# calculating focal length of meniscus lens

As I read about telescope, distance between objective lens and eyepiece must be equal to addition of their focal lengths.

D = F1 + F2

I used one of the eyepiece of 2cm focal length to calculate focal length of a magnifying glass . I seen clear distance view when I kept eyepiece 36cm far from magnifying glass. So it's focal length should be 34cm as per above formula.

I purchased a meniscus lens of -3 power which is used in specs to correct distance view. I kept it approx 4cm far from magnifying lens for clear distance view and concluded that its focal length should be -30cm. I confirmed it with by using magnifying glasses of various focal lengths.

D = F1 + (-F2)

Problem:

Now I tried the same experiment with a specs of -.25 power. I guess its focal length should be -2.5cm as I calculated for above mentioned meniscus lens. But I dint seen any difference in magnification or clear view when I used this specs with any lens.

What's wrong I am doing? How can I confirm its focal length?

• ph.surrey.ac.uk/UserFiles/File/… Feb 26, 2013 at 11:02
• @JohnRennie, my question is related to real time practical and divergent lens. Feb 27, 2013 at 0:56

$$P = \frac{1}{f}$$
Thus a lens of power $-3$ has a focal length of $-1/3$ meter or $-33$ $cm$, not far from the $f= -30$ $cm$ you estimated. Similarly, a power of $-0.25$ Diopters corresponds to $-4$ $m$ or $-400$ $cm$, a very weak negative lens which would scarcely be noticeable when combined with a fairly high powered positive lens (magnifier).