The distance from the lens (actually a combination of the cornea and the crystalline lens) to the retina at the back of the eye is \(2.0 \mathrm{~cm}\). If light is to focus on the retina, a) what is the focal length of the lens when viewing a distant object? b) what is the focal length of the lens when viewing an object \(25 \mathrm{~cm}\) away from the front of the eye?

When viewing a distant object, the object distance is considered to be infinite. Therefore, the lens formula becomes: \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{2.0}\). The term \(\frac{1}{\infty}\) becomes 0, so: \(\frac{1}{f} = \frac{1}{2.0}\). Now, we need to find f. To find f, take the reciprocal of both sides: \(f = \frac{1}{\frac{1}{2.0}}\). The focal length, when viewing a distant object, is \(f = 2.0 \mathrm{~cm}\).

When viewing an object \(25 \mathrm{~cm}\) away from the front of the eye, the object distance, \(d_o\), is \(25.0 \mathrm{~cm}\). Now, we can use the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute the given values into the formula: \(\frac{1}{f} = \frac{1}{25.0} + \frac{1}{2.0}\). To solve for the focal length, f, we first need to find a common denominator for the fractions on the right side of the equation. In this case, the common denominator is 50: \(\frac{1}{f} = \frac{2}{50} + \frac{25}{50}\). Now, combine the fractions on the right side of the equation: \(\frac{1}{f} = \frac{27}{50}\). To find f, take the reciprocal of both sides: \(f = \frac{1}{\frac{27}{50}}\). The focal length, when viewing an object \(25 \mathrm{~cm}\) away from the front of the eye, is \(f \approx 1.85 \mathrm{~cm}\).