Optical Formula Question...

[Pete Hanlin]

Okay, Maria is studying this particular problem in preparation for an upcoming exam.

What power would a single thin lens need to be to replace a system with a +4.00 first lens followed by a +3.00 lens placed 10cm behind?

I know the answer involves vergences and all, but its not a problem we encounter too often in optical practice, so I'm unsure of the exact method needed to find an answer...

Here's my best guess...

First, a +4.00 lens would have a focal point at 25cm to the right... since the second lens is 10cm to the right of the first, that would mean light would come to a focal point 15cm to the right of the second lens if it had no power...

A lens with a focal length of 15cm would be +6.66... Add +3.00 to the vergence and get +9.66... The focal point for a +9.66 lens is 10.34cm... Add the 10cm distance from the first to second lens to the 10.34cm focal length the +3.00 lens will have (considering the light striking it from the first lens already has a vergence of +6.66, and you end up with a total focal length of about +20.34cm (which inverted comes out to be right around +5.00 (+4.91 to be exact).

Like I said, I don't have any texts with this info in it, so I'm quite unsure if this is correct- or way off! How's bout it, Darryl???

Pete

[shanbaum]

I think maybe you're thinking too much: the effective power of the +4.00 lens at a 10cm vertex distance is 6.66 :angry:, which added to the +3.00 lens gives the combined lenses a power of 9.66, so that's the single thin lens you'd need.

That number showed up in your post, but I couldn't follow the rest of it (too sleepy, perhaps). Maybe you were thinking that the single thin lens had to be located where the front lens was, but measured where the back lens was? I don't see that in the problem as stated...

[Maria]

I emailed the leading authority on these matters, and this is what I got back:

Fe = F1 + F2 - t/n * F1 * F2
Fe = 4.00 + 3.00 - 0.1 * 4.00 * 3.00
Fe = +5.80 D


But, if I recall, I got the same as Robert. I imagine Darryl's right, though :)

[Pete Hanlin]

I think the formula would be written:

Fe= F1+F2 - t/n(F1*F2)

5.8= 4+3 - .1/1(4*3)

If Darryl says so, I'm willing to accept it as gospel truth... However, I want to be clear on what we're answering... Yes, I was assuming we were looking for a thin lens that would put the focal point at the same position as the two lenses together- assuming the thin lens was in the same position as the first lens in the first place...

+5.8 it is, though...

Pete

[shanbaum]

Originally posted by Pete Hanlin
Yes, I was assuming we were looking for a thin lens that would put the focal point at the same position as the two lenses together- assuming the thin lens was in the same position as the first lens in the first place...

Do you really think that's the question? I guess it could be. I interpreted it to mean, "what's the back vertex power of the system at the back lens?"

[Maria]

The exact wording of the question is:

A lens system consists of F1 = +4.00D lens and F2 = +3.00D lens, separated by 10cm. Find the equivalent thin lens focal length (fE) of the system

[Darryl Meister]

Hi Guys,

Fe = F1 + F2 - t/n * F1 * F2 is the same as Fe= F1+F2 - t/n(F1*F2) if you perform your order of operations correctly.

You guys are correctly describing "effective power" -- or, as Robert put it, the "back vertex power" -- of the system, but not the "equivalent power." . Although a single +9.67 D lens placed at the plane of the original +3.00 D lens will produce the same focal length when measured from the last lens, the magnification of the system will be different. Note that you guys are assuming that the lens should be placed at the location of the +3.00 D lens. Otherwise you would have to choose a different lens to produce the desired focal length. However, a +5.80 D lens -- which is the "equivalent power" -- placed at the second principal plane of the system will produce both the same focal point and the same magnification.

However, Maria's question is a bit ambiguous, so it might be either one.

Best regards,
Darryl

[Maria]

I'm not called Marie.

As I recall, the answer wanted was at the second lens, but it is fairly hazy where the replacement should go.

[Darryl Meister]

After reading Maria's last clarification (which is how I understood her to ask the question yesterday) I would definitely say that they are looking for the equivalent power of the system in this problem. It's no ABO exam question, is it?!? ;)

Best regards,
Darryl

[Darryl Meister]

Originally posted by Maria
I'm not called Marie.

Sorry about that, Maria. I was typing faster than my brain was working.

Best regards,
Darryl

[Maria]

I'll forgive you if you sit my exam for me. With a wig and a dress and some makeup, not even my own mother would know it's not me. ;)

[Aleyz2020]

I'll go with Shanbaum on the answer. It's the most logical. I don't remember seeing anything like it on the Florida state boards, though.

[Blake]

This is a new topic for me, but I don't have anything better to do! :D

First of all, given the formula

Fe= F1+F2 - t/n(F1*F2)

I'm assuming t is the distance between lenses, in this case 10 cm, and n is the index of refraction of the medium, approx. 1.00 for air. With F1=+4.00 and F2=+3.00, it doesn't take a fancy calculator to tell you that Fe=+5.80.

Darryl - Where is the "second principle plane of the system"?


Blake

[Darryl Meister]

Hi Blake,

The easiest way for me to find the second principal plane of the system is to first find the equivalent focal length (using the equivalent power formula above) and then find the actual back focal length of the system (using back vertex power formula). Once you have the two focal lengths, the distance of the principal plane from the back lens will be equal to the difference in the two focal lengths. The first principal plane can be found by similar calculations using the front vertex power of the system.

