91Ó°ÊÓ

The thin lens equation is given by: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the distance between the object and the lens, and \(d_i\) is the distance between the image and the lens.

To find the focal length when the object is at infinity, we first have to determine the object distance (\(d_o\)) and the image distance (\(d_i\)). When the object is at infinity, the image formed on the retina will be at the focal point. Thus, the object distance is infinity, and the image distance is equal to the distance between the retina and the lens, \(d_i = 2.00 \mathrm{cm}\). Plug these values into the thin lens equation: \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{2.00}\) Solving for \(f\), we get: \(f = 2.00 \mathrm{cm}\) The focal length required to focus on objects at infinity is \(2.00 \mathrm{cm}\).

To find the focal length when the object is at the normal near point \(25 \mathrm{cm}\), first determine the object distance (\(d_o\)) and the image distance (\(d_i\)). The object distance is given as \(d_o = 25 \mathrm{cm}\), and the image distance will be equal to the distance between the retina and the lens, \(d_i = 2.00 \mathrm{cm}\). However, since the retina is fixed and cannot move, we must adjust the object distance (\(d_o\)) to satisfy the thin lens equation. To do this, we use the relation \(d_i=d_r - d_o\), where \(d_r = 2.00 \mathrm{cm}\) is the distance from the retina to the lens. Thus, \(d_o = d_r - d_i = 2.00 - 2.00 = 0.15 \mathrm{cm}\). Now, plug these values into the thin lens equation: \(\frac{1}{f} = \frac{1}{0.15} + \frac{1}{2.00}\) Solving for \(f\), we get: \(f = 1.85 \mathrm{cm}\) The focal length required to focus on objects at the normal near point is \(1.85 \mathrm{cm}\).

We have found that to see objects from \(25.0 \mathrm{cm}\) to infinity, the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\). This matches the result we were asked to show.