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.
[Physics] Why do powers of lenses get added
geometric-opticslensesoptics
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It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.
Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.
Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.
The Magnification is a combination of all of the focal lengths of the picture you have shown above. A real image is created by the objective and tube lens. This creates an image of what you have at the object plane that is magnified by:
$M = \frac {f_{tube lens}}{f_{objective}}$
So, if you were to measure the size of the image, it would be M times larger than the object that is placed to the left of the objective lens.
One common confusion is that many microscope objectives actually create an image all by themselves without a tube lens. There are several standards including objectives that create images 160 mm and 170 mm away from the microscope objective. In your diagram, it implies an infinity corrected objective lens. This means that the image created by the microscope objective is infinitely far from the objective lens. This might lead you to believe that since the light is collimated from the microscope objective, you can place the tube lens anywhere you want. That is not technically correct because of two factors: vignetting and the optical design of the tube lens.
Vignetting means that the light escapes the size of the lens. In your diagram, this would happen if the tube lens is too small. Many infinity corrected objectives are designed for tube lenses that are 180 mm from the objective lens. If the tube lens is not placed at a distance close to 180 mm, you can have vignetting or performance from optical aberrations may cause the image to degrade.
Now, take the final step to the eye of the observer. This is the eyepiece. Your eye prefers (is relaxed) when looking at infinity. Therefore, the eyepiece is typically designed to project the image created by the objective-tube lens pair to infinity. Your diagram actually shows the image at 25 cm instead of at infinity. For this case, the eyepiece is placed at nearly one focal length away from real image (image plane 3 in your diagram).
The final magnification is $ M_{total} = M \times \frac{25 cm}{f_{eyepiece}} $
There are additional considerations including:
- Working distance of the objective (distance between objective and object)
- Eye relief (distance between eyepiece and observer's eye)
- Pupil or eyebox (sometimes you look inside a microscope and it is black until you line up your eye with the microscope)
- Illumination (most objects require some external light source to illuminate them so you can see them!)
One last consideration. If you just want to put the image onto a camera. In which case, you don't need the eyepiece!
Best Answer
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.