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Chapter 2: Problem 69

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal \(25 \mathrm{cm}\). (a) What is the power of the eyes of a woman who can see an object clearly at a

Short Answer

Expert verified
The power of the woman's eyes is \(8 \mathrm{D}\), and she needs a contact lens with a power of \(-4 \mathrm{D}\) to see clearly at a distance of \(25 \mathrm{cm}\).

Step by step solution

01

Understanding Lens Power and Focal Length

The power (P) of a lens is related to its focal length (f) according to the formula: \[P = \frac{1}{f}\] where the power is in diopters (D) and the focal length is in meters (m). Therefore, we can find the power of the lens needed by using this equation.
02

Calculate the Power of Woman's Eyes

According to the problem, the woman can see clearly at a distance of \(12.5 \mathrm{cm}\) (which is her near point distance). We will use this information to find the power of the lens of her eyes. Since the near point distance is given in centimeters, we need to convert it into meters: \(12.5 \mathrm{cm} = 0.125 \mathrm{m}\) Now, using the lens power formula: \(P_\text{eye} = \frac{1}{f_\text{eye}}\) Where \(f_\text{eye}\) represents the focal length of her eyes. Since the near point distance is equal to the focal length: \(P_\text{eye} = \frac{1}{0.125}\) \(P_\text{eye} = 8 \mathrm{D}\) The power of the woman's eyes is \(8 \mathrm{D}\).
03

Calculating the Distance for Contact Lens to See Clearly at \(25 \mathrm{cm}\)

The normal near point distance is given as \(25 \mathrm{cm}\), which we need to convert into meters: \(25 \mathrm{cm} = 0.25 \mathrm{m}\) Now, the focal length of a normal eye is \(0.25 \mathrm{m}\), and we need to find the distance for a contact lens so that she can see clearly at this normal near point distance. Let's denote the power of the contact lens as \(P_\text{lens}\) and the focal length of the contact lens as \(f_\text{lens}\). The total power needed to see clearly at \(25 \mathrm{cm}\) will be the sum of the woman's eye power and the contact lens power. Using the lens power formula for the combined power: \(P_\text{total} = P_\text{eye} + P_\text{lens} = \frac{1}{f_\text{total}}\) Where \(f_\text{total}\) represents the combined focal length to see clearly at \(25 \mathrm{cm}\). Now, we can write the equation in terms of known values: \(\frac{1}{0.25} = 8 + P_\text{lens}\) Solve for \(P_\text{lens}\): \(P_\text{lens} = \frac{1}{0.25} - 8\) \(P_\text{lens} = 4 - 8\) \(P_\text{lens} = -4 \mathrm{D}\) The contact lens has a power of \(-4 \mathrm{D}\) in order for her to see clearly at a distance of \(25 \mathrm{cm}\).

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Key Concepts

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

Lens Power Formula
Understanding the lens power formula is crucial for calculating the strength of a lens required for clear vision. The formula is simple yet profound:

\[P = \frac{1}{f}\]

In this equation, 'P' represents the power of the lens, which is measured in diopters (D), and 'f' represents the focal length of the lens, measured in meters (m). Hence, the power of a lens is the reciprocal of its focal length. A short focal length indicates a stronger lens, as it can bend light rays more sharply, converging them to a focus in a shorter distance. Conversely, a longer focal length corresponds to a weaker lens, with a lesser ability to bend light rays.
Diopters
The diopter is the unit of measurement for the refractive power of a lens, symbolized by 'D'. One diopter represents the power of a lens with a focal length of one meter. In practical terms, regarding the corrective lenses for vision, diopters describe the degree to which a lens will converge or diverge light.

A positive diopter value denotes a converging lens, typically used for farsightedness, while a negative diopter value indicates a diverging lens, like those used to correct nearsightedness. As witnessed in the provided example, a lens power of 8D implies a strong convergent ability, suitable for someone who needs assistance in viewing objects in close proximity.
Focal Length
Focal length, the distance over which a lens focuses light, is a fundamental concept in optics. It determines how sharply a lens can focus light and, in relation to the human eye, translates to the distance at which one can see an object clearly without additional corrective lenses.

A short focal length in the context of vision means that the eye can focus on close objects without effort. That is to say, an individual with a focal length shorter than the average may see objects clearly at distances closer than the normal 25 cm near point. If one has a focal length of 0.125 meters, as our example illustrates, this person's eyes naturally focus at half the usual near point distance, exemplifying strong close-up vision.
Near Point Distance
Near point distance is the closest distance at which the eye can focus on an object. The standard near point distance for a person with normal vision is considered to be about 25 cm. However, this distance can vary quite significantly from one individual to another, influenced by factors such as age and the presence of certain eye conditions.

For instance, the woman in the given example has a near point distance of 12.5 cm, indicating she can focus comfortably at a much closer range than the average person. This ability to focus at a closer distance implies a stronger lens inside the eye or a shorter focal length. As one's near point distance decreases, the necessary lens power increases, demonstrating a direct relationship that helps optometrists determine the correct prescription for eyeglasses or contact lenses.

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Most popular questions from this chapter

What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person's face is \(12.0 \mathrm{cm}\) away? Explicitly show how you follow the steps in the Problem-Solving Strategy: Spherical Mirrors.

An object of height \(3.0 \mathrm{cm}\) is placed at \(25 \mathrm{cm}\) in front of a diverging lens of focal length \(20 \mathrm{cm}\). Behind the diverging lens, there is a converging lens of focal length \(20 \mathrm{cm} .\) The distance between the lenses is \(5.0 \mathrm{cm}\). Find the location and size of the final image.

A certain slide projector has a 100 mm-focal length lens. (a) How far away is the screen if a slide is placed \(103 \mathrm{mm}\) from the lens and produces a sharp image? (b) If the slide is 24.0 by \(36.0 \mathrm{mm}\), what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-Solving Strategy: Lenses.

A near-sighted person has afar point of 80 cm. (a) What kind of corrective lens the person will need if the lens is to be placed \(1.5 \mathrm{cm}\) from the eye? (b) What would be the power of the contact lens needed? Assume distance to contact lens from the eye to be zero.

Let objective and eyepiece of a compound the lege microscope have focal lengths of \(2.5 \mathrm{cm}\) and \(10 \mathrm{cm}\) respectively and be separated by \(12 \mathrm{cm} .\) A 70 - \(\mu \mathrm{m}\) object is placed \(6.0 \mathrm{cm}\) from the objective. How large is the virtual image formed by the objective-eyepiece system?
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