A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.

Let's first understand the information given in the problem. The man cannot see objects clearly beyond a distance of 2.0 m. The distance from the lens to the retina is 2.0 cm. The eye's power of accommodation is 4.0D.

We'll use the lens formula to find the focal length of the lens when the man is unable to see clearly more than 2 meters away. The Lens Formula is given by: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Where: \(f\) is the focal length of the lens, \(d_o\) is the object distance, \(d_i\) is the image distance (distance from the lens to the retina). For the nearsighted man, we know that: \(d_o\) = 2.0 m (maximum clear vision distance) \(d_i\) = 2.0 cm (distance from the lens to the retina)

Now, we'll use the lens formula to find the focal length when the man cannot see clearly beyond 2.0 meters. \(\frac{1}{f} = \frac{1}{2.0} + \frac{1}{0.02} \) Solving for \(f\), we get: \( f = \frac{1}{2.0 + \frac{1}{0.02}} \approx -0.0197 m \) or \(-19.7 \mathrm{cm} \)

Now that we have the lens's focal length, we can calculate the power of the corrective lens. The Power of a lens is given by: \(P = \frac{1}{f} \) So, \(P_{correction} = \frac{1}{-0.0197} \approx -50.8 \mathrm{D} \) However, the eyeglass lens is 2.0 cm away from the eyes, so we need to consider the accommodation effect, which is 4.0D. Therefore, the prescribed power for the corrective lens: \(P_{prescribed} = P_{correction} + P_{accommodation} = -50.8 + 4.0 = -46.8 \mathrm{D} \) So we would prescribe a -46.8 D lens for the near-sighted man.

Now, let's find the nearest object he can see clearly with his glasses. For this, we'll use the Power of Accommodation formula: \(P = P_{prescribed} + P_{accommodation}\) For the man with glasses, \( P = -46.8 + 4.0 = -42.8 D\) And the focal length with glasses: \(f_{glasses} = \frac{1}{P} = \frac{1}{-42.8} \approx -0.0234 m \) or \( -23.4 \mathrm{cm} \) Using the Lens formula again, we can find the nearest object distance (\(d_o\)) with the glasses: \(\frac{1}{-0.0234} = \frac{1}{d_o} + \frac{1}{0.02} \) Solving for \(d_o\), we get: \(d_o = \frac{1}{\frac{1}{0.02} - \frac{1}{-0.0234}} \approx 0.0808 m\) or \(8.08 \mathrm{cm}\) Thus, the nearest object he can see clearly with the glasses is at a distance of 8.08 cm.

Without the glasses, the Power of Accommodation is 4.0D. Using the accommodation and the lens formula, we can find the nearest object distance (\(d_o\)) without the glasses. The focal length without glasses: \(f_{without} = \frac{1}{P} = \frac{1}{4.0} = 0.25 m\) or \(25.0 \mathrm{cm} \) Using the Lens formula, we get: \(\frac{1}{0.25} = \frac{1}{d_o} + \frac{1}{0.02} \) Solving for \(d_o\), we find the nearest object distance without glasses: \(d_o = \frac{1}{\frac{1}{0.25} - \frac{1}{0.02}} \approx 0.0227 m\) or \(2.27 \mathrm{cm}\) The nearest object he can see clearly without glasses is at a distance of 2.27 cm.