Question

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.

Short answer

Answer: The focal length of the corrective lens needed is -0.286 meters (-3.5 diopters). The nearest object he can see clearly with glasses is about 2.0 cm, and without glasses, it is about 1.85 cm.

Step-by-step solution

Find the man's far point without glasses

We have the maximum distance he can see clearly, which is his far point, as \(2.0 \mathrm{m}\). Therefore, the far point without glasses is \(2.0 \mathrm{m}\).

Determine the focal length of the corrective lens needed

To find the corrective lens for this far point, we will use the formula for lens power: \(P = \frac{1}{f}\) where \(P\) represents the power of the lens (in diopters), and \(f\) represents the focal length (in meters). We know the power needed to adjust for the man's far point without glasses: \(P = \frac{1}{2} = 0.5 \mathrm{D}\) Since the corrective lenses are \(2.0\mathrm{cm}\) away from the eyes, the total power of the corrective eyeglasses lenses should be: \(P_{corrective} = P - 4.0\mathrm{D}\) \(P_{corrective} = 0.5\mathrm{D} - 4.0\mathrm{D} = -3.5\mathrm{D}\) Now, we can find the focal length of the corrective lens needed: \(f_{corrective} = \frac{1}{P_{corrective}} = \frac{1}{-3.5\mathrm{D}} = -\frac{1}{3.5} m = -0.286\mathrm{m}\) So, as an amateur optometrist, we would prescribe a corrective lens with a power of \(-3.5\mathrm{D}\) and a focal length of \(-0.286\mathrm{m}\).

Find the nearest objects he can see clearly with and without glasses

First, let's find the maximum power of accommodation the man has with glasses: \(P_{max} = 4.0\mathrm{D}\) With his new glasses, the man's near point (the nearest object he can see clearly) can be found using the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) where \(d_o\) is the object distance and \(d_i\) is the image distance (in this case, the distance from the lenses to the retina). We know that \(d_i = 2.0\mathrm{cm}\) and \(P_{max} + P_{corrective} = 0.5\mathrm{D}\). Therefore, we can find the nearest object distance \(d_o\): \(\frac{1}{d_o} = P_{max} + P_{corrective} - \frac{1}{d_i}\) \(\frac{1}{d_o} = 0.5\mathrm{D} - \frac{1}{0.02\mathrm{m}}\) \(d_o = \frac{1}{0.5\mathrm{D} + 50.0\mathrm{D}} = \frac{1}{50.5\mathrm{D}} = 0.0198\mathrm{m} \approx 2.0\mathrm{cm}\) The nearest object he can see clearly with his glasses is about \(2.0\mathrm{cm}\). Without his glasses, the man can use the full power of accommodation: \(P_{total} = 4.0\mathrm{D}\) Now, finding the nearest object distance without glasses: \(\frac{1}{d'_o} = P_{total} - \frac{1}{d_i}\) \(\frac{1}{d'_o} = 4.0\mathrm{D} - \frac{1}{0.02\mathrm{m}}\) \(d'_o = \frac{1}{4.0\mathrm{D} + 50.0\mathrm{D}} = \frac{1}{54.0\mathrm{D}} = 0.0185\mathrm{m} \approx 1.85\mathrm{cm}\) The nearest object he can see clearly without his glasses is about \(1.85\mathrm{cm}\).