Problem 20 - College Physics

Question

Contact lenses. 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 \(\mathrm{cm}\) from his eyes instead of the usual 25.0 \(\mathrm{cm} .\) (a) Is this person nearson nearsighted or farsighted? (b) What type of lens (con- verging 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: Determine if the person is nearsighted or farsighted

A person who has difficulty seeing nearby objects but can see distant objects is considered farsighted. Since this individual can see distant objects well but has a near point of 45.0 cm instead of the typical 25.0 cm, they are farsighted.

Step 2: Identify the type of lens needed

To correct farsightedness, a converging lens is needed. Converging lenses help focus the light directly on the retina, thus allowing the person to see nearby objects clearly.

Step 3: Calculate the focal length of the lens

We use the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance.For this problem:- Set \(d_o = 25\, \text{cm}\) (the normal near point for a person)- Set \(d_i = -45\, \text{cm}\) (the person's actual near point; negative indicates a virtual image formed at the near point)Substitute:\[ \frac{1}{f} = \frac{1}{25} - \frac{1}{45} \]Calculate:\[ \frac{1}{f} = \frac{45 - 25}{25 \times 45} = \frac{20}{1125} \]Simplifying gives:\[ f = \frac{1125}{20} = 56.25 \text{cm} \]

Step 4: Calculate the power of the lens in diopters

The power of a lens \(P\) in diopters is given by \(P = \frac{1}{f(\text{in meters})}\).First, convert the focal length from centimeters to meters:\[ f = 56.25 \text{cm} = 0.5625 \text{m} \]Now, calculate the power:\[ P = \frac{1}{0.5625} \approx 1.78 \text{ D} \]

Step 5: Conclusion: Recap the solutions

(a) The person is farsighted. (b) A converging lens is needed. (c) The focal length needed is 56.25 cm, and its power is 1.78 diopters.

Key Concepts

Farsightedness

Farsightedness, also known as hyperopia, is a common condition where individuals experience difficulty focusing on nearby objects. This occurs because of the way light enters the eye. In a farsighted eye, the cornea and lens do not bend incoming light enough, causing the image to focus behind the retina instead of directly on it. Consequently, close objects appear blurred, while distant objects remain clear. This condition often appears with aging as the lens in the eye becomes less flexible, making it harder to focus on close objects. Identifying farsightedness involves checking the near point of vision. For the average person, this is approximately 25 cm. If an individual's near point is significantly further than this, for example, 45 cm as seen in some cases, they are likely to be farsighted.

Converging Lens

To correct farsightedness, a converging lens is commonly used. Converging lenses are designed to bend light rays inwards. This adjustment helps the lens of the eye focus images directly on the retina rather than behind it. These lenses are usually convex in shape and can vary in thickness depending on the level of correction needed.

• They gather light from a close object and redirect it to meet at a point that focuses correctly on the retina.
• This adjustment allows the person to see close objects clearly, bringing them into a sharper focus.

When using converging lenses in optical devices such as glasses or contact lenses, they can significantly improve the clarity of near objects for those with farsightedness.

Diopters

Diopters are units used to measure the optical power of a lens. They represent the degree to which a lens can bend light, i.e., how much it converges or diverges light. When we're talking about converging lenses for farsightedness, the value in diopters is positive. The higher the diopter value, the stronger the lens's ability to focus light and help with close vision.

• To calculate diopters, you use the formula: \[ P = \frac{1}{f(\text{in meters})} \]
• The result tells you how much power is required to adjust the focal point to lie on the retina.

Diopters allow optometrists to prescribe the correct lens strength needed to correct issues like farsightedness, ensuring the lens has the correct curvature and focusing power.

Lens Formula

The lens formula is a fundamental equation used in optics to relate the focal length of a lens to the object distance and the image distance. It is expressed as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where:

• \( f \) is the focal length of the lens,
• \( d_o \) is the distance from the object to the lens,
• \( d_i \) is the distance from the image to the lens.

This formula is particularly useful when calculating the necessary focal length of corrective lenses. In the context of farsighted correction, it aids in determining how a lens should be shaped in order to direct light correctly onto the retina. For accurate correction, a person's specific eye measurements, such as their near point, are critical inputs in this formula.