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Chapter 34: Problem 99

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?

Short Answer

Expert verified
(a) 19.7 cm; (b) 17.7 cm

Step by step solution

01

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.
02

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,\[ f = \frac{1}{2.25} \approx 0.444\ m = 44.4\ cm \]
03

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:\[ d' = 17.7 + 2.0 \approx 19.7\ cm \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 shape, offering seamless clarity when transitioning between various focal points.

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Most popular questions from this chapter

A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 cm. Using this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 cm from the lens. What is the index of refraction of the lens?

A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
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