What makes those images positive or negative, or virtual or real? Are they virtual because they are being seen through the lens/mirror? 😕😱👎mad:


If you don't have EK, maybe, if possible, you can add any insight which might help this s*** click.

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[Dr Gerrard]

Just got that focal point only happens when the "photons" from the object are parallel to the ground, then it will converge or diverge to the focal point.

Still though, any helpful tips/links out there to help me understand optics as it will be tested on the MCAT better? Thanks.

 

[Klemptor]

I don't have EK but here is just some basic info as I understand it; hope it helps:

The focal length of a lens or mirror is characteristic of that lens or mirror. The focal point sits on the principal axis.

For spherical mirrors, the focal length is equal to half of the radius of the mirror.

For thin lenses, the focal length is found via the lensmaker's equation, namely:

1/f = (n - 1)(1/r + 1/R)

where r and R are the radii of the lenses comprising the thin lens, and n is the index of refraction of the lens. Be careful to note that this equation gives the inverse of the focal length.

Ray diagrams are key for visualizing what happens and for checking your work. When you do a ray diagram, how you use the focal point depends on the type of lens/mirror you have, and I suspect this is where you're running into a problem.

For each ray diagram, you can draw two rays to find the point of convergence (ie the image location). One of the rays can always be drawn as coming from infinity parallel to the principal axis and refracting through the lens in a direction determined by the focal point. Mirrors are simpler and I'm assuming your problem is with lenses, not mirrors, so I'll explain lens ray diagrams here.

- For converging lenses where o > f, you can draw a ray diagram where the first ray comes from infinity parallel to the principal axis and refracts through the lens toward the backside focal point. The second ray comes from the object, passes through the frontside focal point, and refracts to infinity parallel to the principal axis. The image is virtual and inverted.

- For converging lenses where o < f, the situation is more complicated. As before, the first ray comes from inifinity parallel to the principal axis, and refracts through the lens toward the backside focal point. The second ray is drawn as if it comes from the frontside focal point toward the object, and it travels toward the lens and refracts to infinity parallel to the principal axis. The image forms at the intersection of the backed up refracted rays; it is real and upright. (I believe it's enlarged but I'm not certain about that.)

- For diverging lenses (no matter whether o > f or o < f), as before, the first ray comes from infinity parallel to the principal axis; it refracts as if from the frontside focal point. The second ray travels from the object toward the backside focal point, and refracts to infinity parallel to the principal axis. Again, the image forms at the intersection of the backed up refracted rays. The image is real and upright. (I believe it is reduced but I'm not certain on this point.)

Sign conventions vary between lenses and mirrors. In general, light travels from left to right; the "real" side is on the left of the lens/mirror, and the "virtual" side is on the right of the lens/mirror.

Mirror sign conventions: Incident light approaches from the left; reflected right reflects to the left. On the left (the real side), o, i, R, and f > 0. On the right (the virtual side), o, i, R, and f < 0.

Lens sign conventions: Incident light is on the left; refracted light is on the right. On the incident side, o > 0 and i and R < 0. On the refracted side, o < 0 and i and R > 0.


I hope this helps!

 

[Fort]

It's hard to explain geometrical optics without pictures but I'll try to help.

Basically, an image is real if the light that created that image is real; similarly an image is virtual if the light that created that image is virtual. Our brains can't tell the difference between a virtual and a real image, it's just how we interpret the light traveling through space - a straight line.

Try drawing a ray diagram of an object situated about 2f from a concave mirror. You'll see the light rays that converged to generated the image is "real light" or solid lines. Now try drawing a ray diagram for an object at any distance from a convex mirror. You should have both solid lines and dotted lines in your drawing. If you drew it correctly, you'll notice that the solid lines don't converge at all for the convex mirror, but the dotted lines do; this is the image that we see when looking into a convex mirror and it is virtual because it was formed with "virtual light" or dotted lines.

I know for convex mirrors, and concave lenses all the images formed are virtual no matter what the object distance. For concave mirrors and convex lenses, it depends on the object distance.

Play around with this applet I found, it may help you better understand how lenses and mirrors work. The blue lines and green lines represent real and virtual light, respectively.http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=48

Good luck.

 

[Dr Gerrard]

All right, new questions.

1) What does a negative magnification mean?

2) Where are the two distances used for magnification measured from?

3) Do the object and the image have to be on opposite sides in order to have a positive magnification?

4) How does the focal point of an object appear to me? I still know it means that is where light parallel to the ground converges/diverges to after it hits a mirror or lens, but I cannot see what this means in real life.

