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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?

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John, your method for determining approximate lens focal lengths is a valid one, although since your eye will automatically accommodate to some extent, accuracy will be limited.

The apparent discrepancies follow from the rather obscure convention of labeling lens "Power" in units of Diopters, which are inverse meters.

The power P of an optic is defined as the inverse of the focal length f in meters:

$$ P = \frac{1}{f} $$

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).

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  • $\begingroup$ So if I use .25 diopters power lens with 6mm eyepiece(to make a telescope), I can get magnification up to 6667. But I have to keep 40006 mm distance between both lenses which is not recommended. $\endgroup$ – Amit Kumar Gupta Mar 2 '13 at 14:41

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