# Why do powers of lenses get added?

In our physics class, we were told that a power of a lens is defined as the reciprocal of its focal length. Also, the powers of multiple lenses get added up. But I didn't understand why this is. I mean consider a series of two convex lenses with focal lengths $f_1, f_2$. Clearly, the resultant image distance is going to depend upon the distance between the lenses (that is how a compound microscope works; by adjusting the distance between the objective and the eyepiece different magnifications can be obtained). So, I'm guessing, the resultant focal length of the system (whatever that is), will also depend upon the distance between the lenses. How does the power of the lens fit into all this? Also, what do they mean by the 'resultant power of the lenses' in the first place? The resultant system of lenses is not going to be like these lenses which have negligible thickness. Where are you going to measure the focal length from? The system of lenses does not have a fixed pole.

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You are correct in stating that the resultant effect on incoming rays and their focussing depend on the two lenses and the distance between them. To reduce the two lens system in general we use the principal planes method and then can describe the situation in terms of two parallel planes only. But when they say that the powers are added or the "resultant" focal length is given by $$1/F=1/f_1+1/f_2$$ is valid only in case of thin lenses very close together such that their combined thickness can be ignored. You can easily get this equation by assuming their (lenses') centres two be coincident and then applying some geometry or by using the general refraction -through-curved-surfaces equation and trace your image based on the object distance. In any case, the finiteness of the thickness has to be eliminated to get this equation. If we assume two very thin lenses in contact, than this equation holds true and the "resultant" power or the focal length is just the power/focal length of the single lens that can replace the two-lens system without any change in its optical properties (in paraxial approximation).

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These are all good questions

So, I'm guessing, the resultant focal length of the system (whatever that is), will also depend upon the distance between the lenses. How does the power of the lens fit into all this?

That is correct. In the limit where your lenses are thin the powers of two lenses is added as follows: $$\phi_\text{tot} = \phi_1 + \phi_2 - \phi_1\phi_2\tau$$ where the individual powers are given by $\phi_1 = 1/f_1,\phi_2 = 1/f_2$, and $\tau = t/n$. $t$ is the spacing between the two lens elements and $n$ is the index of refraction of the medium between your two lenses ($n=1$ in vacuum). The effective focal length is then $f_\text{effective} = 1/\phi_\text{tot}$

There is a distinction between the effective focal length $f_\text{effective}$, the rear focal length $f_\text{R}$, and the front focal length $f_\text{F}$.

$$f_\text{R} = n_\text{R}f_\text{effective}\quad \quad f_\text{Front} = -n_\text{Front}f_\text{effective}$$ I.e. the front focal length is measured from the front principal plane to the front focal point (negative distance for a positive effective focal length) and the rear focal length is measured from the rear principal plane to the rear focal point (positive distance for a positive effective focal length).

Also, what do they mean by the 'resultant power of the lenses' in the first place? The resultant system of lenses is not going to be like these lenses which have negligible thickness. Where are you going to measure the focal length from?

This is a very important point. For a system of multiple lenses you measure the front and rear focal lengths from what are called the front and rear principal planes (rather than measuring directly from a lens element). For the case of a single lens the front and rear principal planes are located at the lens. For two lenses, the front principal plane is shifted from the first lens element a distance

$$d_\text{F} = + \frac{\phi_1}{\phi_\text{tot}} \frac{n_\text{F}\;t}{n}$$

the rear principal plane is shifted from the second lens element a distance.

$$d_\text{R} = -\frac{\phi_1}{\phi_\text{tot}}\frac{n_\text{R}\;t}{n}$$

For the above two equations I am defining the first lens to be on the left and the second lens on the right. The separation is $t$ and the index of the medium between the two lenses is $n$. The index of the medium to the right of the two lenses is $n_\text{R}$ and the index of the medium to the left of the two lenses is $n_\text{F}$. A value of $d_\text{F}$ or $d_\text{R}$ less than zero indicates a shift of the respective principal plane to the left; a value greater than zero indicates a shift to the right.

Disclaimer: this response is only a very brief introduction to the ideas of gaussian optics and the cardinal points. This stuff can be really confusing for anybody, especially when you consider all of the sign conventions. Also, these equations are really only valid in the paraxial limit for rotationally symmetric systems. Having said that, these basic formulations can be expanded to a system of any number of lenses -- not just 2. If you really want to understand this stuff there are surely some good books on this material. Hecht seems to be the book for intro optics, although I haven't read him. Check the table of contents to make sure gaussian optics is covered (it should be) because that's quite an expensive text.

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+1 for a very interesting answer. But, I'm afraid, Satwik Pasani mentioned what I needed: that the powers can only be added if the lenses are very close together. Gaussian optics seems to be intriguing. I'll check it out. – Gerard Aug 18 '13 at 1:51
@user1305192 Satwik's equation is a special case of the one I posted. Note that if $t = 0$ (very close together lenses) the equation reduces to Satwik's. – Joe Aug 18 '13 at 3:42