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Least distance of distinct vision is the minimum object's distance that is able to produce a distinct image on the retina. Yet, it is treated as image's distance while applying lens equation.

Say, for an elderly person, least distance of distinct vision is $50$ cm. SO, if I am to calculate the focal length of the corrective lens, would I consider the object's distance as $50$ cm, or would it be $25$ cm - the normal 'near point' for a healthy eye?

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  • $\begingroup$ Detailed explanation would be much appreciated. $\endgroup$ – Swami Jun 9 '14 at 6:07
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    $\begingroup$ What do you you need the corrective lens to do? Are you trying to enable the user to focus on things nearer than 50cm? $\endgroup$ – John Rennie Jun 9 '14 at 6:16
  • $\begingroup$ Yes, you are right. $\endgroup$ – Swami Jun 9 '14 at 6:19
  • $\begingroup$ I want to know, what has the least distance of distinct vision got to do with the distance of image formation? In a general sense. $\endgroup$ – Swami Jun 9 '14 at 6:21
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The resting focal length of the eye is around $17$mm - you see various different figures but let's assume $17$mm is correct as the exact value doesn't affect what follows. Anyhow, that means the distance from the optical centre of the lens/cornea to the retina is $17$mm so in the lens formula we require $v = 17$mm for a clear image.

The eye focuses by deforming the lens to change the focal length. If we take the object distance to be 250mm i.e. the least distance of distinct vision and use the lens equation:

$$ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} $$

We find that for a clear image at $u = 250$mm and $v = 17$mm the focal length of the lens has to be reduced to about $15.9$mm.

As we age the lens gets less flexible so our ability to reduce its focal length decreases. If we use your value of $500$mm for the least distance of distinct vision of an elderly person, i.e. $u = 250$mm and $v = 17$mm, the focal length of the lens works out to be $16.44$mm.

So if you want to allow the position of closest focus to be moved in to $250$mm you need spectacles that increase the ability to focus. Since there is quite a large distance between the lens/cornea and the spectacles you'll need to do the calculation for two separate lenses, but this is straightforward. Calculate the image position from the first lens then use that as the object position for the second lens.

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The near point of a far sighted person is farther away (50 cm) than what it would be for a normal person (25 cm) Meaning they cannot see things closer than 50 cm, where normal people cannot see closer than 25 cm.

So the aim of a corrective lens is to make the resultant vision have a near point at 25 cm. So what they do is they use a lens that can form the image of an object at 25 cm, 50 cm from the eye. Then the eye's own lens will see that image clearly, as 50 cm is its near point. The final result is that objects at 25 cm are seen clearly.

So, in New Cartesian sign convention, for the corrective lens, v = +50 cm u = +25 cm

( Near point and LDDV have the same meaning )

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