everyday-lifegeometric-opticsvision
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?
First, let me try to clarify the question, because not everyone seems to get it ("where would the photons come from?"). If I'm buying binoculars, I might choose from 8x24, 8x36, or 8x50 binoculars. The first number is the magnification; 8x for all the examples. The second number is the size of the objective in millimeters, 24 to 50 millimeters in the example. All else being equal, the larger objectives will gather more light, and produce a brighter image. The 8x24 and 8x50 binoculars will deliver the same 8x magnification, but the 8x50 pair will be brighter.
So it's reasonable to ask if you could make 1x binoculars that have a big objective that brightens the scene in front of you. There's no conservation of energy issue here; the larger front objective would gather more light. The problem is that there's a relationship between the magnification and the sizes of the entrance and exit pupils:
magnification = entrance_pupil_size / exit_pupil_size
If you create a binocular with an entrance pupil (objective, basically) that's larger than your eye's pupil, so that it can gather more light than your eye's pupil, then at 1x magnification, the exit pupil is going to be just as big as the objective lens. And since that's bigger than your eye's pupil, the "extra" light is going to run into your iris instead of going through the pupil, and you will gain no advantage from it.
If the object is in the focal plane of the lens then the upright, virtual image is seen at infinity - parallel rays of light enter the eye.
Bringing the object closer to the lens also brings the virtual image closer and the angular magnification also increases.
With the unaided eye the3 greatest magnification is obtained when the virtual image is as close as it can be whilst the optical surfaces within the eye still produce a focussed image on the retina.
That distance is called the least distance of distinct vision and usually is taken to be 25 cm.
So the position of the virtual image as seen by the eye varies as the position of the object varies.
You might find using the Phet Geometrical Optics simulation informative?
Here are some screen shots:
Best Answer
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.