Problem 59 - College 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 hear point be if his old glasses were contact lenses instead?

Step-by-Step Solution

Answer

(a) Near point with glasses: 30 cm. (b) Near point with contacts: 30.7 cm.

Step 1: Understand the Problem

A person with a near point of 85 cm usually needs glasses to see things up close. We are given the power of old glasses as +2.25 diopters, and asked to find the new near point both when the glasses are 2.0 cm in front of the eye and when they are contacts.

Step 2: Identify the Lens Formula

The lens formula is given by \( \,\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \,\), where \(\, u\) is the object distance (near point), \(\, v\) is the image distance, and \(\, f\) is the focal length of the lens.

Step 3: Calculate Focal Length from Diopters

The power \( P \) of a lens in diopters is related to its focal length \( f \) in meters by \( P = \frac{1}{f} \). For the given lens power of +2.25 diopters, calculate: \[ f = \frac{1}{2.25} \approx 0.444 \text{ meters} \]

Step 4: Adjust Object Distance for Lens Position (Glasses)

When wearing glasses, the lens is 2 cm in front of the eye. If the eye's near point is 85 cm, the near point with glasses (object distance \( u \) from the lens) is \( u = 85 \text{ cm} - 2 \text{ cm} = 83 \text{ cm} = 0.83 \text{ m}. \)

Step 5: Apply Lens Formula for Glasses

Using the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), solve for \( v \) when \( f = 0.444 \) and \( u = 0.83 \):\[ \frac{1}{0.444} = \frac{1}{v} - \frac{1}{0.83} \]\[ \frac{1}{v} = \frac{1}{0.444} + \frac{1}{0.83} \approx 2.25 \text{ diopters} \]Solve for \(v\): \( v \approx 0.30 \text{ m} \).

Step 6: Calculate Near Point with Contacts

For contact lenses, \(\, v \) equals the new near point as the contacts rest on the eye. Use \( f = 0.444 \):\[ \frac{1}{0.444} = \frac{1}{v} - \frac{1}{0.85} \]\[ \frac{1}{v} = \frac{1}{0.444} + \frac{1}{0.85} \]Solve for \(v\) to find the effective near point with contacts.

Step 7: Solve for Near Point with Contacts

Solving the equation, we find:\[ \frac{1}{v} \approx 2.25 \, + \, 1.176 \]\[ v \approx 0.307 \text{ m} \approx 30.7 \text{ cm} \] when wearing contact lenses.

Key Concepts

Lens Formula

The lens formula is essential in understanding how lenses form images. It's expressed as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f \) stands for the focal length of the lens, \( v \) is the image distance, and \( u \) is the object distance. This formula helps us calculate the location where the image of an object will form based on the position of the object and the focal length of the lens.
This formula is a crucial part of optics because it describes how lenses bend light to aid our vision or enhance instruments like cameras and telescopes. By rearranging this formula, you can solve for any of the missing variables when the other two are known, making it a versatile tool in lens-related calculations.

Diopters

Diopters measure the optical power of lenses. They’re expressed in inverse meters (m⁻¹). It's calculated as the reciprocal of the focal length in meters: \( P = \frac{1}{f} \). For instance, a lens with a power of +2.25 diopters has a focal length of approximately 0.444 meters.
Adding or subtracting diopters from lenses directly affects how they focus light. Positive diopters are used to correct farsightedness by converging light rays, and negative diopters are for nearsightedness, diverging the light rays. Understanding diopters helps in selecting the right lens for vision correction or optical tasks.

Focal Length

Focal length is the distance from the lens at which it converges or diverges light to a point. It's a crucial measurement that influences how optical devices magnify or minify objects. Focal length is measured in meters or centimeters and is inversely related to diopters. Shorter focal lengths result in stronger converging power and higher diopters.
In practical scenarios, a lens with a focal length of 0.444 meters, as derived from its diopter value (+2.25 diopters), helps us calculate the optical effects on image distances, which is essential when designing things like glasses and cameras for optimal focus.

Near Point

The near point is the closest distance at which the eye can focus an object clearly. It’s an important aspect in determining the need for corrective lenses. A normal near point is about 25 cm for a typical young adult. When an individual’s near point extends further, it indicates a possible need for correction to aid near vision.
In the problem, the person has a near point of 85 cm. Using corrective lenses, the near point changes based on the lenses' power and position relative to the eyes, allowing the person to comfortably read or see objects up close when wearing glasses or contact lenses.

Corrective Glasses

Corrective glasses adjust how light enters the eye, compensating for issues like nearsightedness or farsightedness. When worn, these glasses bring the near point closer or make faraway objects clear, adapting to the visual requirements of the wearer.
In the given physics problem, the glasses with a power of +2.25 diopters adjust the near point from 85 cm to much closer distances, helping the person focus on nearby objects. Unlike regular glasses, contact lenses sit on the eye and directly adjust the effective distance light travels into the eye, generally offering a slightly different correction magnitude from spectacles.