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A pair of eyeglasses is designed to allow a person with a farpoint distance of \(2.50 \mathrm{m}\) to read a road sign at a distance of \(25.0 \mathrm{m}\). Find the focal length required of these glasses if they are to be worn (a) \(2.00 \mathrm{cm}\) or (b) \(1.00 \mathrm{cm}\) from the eyes.

Short Answer

Expert verified
(a) Focal length is -2.75 m, (b) Focal length is -2.77 m.

Step by step solution

01

Define the Problem

The problem requires us to find the focal length for glasses which are to be worn at two different distances from the eyes. This person needs assistance to see objects at a distance (such as a road sign) clearly. Given data includes a far-point distance of their naked eye (2.50m) and the desired clear vision distance (25.0m).
02

Understanding Farpoint and Corrective Lens Formula

Since the person can see objects clearly only up to 2.50m, the glasses must make the road sign, which is 25.0m away, appear to be at this far-point distance. The lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) is used, where \( f \) is the focal length, \( v \) is the image distance (the far-point, -2.50m), and \( u \) is the object distance (-25.0m, negative because distance is measured towards the object).
03

Calculating Focal Length for Part (a)

Apply the lens formula for glasses placed 2.00 cm (0.02 m) from the eyes. Correct the far-point distance: \( \text{corrected } v = -2.50 + 0.02 = -2.48 \) m.Substitute this into the lens formula:\[ \frac{1}{f} = \frac{1}{-2.48} - \frac{1}{-25} \]This results in: \( f = \frac{-2.48 \, \times \, -25}{2.48 - 25} \).
04

Calculating Specific Focal Length

Calculate \( f \) using the substitution:\[ \frac{1}{f} = \frac{1}{-2.48} + \frac{1}{25} \]\[ \Rightarrow \frac{1}{f} = -0.4032 + 0.04 \]\[ \Rightarrow \frac{1}{f} = -0.3632 \]Thus, \( f \approx -2.75 \) m.
05

Calculating Focal Length for Part (b)

Adjust the image distance for glasses placed 1.00 cm (0.01 m) away. New far-point: \( -2.50 + 0.01 = -2.49 \) m.Apply the lens formula:\[ \frac{1}{f} = \frac{1}{-2.49} - \frac{1}{25} \].
06

Final Calculation for Focal Length in Part (b)

Perform the calculation:\[ \frac{1}{f} = -0.4016 + 0.04 \]\[ \Rightarrow \frac{1}{f} = -0.3616 \]\( f \approx -2.77 \) m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

focal length
The concept of focal length is essential in optics, especially when dealing with lenses. Focal length is the distance between the center of a lens and its focus point, where parallel rays of light converge.
In terms of corrective lenses, focal length determines how the lens will bend light to aid in vision correction. When the lens has a short focal length, it means it is stronger at bending light. Conversely, a longer focal length indicates a weaker lens.
  • If someone is nearsighted, their corrective lenses will have a focal length tailored to extend their far point, allowing them to see distant objects clearly.
  • For farsighted individuals, lenses with a focal length that shortens their near point will bring nearby objects into focus.
The precise focal length needed depends on several factors, including the desired improvement in vision and the location where the glasses are positioned relative to the eyes. Adjusting this parameter is key in designing effective corrective eyewear.
lens formula
The lens formula is a mathematical representation used to determine the relationship between the focal length of a lens and the distances of the object and the image.
It is expressed as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where:
  • \( f \) is the focal length of the lens.
  • \( v \) is the image distance, which in the context of corrective lenses, corresponds to the far-point of the unaided eye.
  • \( u \) is the object distance, representing the distance to the object the person desires to see clearly, like a road sign or a book.
This formula allows opticians to calculate the necessary focal length for lenses to correct vision to a specified distance. By substituting the known values into this equation, one can solve for the focal length \( f \), determining the lens strength required to bring images into focus at the intended distance.
corrective lenses
Corrective lenses are designed to improve vision by compensating for deficiencies in the eye's focusing ability. They are crafted to alter the path of incoming light so that it focuses correctly on the retina, producing a clear visual image.
Here are the primary types of vision impairments corrected by lenses:
  • Nearsightedness (Myopia): The focal point of images is in front of the retina, making distant objects appear blurry. Concave lenses, which diverge light rays, shift the focus back onto the retina.
  • Farsightedness (Hyperopia): Light focuses behind the retina, causing close objects to look blurred. Convex lenses help converge the light sooner, correcting the focus onto the retina.
  • Astigmatism: This occurs when the curvature of the cornea is uneven, causing distorted images. Cylindrical lenses are used to correct this by focusing light more evenly.
Corrective lenses are specially tuned by adjusting their curvature and thickness to align with each person's unique vision needs, ensuring that both near and far objects are clear and in focus. This customization includes determining the optimal distance the lenses can be comfortably worn from the eyes to maximize correction efficiency.

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Most popular questions from this chapter

You are taking a photograph of a poster on the wall of your dorm room, so you can't back away any farther than \(3.0 \mathrm{m}\) to take the shot.The poster is \(0.80 \mathrm{m}\) wide and \(1.2 \mathrm{m}\) tall, and you want the image to fit in the \(24-\mathrm{mm} \times 36-\mathrm{mm}\) frame of the film in your camera. What is the longest focal length lens that will work?

IP You are taking a photograph of a horse race. A shutter speed of 125 at \(f / 5.6\) produces a properly exposed image, but the running horses give a blurred image. Your camera has \(f\) -stops of \(2,2.8,4,5.6,8,11,\) and \(16 .\) (a) To use the shortest possible exposure time (i.e., highest shutter speed), which \(f\) -stop should you use? (b) What is the shortest exposure time you can use and still get a properly exposed image?

A photograph is properly exposed when the aperture is set to \(f / 8\) and the shutter speed is \(125 .\) Find the approximate shutter speed needed to give the same exposure if the aperture is changed to \(f / 2.4\)

A pirate sights a distant ship with a spyglass that gives an angular magnification of \(22 .\) If the focal length of the eyepiece is \(11 \mathrm{mm},\) what is the focal length of the objective?

Landing on an Aircraft Carrier The Long-Range Lineup System (LRLS) used to ensure safe landings on aircraft carriers consists of a series of Fresnel lenses of different colors. Each lens focuses light in a different, specific direction, and hence which light a pilot sees on approach determines whether the plane is above, below, or on the proper landing path. The basic idea behind a Fresnel lens, which has the same optical properties as an ordinary lens, is shown in Figure \(27-24\), along with a photo of the LRIS. Suppose an object (a lightbulb in this case) is \(17.1 \mathrm{cm}\) behind a Fresnel lens, and that the corresponding image is a distance \(d_{i}=d\) in front of the lens. If the object is moved to a distance of \(12.0 \mathrm{cm}\) behind the lens, the image distance doubles to \(d_{1}=2 d\). In the L.RLS, it is desired to have the image of the lightbulb at infinity. What object distance will give this result for this particular lens?

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