Problem 52 - University Physics with Modern Physics

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. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

Step-by-Step Solution

Answer

(a) Farsighted; (b) Converging lens; (c) Focal length: 56.25 cm, Power: +1.78 D.

Step 1: Understanding Nearsightedness vs. Farsightedness

Nearsightedness (myopia) occurs when a person can see nearby objects clearly but has difficulty with distant objects. Farsightedness (hyperopia) is the opposite—distant objects are clear, but nearby objects are blurred. Since the person's near point is 45.0 cm rather than the usual 25.0 cm, they are farsighted.

Step 2: Identifying the Required Lens Type

Farsighted individuals need converging lenses to correct their vision. This lens type helps focus images closer, reducing the near point distance to normal (25.0 cm).

Step 3: Calculating the Focal Length of the Lens

To find the focal length of the lens needed, use the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( v \) is the desired image distance (25.0 cm) and \( u \) is the given near point (45.0 cm). Substituting these values, we get: \( \frac{1}{f} = \frac{1}{25} - \frac{1}{45} \). Solving this gives \( f \approx 56.25 \) cm.

Step 4: Converting Focal Length to Diopters

The power of a lens in diopters (D) is given by the formula \( P = \frac{100}{f} \), where \( f \) is the focal length in centimeters. Substituting the focal length from the previous step, \( P = \frac{100}{56.25} \approx +1.78 \). Thus, the power of the lens needed is +1.78 diopters.

Key Concepts

Farsightedness

Farsightedness, also known as hyperopia, occurs when a person finds it difficult to see close objects clearly. This condition is contrasted with nearsightedness, where distant objects become blurry. For farsighted individuals, the near point—the closest point they can see clearly—is farther than it should be. In our exercise, we determined the person is farsighted because their near point extends to 45.0 cm rather than the usual 25.0 cm.
Farsightedness happens because the light entering the eye focuses behind the retina instead of on it, often due to a shorter-than-normal eyeball or less curvature of the cornea. This causes difficulty in reading or focusing on nearby tasks without corrective lenses.

Converging Lenses

Converging lenses, often known as convex lenses, are the solution for correcting farsightedness. These lenses bend parallel light rays toward a focal point, thereby focusing light more effectively to that point.
By employing converging lenses, images that once focused behind the retina can now be shifted onto the retina, which aids in clear vision for close objects. In the context of our exercise, converging lenses help reduce the near point of the farsighted individual to a normal distance, making it possible for them to see close up without strain. These lenses are characteristically thicker in the middle than at the edges.

• Convex shape
• Focus light directly on the retina
• Corrects hyperopia by adjusting the image focus point

Lens Formula

The lens formula is a fundamental part of optics that relates the object distance, image distance, and focal length of a lens. It's given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where:
- \( f \) is the focal length of the lens
- \( v \) is the image distance
- \( u \) is the object distance.
This formula is critical in finding out how powerful a lens needs to be in order to correct vision issues. In the case of farsightedness, we aim to bring the near point back to a standard distance, which requires calculating the necessary focal length using this formula.
In our problem, using given near point (\( u = 45 \) cm) and desired image distance (\( v = 25 \) cm), we solve the lens formula to find \( f \approx 56.25 \) cm.

Diopters

Diopters are units used to measure the optical power of a lens. They indicate how strongly a lens converges (or diverges) light. The power is calculated as the reciprocal of the focal length (in meters). The formula converts focal length to diopters: \[ P = \frac{100}{f} \]where \( f \) is the focal length in centimeters.
For our particular exercise, after calculating the focal length as approximately 56.25 cm, the power in diopters is computed to be approximately +1.78. This positive value signifies a converging lens, suitable for correcting farsightedness. Diopters not only help in quantifying lens power but also assist optometrists in determining the correct lens type needed for vision correction.