In our example, the back focal length is equal to 1 / 9.67 = 10.3 cm, while the equivalent focal length is equal to 1 / 5.80 D = 17.2 cm. This gives the position of the second principal plane at 17.2 - 10.3 = 6.9 cm in front of the back lens. If we place a single +5.80 D lens at this new location, it will produce the same magnification and focal point location as the original +4.00/+3.00 system.

Best regards,
Darryl

[shanbaum]

Had the question been phrased something like, "a single thin lens, positioned where, replaces the following system", I would understand it to be looking for equivalent power. As phrased, I'd say it appears to be looking for the effective power of the system. In any case, the question could be expressed so as to be unambiguous, which it obviously is not.

[shanbaum]

...also, Darryl me boyo, isn't the total magnification dependent on the shapes of these lenses? No mention is made of that...

[Pete Hanlin]

Okay, I don't want to beat a dead horse (well, maybe just a little ;) ), but as you mention, the focal point from the second lens (using a back vertex power of 9.67) is 10.34cm. This means (measuring from the front of the first lens), that the total focal length is 20.34cm. If you placed a single +4.91 lens at the same point as the first lens, wouldn't it also bring light to a focus at 20.34cm? Isn't everything relative to the focal length of the system? In a vague question like this, is one supposed to assume that the single lens is to be positioned at the back of the system?

Pete

[Pete Hanlin]

Okay, I'm re-re-re-reading your answer (no, I'm not stuttering, I'm repeatedly reading your explanation ;) ), and it seems to me that- if you increase the vertex distance from the focal point by 3.1cm, my answer is correct for where I positioned my lens (at the plane of the original first lens in the system). So wouldn't the equivalent power be the same (even though the magnification will not be)?

As I understand things now, what you are saying is that the formula you gave results in a single lens with the same magnification and focal length properties as the original two lens system?

Just to be a pain in the neck, would that mean if I made up the following question...

"To replace a system of three lenses that includes a +2.00 lens, followed by a +4.00 lens placed 5cm behind, followed by a +5.00 lens placed 3cm behind- what single lens would be required in order to maintain an equivalent focal length and amount of magnification?"

...I would say

1.) 2.00 + 4.00 -(.05 * (2.00*4.00))
2.) 6.00- 0.4= +5.40 power for lens to replace first set of lenses placed 2.44cm in front of second lens... (because the effective back vertex power of the first set of lenses is +6.22)
3.) +5.40 + 5.00 -(.0544703 * (5.40 * 5.00)
4.) +10.40 - 1.47 = +8.93 power for lens to replace result of first set plus third lens placed 3.29cm in front of third lens... (because the effective back vertex power of the replacement for the first set plus the back lens is +12.65)

So the equivalent single vision lens for the three lens system would be a +8.93 lens placed 3.29cm in front of the third lens?

If I try the original formula given for two lenses and try to do all three lenses at once (using the distance between the first and last lenses as "t," the answer comes up to +7.80... but I don't think this would work, because taking all three lenses into account at the same time doesn't allow for different placement of the middle lens).

Sorry for all the extra brain work (though I'm pretty sure Darryl can do this in his sleep- probably computes stuff like this on long boring drives just to keep his mind busy ;) ), but I just want to see if I'm understanding the process.

Thanks,
Pete

[Darryl Meister]

Originally posted by shanbaum
...also, Darryl me boyo, isn't the total magnification dependent on the shapes of these lenses? No mention is made of that...

Remember, the problem specifically asked for the equivalent focal length, so we don't have to guess anymore about what the authors wanted.

You are right about the magnification to the extent that they provide shape information about the lenses... However, since they didn't, you have to assume that each lens is infinitely "thin" so that only the power is consequential. This is pretty common for these problems, since it would require a great deal more calculation otherwise. Had they provided "thick" lenses, you would probably have to calculate the equivalent power and principal plane position of each lens first, then use those new equivalent lenses and positions to determine the overall equivalent power of the system.

Equivalent power isn't something we deal directly with much in ophthalmics -- except when using low vision products and such. However, even though we don't use it much, it is still very applicable to things like spectacle magnification. The spectacle magnification formula, for instance, can be derived on the basis of equivalent power. As you change the shape (base curve and thickness) of a lens, you shift the location of the principal planes. This, in turn, changes the spectacle magnfication without changing the back vertex power. This is why the spectacle magnification formula has both a shape and a power component to it.

Best regards,
Darryl

[Darryl Meister]

I'm going to submit this graphic, which I've drawn to illustrate the concept of equivalent power. When positioned correctly, the +5.80 D lens will produce the same focal point and magnification as the original +4.00/+3.00 system.

Best regards,
Darryl

[Darryl Meister]

Originally posted by Pete Hanlin
If you placed a single +4.91 lens at the same point as the first lens, wouldn't it also bring light to a focus at 20.34cm? Isn't everything relative to the focal length of the system?

Hi Pete,

The system really doesn't have a "total" focal length, though the back vertex focal length and equivalent focal length are probably the closest thing to one. You are still thinking in terms of the focal point only, and not magnification. Technically, I could put any plus lens in there and still bring light to a focus at the same point if I position the lens at the correct distance. (You have just arbitrarily selected two positions already, the locations of the first and second lenses.)

There is really no reason to assume that the single new lens is going to be positioned at the same location as the original second lens -- anymore than you should assume that it will be positioned at the same location as the first lens. I believe that you're just getting hung up on what you consider to be the vagueness of the question. ;) The author of the question wasn't vague at all, he/she asked for the "equivalent focal length" of the system. This is a very specific value, the reciprocal of the equivalent power.

Best regards,
Darryl