A man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. Assume the glasses are $2.0 \mathrm{cm}$ from his eyes. (a) What is his uncorrected far point? (b) What refractive power lenses should he use for distance vision? (c) His uncorrected near point is \(1.0 \mathrm{m} .\) What should the refractive powers of the two lenses in his bifocals be to give him clear vision from \(25 \mathrm{cm}\) to infinity?

The given problem states that a man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. The power of the glasses is given by the formula \(P = 1/f\), where \(f\) is the focal length. We can now calculate the focal length of the lens, \(f = \frac{1}{P} = \frac{1}{+2.0 \mathrm{D}} = +0.5\mathrm{m}\). Now, we can use the lens formula, which is given as follows: \(\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_i}\), where \(d_0\) is the object distance, \(d_i\) is the image distance, and \(f\) is the focal length of the lens. Since the glasses are \(2.0 \mathrm{cm}\) away from his eyes, and the book is \(40.0 \mathrm{cm}\) away, the object distance \(d_0 = 40\mathrm{cm} - 2\mathrm{cm} = 38\mathrm{cm} = 0.38\mathrm{m}\). The image distance \(d_i\) can be found due to the relaxed eye and be considered as uncorrected far point, substituting the values in the lens formula: \(\frac{1}{0.5}=\frac{1}{0.38}+\frac{1}{d_i}\). Solving for \(d_i\), we get: \(d_i = \frac{1}{\frac{1}{0.5} - \frac{1}{0.38}} \approx 4.75\mathrm{m}\). So, his uncorrected far point is \(4.75\mathrm{m}\).

To find the refractive power lenses for distance vision, we need the lens power that can produce an image at infinity. We can achieve this if the image distance \(d_i = \infty\). In that case, the lens formula becomes: \(\frac{1}{f}=\frac{1}{d_0}\). Rearranging and plugging in the uncorrected far point value: \(f = d_0 = 4.75\mathrm{m}\). Now, we can find the lens power for distance vision using the lens power formula: \(P = \frac{1}{f} = \frac{1}{4.75} \approx 0.21 \mathrm{D}\). The man should use lenses with \(0.21 \mathrm{D}\) power for distance vision.

We need to find the lens powers for clear vision from \(25\mathrm{cm}\) to infinity. The lenses in bifocals should have two separate powers, one for near vision and one for distance vision. To correct near vision (from \(25\mathrm{cm}\)), given the uncorrected near point is \(1.0 \mathrm{m}\), let's apply the lens formula: \(\frac{1}{f_\text{near}} = \frac{1}{1.0} - \frac{1}{0.25}\). Solving for \(f_\text{near}\), \(f_\text{near} = \frac{1}{\frac{1}{1.0} - \frac{1}{0.25}} \approx -0.333\mathrm{m}\). Now, finding the power for the near lens: \(P_\text{near} = \frac{1}{f_\text{near}} \approx -3.0\mathrm{D}\). Since he already has \(+2.0\mathrm{D}\) lenses for reading, he would need a power of \((-3.0\mathrm{D}) + (+2.0\mathrm{D}) = -1.0\mathrm{D}\) for near vision in his bifocals. For distance vision, we already found the correct lens power in part (b) to be \(0.21\mathrm{D}\). So, the two lenses in his bifocals should have refractive powers of \(-1.0\mathrm{D}\) and \(0.21\mathrm{D}\) for clear vision from \(25\mathrm{cm}\) to infinity.