Question

For a person whose near point is \(115 \mathrm{~cm},\) so that he can read a computer monitor at \(55 \mathrm{~cm},\) what power of read ing glasses should his optician prescribe, keeping the lenseye distance of \(2.0 \mathrm{~cm}\) for his spectacles?

Short answer

Answer: The optician should prescribe reading glasses with a power of approximately 1.31 diopters.

Step-by-step solution

Understand the lens formula

The lens formula for thin lenses is given by: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ where f is the focal length, v is the image distance (near point), and u is the object distance (distance to the computer monitor).

Convert the near point and monitor distance to the same units

In the problem, we are given the near point in cm. We need to convert it to meters since most standard optics problems use meters. The near point is 115 cm, which is equal to 1.15 meters. The monitor distance is given as 55 cm, which is equal to 0.55 meters.

Find the image distance with the lense-eye distance

We are given the lense-eye distance as 2.0 cm, which is equal to 0.02 meters. To find the image distance (near point) considering the lense-eye distance, we can subtract the lense-eye distance from the desired monitor distance: $$v = 0.55 - 0.02 = 0.53 \text{ meters}$$

Use the lens formula to find the focal length

We can now plug the image distance (v) and object distance (u) into the lens formula to find the focal length (f): $$\frac{1}{f} = \frac{1}{0.53} - \frac{1}{1.15}$$ Solve for f: $$f = \frac{1}{(\frac{1}{0.53} - \frac{1}{1.15})} \approx 0.766 \text{ meters}$$

Calculate the lens power

The power (P) of a lens in diopters is the reciprocal of the focal length in meters: $$P = \frac{1}{f}$$ Plug the focal length into the formula: $$P = \frac{1}{0.766} \approx 1.305 \text{ diopters}$$ The optician should prescribe reading glasses with a power of approximately 1.31 diopters for the person to be able to read a computer monitor at a distance of 55 cm.

Key concepts

Lens Formula

In optics, one of the essential tools for understanding how lenses work is the lens formula.
It is primarily used with thin lenses, and it is expressed as:

• \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)

Here, \(f\) is the focal length of the lens, \(v\) is the image distance, and \(u\) is the object distance.
In our specific problem, the image distance is related to the near point, and the object distance is the distance from the person to the computer monitor.
This formula helps us calculate the focal length needed for the person to clearly see the monitor with the correct glasses.

Diopters

Diopters are a unit of measurement for the optical power of a lens.
The optical power indicates how strongly a lens converges or diverges light.
The formula to calculate lens power in diopters \(P\) is:

• \(P = \frac{1}{f}\)

Where \(f\) is the focal length of the lens in meters. By using diopters, opticians can easily determine the strength of glasses needed for vision correction.
In this exercise, the final prescription of 1.31 diopters means that the lenses will help bend light appropriately for reading at the desired distance.

Focal Length

The focal length of a lens is a measure of how strongly it converges or diverges light.
It is the distance over which initially collimated rays are brought to a focus.
For lenses with positive focal lengths, light is brought to a point of focus; for negative focal lengths, light is spread out.
In the exercise, we calculated a focal length of approximately 0.766 meters.
This indicates the distance from the lens where light rays from the monitor should converge for the individual to see clearly, given their initial near point of 115 cm.

Near Point

A person's near point is the shortest distance at which their eyes can focus comfortably on an object.
For someone who needs glasses, this distance is often longer than the typical near point.
In the exercise, the user's near point is given as 115 cm.
The task is to calculate the necessary lens correction to adjust this near point to the required 55 cm, allowing for comfortable reading.
Adjusting the near point through optical aids, like the prescribed glasses, makes tasks like reading or working on a computer easier and more comfortable.

Thin Lenses

Thin lenses are an idealization that simplifies the study of lens systems in optics.
They are assumed to have negligible thickness, focusing only on the surfaces that refract light.
This assumption allows us to use the simplified lens formula to calculate focal lengths and other optical properties.
The concept of thin lenses is crucial for understanding the optical design of spectacles, where the goal is to improve or correct vision by altering the path of light rays as they enter the eye.
In this problem, using the thin lens formula helps in calculating the required lens power for vision correction with minimal complexity.