5) In the case of a diverging lens, why does the image appear upright? This is the picture EK draws: Eye -> lens -> and the image, focal point, and object are all on the same side. Is this upright because the focal point is between the image and object?



Just came up with these questions, haven't read the two responses yet, but they appeared while I was typing these up.

Sorry if these are already answered.

Thank you so much for typing all that, I appreciate it a lot.

 

[Klemptor]

1) What does a negative magnification mean?

A negative magnification indicates an inverted image.

2) Where are the two distances used for magnification measured from?

For lenses, distances are always measured from the center of the lens. For mirrors, distances are always measured from the back of the mirror.

Do the object and the image have to be on opposite sides in order to have a positive magnification?

Magnification is calculated by

M = -i/o = h'/h where h' is the height of the image and h is the height of the object. If both are on the same side of the lens, magnification should be positive. But this says nothing about the relative sizes of object and image. If M < 1, your object is reduced; if M > 1, your object is enlarged.

As for your other two questions, I'm not sure I understand them. Maybe someone else can help?

 

[Dr Gerrard]

A negative magnification indicates an inverted image.



For lenses, distances are always measured from the center of the lens. For mirrors, distances are always measured from the back of the mirror.



Magnification is calculated by

M = -i/o = h'/h where h' is the height of the image and h is the height of the object. If both are on the same side of the lens, magnification should be positive. But this says nothing about the relative sizes of object and image. If M < 1, your object is reduced; if M > 1, your object is enlarged.

As for your other two questions, I'm not sure I understand them. Maybe someone else can help?

Thanks a lot!! Makes sense, except the 3rd one.

So if both objects are on the same side, you say magnification should be positive, but that doesn't make sense according to the equation M=-i/o. The sign of both i and o would be the same if they were on the same side, and thus the magnification would be negative and the image would be inverted, correct?

This is what EK says, but does not appear to correlate with their pictures, let me see if I can find them online somewhere.

Also, the answers to the last two questions would still be helpful, if anyone else knows them.

 

[Klemptor]

So if both objects are on the same side, you say magnification should be positive, but that doesn't make sense according to the equation M=-i/o. The sign of both i and o would be the same if they were on the same side, and thus the magnification would be negative and the image would be inverted, correct?

Remember the sign conventions are different for lenses and mirrors. The sign of both i and o are the same if you're talking about mirrors but for lenses the conventions are different. This means that, for lenses:

- If the object and image are both on the real side, o > 0 and i < 0; therefore, -i/o is positive.

- If the object and image are both on the virtual side, o < 0 and i > 0; therefore -i/o is positive.

 

[Klemptor]

One other thing I want to clarify; I'm not sure I was clear about this in my first post: for a diverging lens, the focal length is taken to be negative; for a converging lens, the focal length is taken to be positive.

 

[Dr Gerrard]

I think I got it, but won't know until I try out some problems.

Anyone know where I can find any?

 

[lDanny]

Quick question:

For a concave mirror and converging lens, if the object is moved between the focal point and len/mirror then it would be negative, virtual, reduced upright image. Instead of positive, real, enlarged inverted image?

thanks

 

[Myster Ycongo]

sorry to bring this back from the grave, but for those who have no idea how to properly learn optics, what do you suggest I read? I tried going over the EK method and it made my head spin. Is EK the best method? If so, I'll give it another go. I'm not having too much trouble with any other review section except for optics. Thanks for the time.

 

[BerkReviewTeach]

sorry to bring this back from the grave, but for those who have no idea how to properly learn optics, what do you suggest I read?

If you have access to the new BR physics book, pages 247 through 250 give a series of examples that summarize to make a checklist method for determining image position. The idea is that diverging lenses and mirrors do the same thing every time (they give an image that is smaller, upright, and virtual), so you don't need to do much more than visualize the system. For converging lenses and mirrors, the image position is relative to the object position and it starts by knowing that if one is at the R, then the other must also be at R. If one is inside of R, then the other must be beyond R. If you try the examples (10.7a - 10.9b, and especially 10.9a), the pattern should become clear and make these questions easy.

There's also a different way of looking at the equation that makes it easy to get the image position in about ten seconds, even when you can't visualize what's going on.

 

[Myster Ycongo]

If you have access to the new BR physics book, pages 247 through 250 give a series of examples that summarize to make a checklist method for determining image position. The idea is that diverging lenses and mirrors do the same thing every time (they give an image that is smaller, upright, and virtual), so you don't need to do much more than visualize the system. For converging lenses and mirrors, the image position is relative to the object position and it starts by knowing that if one is at the R, then the other must also be at R. If one is inside of R, then the other must be beyond R. If you try the examples (10.7a - 10.9b, and especially 10.9a), the pattern should become clear and make these questions easy.

