Problem 99 - University Physics with Modern Physics

Question

A person with a near point of \(85 \mathrm{~cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

Step-by-Step Solution

Answer

(a) 19.7 cm; (b) 17.7 cm

Step 1: Understand the problem

We need to find the new near point when the person wears glasses with a power of "+2.25 diopters". The glasses are positioned 2.0 cm in front of the eyes. We will also find the near point when these glasses are used as contact lenses.

Step 2: Convert the power to focal length

The power of a lens is related to its focal length by the formula: \[ f = \frac{1}{P} \]where \(P = +2.25\ diopters.\) Thus,\[ f = \frac{1}{2.25} \approx 0.444\ m = 44.4\ cm \]

Step 3: Find the new near point with the glasses

The new near point \(d'\) can be found using the lens formula and the knowledge that the image formed by the lens must correspond to his normal near point \(d = 85\ cm.\)Using lens formula:\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]where \(v\) is the image distance. Since the image formed should be at the near point:\[ \frac{1}{v} - \frac{1}{85} = \frac{1}{44.4} \]Solving for \(v\),\[ \frac{1}{v} = \frac{1}{44.4} + \frac{1}{85} \approx 0.0565 \Rightarrow v \approx 17.7\ cm \]The glasses are 2.0 cm in front of the eyes, so the near point from the eyes is:\[ d' = 17.7 + 2.0 \approx 19.7\ cm \]

Step 4: Calculate near point if glasses were contacts

If the glasses were contact lenses, they would be at 0 distance from the eyes. We use the same lens formula for contact lenses with \(f = 44.4\ cm\).\[ \frac{1}{v} - \frac{1}{85} = \frac{1}{44.4} \]Solving for \(v\),\[ \frac{1}{v} = \frac{1}{44.4} + \frac{1}{85} \approx 0.0565 \Rightarrow v \approx 17.7\ cm \]This means the near point with contact lenses is \(v = 17.7\ cm\) since there's no 2.0 cm additional distance.

Key Concepts

Lens Power

Lens power is a measure of how much a lens can converge or diverge light. It is crucial for correcting vision. The power of a lens is measured in diopters (D) and is the inverse of its focal length (in meters).
The formula is \[ P = \frac{1}{f} \]where \( P \) is the lens power and \( f \) is the focal length.
The more powerful the lens, the shorter its focal length. In this exercise, glasses with a +2.25 diopter power can significantly alter the focus of light, helping to correct the user's near vision.

• Positive power lenses (plus lenses) are used for farsightedness (hyperopia).
• Negative power lenses (minus lenses) are used for nearsightedness (myopia).

Focal Length

The focal length of a lens is the distance from the lens to the point where it focuses light is converging or diverging. A shorter focal length indicates a stronger lens, as it bends light more sharply.
We calculate focal length using the lens power formula, where:\[ f = \frac{1}{P} \]For the lens power \(+2.25\) diopters, the focal length is approximately \(44.4\ cm\) or \(0.444\ m\).
Focal length is a cornerstone in determining how lenses change the perceived distance of objects, crucial for calculating precise vision corrections.

Near Point

The near point is the closest distance at which the eye can focus on an object clearly. It's significant for those needing reading glasses or bifocals.
For instance, in this problem, the person has a near point of \(85\ cm\). That means, without glasses, they can comfortably see objects only from that distance and further away.
By altering the lens type or position, like using different glasses or contact lenses, the near point can shift closer, resulting in clearer vision at shorter distances. With the glasses worn in this scenario, the near point becomes approximately located at \( 19.7\ cm\), showing a marked improvement in closeness.

Contact Lenses

Contact lenses differ from glasses in that they sit directly on the surface of the eye, eliminating any gap that glasses introduce.
This closeness significantly affects how corrective lens power is perceived and applied.

• Since contact lenses do not have a distance from the eye, calculations that depend on lens-to-eye distance are altered.
• In this exercise, for contact lenses, the near point is calculated to be approx \( 17.7\ cm\), removing the gap induced by glasses worn 2 cm away from the eyes.

Contact lenses provide a more natural correction by closely matching the eye’s shape, offering seamless clarity when transitioning between various focal points.