Problem 72 - Physics

Question

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper \(42.00 \mathrm{~cm}\) from her eyes in order to read it. She wears the eyeglasses \(2.00 \mathrm{~cm}\) from her eyes. How far is her near point from her eyes?

Step-by-Step Solution

Answer

Her near point is 1.21 meters from her eyes.

Step 1: Understand the problem

We need to find how far the woman's near point is from her eyes while wearing an old pair of glasses with a known refractive power. We know the distance from her eyes to the newspaper when she holds it to read, and the distance from her eyes to the glasses.

Step 2: Identify the given values

The refractive power of the eyeglasses is given as \(1.660\) diopters. The woman holds the newspaper \(42.00\, \text{cm}\) from her eyes to read it. The glasses are \(2.00\, \text{cm}\) from her eyes.

Step 3: Calculate the distance from the glasses to the newspaper

Convert all measurements to meters for consistency. The distance from the eyes to the newspaper is \(0.42\, \text{m}\), and the distance from the eyes to the glasses is \(0.02\, \text{m}\). The distance from the glasses to the newspaper is \(0.42\, \text{m} - 0.02\, \text{m} = 0.40\, \text{m}\).

Step 4: Use the lens formula

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length (reciprocal of refractive power), \( d_o \) is the object distance (from glasses to newspaper), and \( d_i \) is the image distance (from glasses to near point).

Step 5: Calculate the focal length

Refractive power \( P = 1.660 \, \text{diopters} \) means \( f = \frac{1}{1.660} \, \text{m} \approx 0.6024 \, \text{m} \).

Step 6: Solve for image distance

Use the lens formula with \( d_o = 0.40 \, \text{m} \): \[ \frac{1}{0.6024} = \frac{1}{0.40} + \frac{1}{d_i} \].Solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{0.6024} - \frac{1}{0.40} \).

Step 7: Calculate near point distance from glasses

Calculate \( \frac{1}{d_i} \approx 1.6598 - 2.5 = -0.8402 \). Thus, \( d_i \approx -\frac{1}{0.8402} \approx -1.19 \, \text{m} \).The negative sign suggests the near point is on the same side as the object, which is expected here.

Step 8: Convert near point distance to distance from eyes

Since the near point is \(1.19\, \text{m}\) from the glasses and the glasses are \(0.02\, \text{m}\) from the eyes, the near point from her eyes is \(1.19\, \text{m} + 0.02\, \text{m} = 1.21\, \text{m}\).

Key Concepts

Refractive Power

Refractive power is an essential concept in optics, especially when dealing with lenses. It measures a lens's ability to bend light, typically gauging how much it can converge or diverge light rays. Refractive power is denoted by the letter "P" and is expressed in units called diopters. Diopters are calculated as the reciprocal of the focal length measured in meters. For example, if a lens has a refractive power of 1.660 diopters, this means its focal length is approximately 0.602 meters. This measurement can guide us in understanding how effectively a lens corrects vision problems like farsightedness. A higher refractive power indicates a stronger bending ability, which is crucial for adjusting the focus of light directly onto the retina.

Lens Formula

The lens formula is a critical equation in optics, helping to connect important variables of a lens's functionality. It is expressed as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, \( f \) is the focal length of the lens, \( d_o \) is the object distance (the distance from the lens to the object being observed), and \( d_i \) is the image distance (the distance over which the lens forms the image).
This formula is particularly useful when needing to find one of these variables, provided the others are known. For instance, it can determine where an image will form (image distance) once the object distance and focal length are available, as seen in our eyeglasses scenario.

Focal Length

The focal length of a lens is a key optical feature that tells us the distance over which initially parallel rays of light are brought to a focus. For any given lens, the focal length is the distance from the lens where it converges the light rays. The shorter the focal length, the more powerful the lens in terms of bending light. In our example with the refractive power of 1.660 diopters, we found the focal length to be approximately 0.602 meters.
Focal length plays a pivotal role in determining how well a lens can adjust for vision problems. Understanding this helps in designing corrective lenses that can precisely alter the focus of light.

Image Distance

The image distance is defined as the distance from the lens to the point where an image forms. This is dependent on both the object distance and the focal length, and it can be calculated using the lens formula. In working with corrective lenses, such as those of eyeglasses, determining image distance helps understand where the lens will cause light to converge.
In the textbook scenario, solving for image distance provides insights into the near point of vision for the farsighted individual, effectively indicating how far she can clearly see an object while wearing a particular lens. A negative image distance suggests that the image forms on the same side as the object, a common occurrence in virtual images formed by converging lenses.

Diopters

Diopters are units of measurement for refractive power and are pivotal in determining the strength of lenses. They provide a straightforward numerical value representing how much a lens can converge (positive diopters) or diverge (negative diopters) light rays. In corrective eyewear, the diopter value indicates the lens's capability to rectify refractive errors, such as myopia or hyperopia. For instance, lenses with a diopter value of 1.660 can precisely focus light to enhance clarity for people with vision shortcomings.
Calculating diopters involves taking the reciprocal of the focal length in meters. This conversion aids in designing and choosing the proper lens for specific optical needs, ensuring optimal vision correction.