If an object is placed between focal length and optical center of a convex lens, a virtual, magnified and erect image is formed. But if we change the position of object within the focal length of lens, does the image still remains at 25 cm from eye? In other words is image always formed at least distance of distinct vision regardless of the position of object as long as it is within the focal length?
[Physics] Does a magnifying glass always make image at least distance of distinct vision
geometric-opticsoptics
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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.
The lens equation and the ray tracing does agree. Let's denote the distance from the object and image distances to the first convex lens as $o_1$ and $i_1$ respectively. Let's denote the object and image distances to the second concave lens as $o_2$ and $i_2$. The focal lengths are $f_1=2 \text{ cm}$ and $f_2=-2 \text{ cm}$.
The lens equation reads $\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$.
For the first lens, $\frac{1}{3\text{ cm}}+\frac{1}{i_1}=\frac{1}{2\text{ cm}}$.
Solving for the image distance yields $i_1=6\text{ cm}$.
The image of the first lens becomes the object for the second lens, and since the two lenses are $4\text{ cm}$ apart, we have $o_2 = 4\text{ cm}-i_1 = -2\text{ cm}$.
The lens equation for the second lens reads as $\frac{1}{-2\text{ cm}}+\frac{1}{i_2}=\frac{1}{-2\text{ cm}}$.
Thus $1/i_2 = 0$, and $i_2=\infty$. There is neither a real image (converging rays in front of the second lens) nor a virtual image that is formed (diverging rays). Rather the rays are parallel.
Well, we can see this agrees with the ray tracing diagram (made with OpticalRayTracer software ver. 9.6):
Best Answer
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: