Question

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?

Short answer

The uncorrected far point of the man's eye is 4.75 meters. (b) What is the refractive power of the lenses required for distance vision? The refractive power of the lenses required for distance vision is 0.21 D. (c) What are the refractive powers of the two lenses required in bifocals for clear vision from 25 cm to infinity? The refractive powers of the two lenses required in bifocals for clear vision from 25 cm to infinity are -1.0 D and 0.21 D.

Step-by-step solution

(a) Finding the uncorrected far point

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 \(4.75\mathrm{m}\).

(b) Finding the refractive power lenses for distance vision

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.

(c) Finding the refractive powers of two lenses in bifocals

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.