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Ch 34: Geometric Optics
All textbooksYoung & Freedman Calc 14th EditionCh 34: Geometric OpticsProblem 54c
Chapter 34, Problem 54c

BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. What focal length contact lens is needed, and what is its power in diopters?

Verified step by step guidance
1
Determine the type of lens needed: Since the person cannot focus on objects beyond 75.0 cm, they are farsighted. This means a diverging lens is required to correct their vision.
Identify the far point of the person's vision: The far point is the maximum distance at which the person can see clearly without correction. Here, it is given as 75.0 cm, or 0.75 m.
Set up the lens formula: The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length of the lens, \( v \) is the image distance (the far point of the corrected vision, which is infinity for normal vision), and \( u \) is the object distance (the far point of the person's uncorrected vision, which is -0.75 m since it is virtual).
Simplify the lens formula for this case: Since \( v \to \infty \), \( \frac{1}{v} = 0 \). The formula simplifies to \( \frac{1}{f} = -\frac{1}{u} \). Substitute \( u = -0.75 \) m to find \( f \).
Calculate the power of the lens: The power \( P \) of a lens in diopters is given by \( P = \frac{100}{f} \), where \( f \) is in meters. Use the focal length obtained in the previous step to calculate the power.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Focal Length

Focal length is the distance from the lens to the point where parallel rays of light converge or appear to diverge. For corrective lenses, the focal length is crucial as it determines how well the lens can focus light onto the retina. In this case, the focal length needed can be calculated based on the distance at which the person can see clearly.
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Lens Power

Lens power, measured in diopters, is the reciprocal of the focal length in meters. It indicates the strength of the lens required to correct vision. A positive power indicates a converging lens, which is typically used for hyperopia (farsightedness), while a negative power indicates a diverging lens for myopia (nearsightedness).
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Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens, expressed as 1/f = 1/v - 1/u. This formula is essential for determining the necessary focal length of the contact lens based on the person's vision limitations. By applying this formula, one can find the focal length needed to correct the person's inability to focus on distant objects.
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Related Practice
Textbook Question

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. Is this person nearsighted or farsighted?

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Textbook Question

BIO Ordinary Glasses. Ordinary glasses are worn in front of the eye and usually 2.0 cm in front of the eyeball. Suppose that the person in Exercise 34.52 prefers ordinary glasses to contact lenses. What focal length lenses are needed to correct his vision, and what is their power in diopters?

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Textbook Question

BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. Is she nearsighted or farsighted?

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Textbook Question

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. What angular magnification is obtainable with the lens if the object is at the focal point?

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Textbook Question

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

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Textbook Question

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. What is the magnitude of the linear magnification produced by the objective?

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