There's also a different way of looking at the equation that makes it easy to get the image position in about ten seconds, even when you can't visualize what's going on.

I have the 2009 version, I will try my best to go through it. What is the 10 second image positioning method you're referring to?

 

[Mcat35]

I have the 2009 version, I will try my best to go through it. What is the 10 second image positioning method you're referring to?

there are just certain rules you have to know.

first and foremost lens maker's equation

1). 1/o + 1/i = 1/f = 2/R (r = radius of curvature)
2). focal length is + for converging lenses, - for diverging lenses
3). real image is on the opposite side of the lens as the object example: you are looking at an apple through your spectacles. what you see is a real image. (you don't actually have to wear glasses since the lens in your eye is also an converging lens) (image thats on your retina = real image) (wearing glasses actually makes it a 2 lens system which is just a bit more complicated but image formed by the first lens becomes object for the 2nd lens (the lens in your eye)
4.) virtual image is on the same side of the object.... (note for lenses this is the case, but its opposite for mirrors) simple enough right( it is on the same side as the apple)
5.) DIVERGING lens always form virtual images.
6.) converging lens forms real image when the object is outside the focal point, forms a virtual image when object is placed inside the focal point. NO image when the object is placed at the focal point. (try the lens masters equation and plug in 1/o + 1/i = 1/f ) with the same F and same O what happens you get 1/i = zero. so what does i = thats right INFINITY.
7.) RIP VUN (stands for *Real Inverted Positive *Virtual Upright Negative.
8.) based on that because magnification M= -i/o a positive magnification woudl indicate that image is virtual (based on RIP VUN) (since object distance is always positive Image distance is what matters)

As for the rest just plug and chug into the lens maker's equation and you should find out what it is.. Knowing these rules just saves a lot of time.

Took me a while to get all this too.. and a lot of googling.

Keep on working at it and you will get it.

 

[Mcat35]

I don't have EK but here is just some basic info as I understand it; hope it helps:

The focal length of a lens or mirror is characteristic of that lens or mirror. The focal point sits on the principal axis.

For spherical mirrors, the focal length is equal to half of the radius of the mirror.

For thin lenses, the focal length is found via the lensmaker's equation, namely:

1/f = (n - 1)(1/r + 1/R)

where r and R are the radii of the lenses comprising the thin lens, and n is the index of refraction of the lens. Be careful to note that this equation gives the inverse of the focal length.

Ray diagrams are key for visualizing what happens and for checking your work. When you do a ray diagram, how you use the focal point depends on the type of lens/mirror you have, and I suspect this is where you're running into a problem.

For each ray diagram, you can draw two rays to find the point of convergence (ie the image location). One of the rays can always be drawn as coming from infinity parallel to the principal axis and refracting through the lens in a direction determined by the focal point. Mirrors are simpler and I'm assuming your problem is with lenses, not mirrors, so I'll explain lens ray diagrams here.

- For converging lenses where o > f, you can draw a ray diagram where the first ray comes from infinity parallel to the principal axis and refracts through the lens toward the backside focal point. The second ray comes from the object, passes through the frontside focal point, and refracts to infinity parallel to the principal axis. The image is virtual and inverted.

- For converging lenses where o < f, the situation is more complicated. As before, the first ray comes from inifinity parallel to the principal axis, and refracts through the lens toward the backside focal point. The second ray is drawn as if it comes from the frontside focal point toward the object, and it travels toward the lens and refracts to infinity parallel to the principal axis. The image forms at the intersection of the backed up refracted rays; it is real and upright. (I believe it's enlarged but I'm not certain about that.)

- For diverging lenses (no matter whether o > f or o < f), as before, the first ray comes from infinity parallel to the principal axis; it refracts as if from the frontside focal point. The second ray travels from the object toward the backside focal point, and refracts to infinity parallel to the principal axis. Again, the image forms at the intersection of the backed up refracted rays. The image is real and upright. (I believe it is reduced but I'm not certain on this point.)

Sign conventions vary between lenses and mirrors. In general, light travels from left to right; the "real" side is on the left of the lens/mirror, and the "virtual" side is on the right of the lens/mirror.

Mirror sign conventions: Incident light approaches from the left; reflected right reflects to the left. On the left (the real side), o, i, R, and f > 0. On the right (the virtual side), o, i, R, and f < 0.

Lens sign conventions: Incident light is on the left; refracted light is on the right. On the incident side, o > 0 and i and R < 0. On the refracted side, o < 0 and i and R > 0.


I hope this helps!

there is all sorts of mistakes in this one, I wouldn't read this ya'll lol...

Real image is always inverted..... not upright