You’ve got it right. The dioptres = 1/focal length, so focal length = 1/dioptres.
e.g. 3.3 dioptres = 1/(3.3) = 0.30303m focal length
-2 dioptres = 1/(-2) = -0.5m focal length

http://medical-dictionary.thefreedictionary.com/Dioptres for more info.

@jahono
The book I’m working on says the answer should be 5.0m not -5m, meh.

Hmmm I call typo, or slackness with signs. It would be 5m but in the negative direction (on the side of the lens where the parrallel light rays are coming from) so yeah should be -5m (for a -1/5 dioptre lens).

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Dioptre=1/Focal length.
Let’s take an example.
A person can see things clearly upto a distance of 50 cms and not beyond that, what would be the power in D of the lenses he’d use and what would be the type of lens.If another person can’t see near objects clearly within a distance of 50 cms,what would be the type of lenses he’d be using to rectify his vision and what would be the power in dioptre?
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Sounds interesting, but the lenses that are used for near vision and far vision are different.For short-sighted vision concave lenses are used and for distant vision(the name is confusing,but distance vision means ability to see distant objects better and have a hazy near vision.This is seen in middle age people generally who have problem reading books,but can see distant things clearly) convex lenses are used.In case of difficulty to see distant objects,the power of the lenses are in + and in shortsightedness it’s -.
The reson for this is, in short vision,the image is formed before the retina and in distant vision it’s formed behind the retina.Anything behind the retina is termed as plus and infront of the retina is minus.
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So the modified formulas for short-sightedness is D=-1/f and for Hypermetropia or longsightedness D=1/f
In the above examples,f=50 cm
So here 1 f is taken in meters
So for teh first part of the question it’d be -1/0.5= -2D and concave lens and for the second part it’s 1/0.5= 2D and convex lens.
Thank you all
ETW.