Problem 46 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A person has a far point of \(14 \mathrm{~cm}\). What power glasses would correct this vision if the glasses were placed \(2.0 \mathrm{~cm}\) from the eye? What power contact lenses, placed on the eye, would the person need?

Step-by-Step Solution

Answer

Glasses: 8.33 D; Contacts: 7.14 D.

Step 1: Understanding the Problem

The far point of a person is the farthest distance at which they can see clearly without any corrective lenses. In this case, it's given as 14 cm. We need to find the power of eyeglasses and contact lenses that would enable the person to see objects at infinity. The eyeglasses are placed 2 cm from the eye.

Step 2: Setting Up the Glasses Problem

To correct vision, glasses need to create an image at the person's far point (14 cm away) of an object at infinity. The lens-eye distance affects this, so the actual distance from the lens to the far point is 14 cm - 2 cm = 12 cm.

Step 3: Using Lens Formula for Glasses

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For an object at infinity, \( u \to \infty \), so \( \frac{1}{u} = 0 \) and \( \frac{1}{f} = \frac{1}{v} \). Therefore, \( v = 12 \) cm. So \( \frac{1}{f} = \frac{1}{12} \) cm.

Step 4: Calculating Power for Glasses

The power of a lens \( P \) is given by \( P = \frac{1}{f(m)} \), where \( f(m) \) is the focal length in meters. Convert 12 cm to meters: 12 cm = 0.12 m. Thus, \( P = \frac{1}{0.12} \approx 8.33 \) diopters.

Step 5: Setting Up the Contact Lens Problem

For contact lenses placed directly on the eye, the effective distance to the far point is 14 cm, as there is no distance between the lens and the eye.

Step 6: Using Lens Formula for Contact Lenses

Using the same approach as above, \( v = 14 \) cm, so \( \frac{1}{f} = \frac{1}{14} \).

Step 7: Calculating Power for Contact Lenses

Convert \( f = 14 \) cm to meters: 14 cm = 0.14 m. The power required is \( P = \frac{1}{0.14} \approx 7.14 \) diopters.

Key Concepts

Vision Correction

Vision correction is all about helping people see the world more clearly.
When someone has a vision problem, it usually means their eyes can't focus light directly on the retina for a clear image.
This leads to blurry vision. Typically, vision corrections are achieved through devices like glasses and contact lenses.

• Glasses: These are lenses mounted in frames that are worn a short distance from the eyes.
• Contact Lenses: These are lenses placed directly on the surface of the eye.

For someone whose far point is too close—like in this problem where the person can only see clearly up to 14 cm—special lenses are needed.
These lenses help refocus light onto the retina so that distant objects appear clear.
By understanding these concepts, we learn how corrective lenses bring distant objects into focus, alleviating any visual discomfort.

Lens Formula

The lens formula is a crucial tool for calculating the relationships between the distance of an object, the image, and the focal length of a lens. The formula is expressed as:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where:

• \( f \): Focal length of the lens
• \( v \): Image distance (distance from the lens to the image)
• \( u \): Object distance (distance from the lens to the object)

In scenarios like vision correction, usually, we deal with objects that are far away, essentially at infinity.
This simplifies our calculations as \( \frac{1}{u} \) becomes 0, making the lens formula easier to use:\[ \frac{1}{f} = \frac{1}{v} \]For the eyeglasses in the problem, the image distance required is 12 cm.
For contact lenses, since they sit right on the eye, we work with the full 14 cm.
Thus, applying the lens formula helps us determine the needed power of lenses to correct vision at specific distances.

Diopters

Diopters are the unit of measurement used to express the optical power of a lens.
The higher the diopter number, the stronger the lens.A lens's power in diopters is calculated as the reciprocal of its focal length in meters:\[ P = \frac{1}{f(m)} \]Where:

• \( P \): Power of the lens in diopters
• \( f(m) \): Focal length in meters

In the provided problem, we determined the power of eyeglasses by converting 12 cm to 0.12 m, resulting in about 8.33 diopters.
For contact lenses, 14 cm converts to 0.14 m, and the calculated power is approximately 7.14 diopters.Understanding diopters is essential for anyone needing corrective lenses, as it directly relates to how much clearer their vision will become with a particular lens.
Diopters dictate how well a lens converges or diverges light, impacting clarity and visual comfort.