Question

What must be the position of an object in order that it may be seen distinctly through a +9.0 D lens placed 2.9 cm in front of an eye, the eye being accommodated for a distance of 35.5 cm? Give your answer in cm as a distance to the lens.

Answer 1

First, we need to understand the concepts. Since the eye is accommodated for a distance of 35.5 cm this means that the IMAGE of the object, generated by the artificial lens must be at that distance from the eye. (Because after the image is formed, it will be processed by the eye as the new objetc). I'm drawing the needed image in red here:

We also see that this distance implies that the distance from this image to the artificial lens is should be 32.6cm.

Now we have this important information: This is the distance from the image of the object to the artificial lens. However, since the image should be on the same side as the object (see figure) it means that this distance should be put with a negative sign:

with optical power +9.0 D. From the optical power we know (by definition) thatt is the reciprocal of the focal length in metres: 1/f=+9.0 D which is

. Now we use the thin lens equation:

knowing already what the values are for 1/f and for d(image)

We solve for (which is positive, as expected).