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A person's prescription for her new bifocal glasses calls for a refractive power of -0.445 diopter in the distance-vision part, and a power of +1.85 diopters in the close-vision part. What are the near and far points of this person's uncorrected vision? As. sume the glasses are \(2.00 \mathrm{cm}\) from the person's eyes, and that the person's near-point distance is \(25.0 \mathrm{cm}\) when wearing the glasses.

Short Answer

Expert verified
Compute the focal lengths, solve the lens equations, and convert to determine uncorrected near and far points.

Step by step solution

01

Understand the Problem

The problem provides two focal powers related to bifocal lenses, and we need to determine the person's near and far point vision without the glasses. The prescription for the distance vision is -0.445 diopters, and for the close vision, it is +1.85 diopters. The glasses are worn 2.00 cm away from the eyes, and the individual's near point with glasses is 25.0 cm. We need to use this information to find the corresponding uncorrected vision points.
02

Convert Diopters to Focal Lengths

The formula to convert diopters to focal length in meters is given by:\[ f = \frac{1}{P}, \]where \(f\) is the focal length and \(P\) is the power in diopters. Therefore:For distance vision: \[ f_d = \frac{1}{-0.445} \approx -2.247 \text{ m}. \]For close vision:\[ f_c = \frac{1}{1.85} \approx 0.541 \text{ m}. \]
03

Calculate Image Distances

To calculate the image distances, consider the glasses being 2.00 cm (0.02 m) away from the eye:For distance vision, using the lens formula:\[ \frac{1}{v} = \frac{1}{0.02} + \frac{1}{f_d}. \]Solving this for \(v\), the viewer's far point, we use:\[ \frac{1}{v} = \frac{1}{0.02} - \frac{1}{2.247}. \]Find \(v\) by solving the equation.
04

Adjust for Near Point with Glasses

The near point with the glasses is 25.0 cm (0.25 m). Using the initial formula with the close vision diopter:\[ \frac{1}{v_n} = \frac{1}{0.25} - \frac{1}{f_c}. \]Solve for \(v_n\), which gives the individual's accurate near point when not using the glasses.
05

Compute Near and Far Points Without Glasses

After solving the above equations, the near and far points of the uncorrected vision can be calculated from the respective image distances derived for the far and near points. This involves using the reverse of the transformations applied by wearing glasses. Calculate these values in meters to determine the uncorrected vision.

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Key Concepts

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

Refractive Power
The refractive power of a lens is a measure of how much it can bend light to form an image. This property is essential in vision correction, as it determines how well lenses can adjust focus for clear sight at different distances. Refractive power is expressed in diopters, which represent the reciprocal of the focal length. For example, a lens with a refractive power of +2 diopters has a focal length of 0.5 meters. Similarly, a negative diopter value indicates a lens used for correcting farsightedness, by shifting the focus of distant objects onto the retina. Understanding refractive power helps in selecting the correct lenses to address vision issues like nearsightedness and farsightedness.
Diopter
Diopters are the units used to describe the refractive power of lenses. The concept is straightforward: a diopter value indicates how strongly a lens can converge or diverge light. One diopter equals a lens with a focal length of one meter. A positive diopter number, like +1.85, is typical for converging lenses used in reading glasses, which help with seeing close objects. Conversely, negative diopters, such as -0.445, correspond to diverging lenses, which assist in seeing far objects by correcting nearsightedness. Diopters allow optometrists to precisely prescribe lenses that accommodate individual vision requirements.
Focal Length Conversion
Converting diopters to focal length is critical for understanding how lenses influence vision. The focal length, often given in meters, is the distance at which parallel rays of light are focused by the lens. The conversion formula is:\[ f = \frac{1}{P} \]where \(f\) is the focal length in meters and \(P\) is the lens's power in diopters. For example, a lens prescribed with +1.85 diopters will have its focal length calculated as \(f = \frac{1}{1.85}\), which is approximately 0.541 meters. This calculation is central to understanding the lens's capacity to adjust vision, enabling one to quantify the extent to which a lens can bring clarity to the visual field.
Lens Formula
The lens formula is a critical equation in optics that relates object distance, image distance, and focal length. It is expressed as:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. By rearranging this formula, one can solve for any of the three variables given the other two. In practical situations, such as the study of bifocal glasses, the formula helps calculate uncorrected near and far points of vision. For instance, adjusting for the glasses being worn 2 cm from the eyes, one can use this formula to determine the adjusted image distances and thus reveal uncorrected vision metrics. Understanding the lens formula is crucial for accurately addressing vision correction needs.

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Most popular questions from this chapter

To engrave wishes of good luck on a watch, an engraver uses a magnifier whose focal length is \(8.65 \mathrm{cm}\). If the image formed by the magnifier is at the engraver's near point of \(25.6 \mathrm{cm}\) find (a) the distance between the watch and the magnifier and

A Big Eye The largest eye ever to exist on Earth belonged to an extinct species of ichthyosaur, Temmodontosaurus platyodon. This creature had an eye that was \(26.4 \mathrm{cm}\) in diame ter. It is estimated that this ichthyosaur also had a relatively large pupil, giving it an effective aperture setting of about \(f / 1.1\). (a) Assuming its pupil was one-third the diameter of the eye, what was the approximate focal length of the ichthyosaur's eye? (b) When the ichthyosaur narrowed its pupil in bright light, did its \(f\) -number increase or decrease? Explain.

A converging lens of focal length \(8.000 \mathrm{cm}\) is \(20.0 \mathrm{cm}\) to the left of a diverging lens of focal length \(-6.00 \mathrm{cm} .\) A coin is placed \(12.0 \mathrm{cm}\) to the left of the converging lens. Find (a) the location and (b) the magnification of the coin's final image.

Fitting Contact Lenses with a Keratometer When a patient is being fitted with contact lenses, the curvature of the patient's cornea is measured with an instrument known as a keratometer. A lighted object is held near the eye, and the keratometer measures the magnification of the image formed by reflection from the front of the cornea. If an object is held \(10.0 \mathrm{cm}\) in front of a patient's eye, and the reflected image is magnified by a factor of \(0.035,\) what is the radius of curvature of the patient's cornea?

IP A person needs glasses with a refractive power of -1.35 diopters to be able to focus on distant objects. (a) Is this person nearsighted or farsighted? Explain. (b) What is this person's (unaided) far point?

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