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Chapter 33: Problem 2

(1) Sunlight is observed to focus at a point 18.5 \(\mathrm{cm}\) behind a lens, \((a)\) What kind of lens is it? \((b)\) What is its power in dionters?

Short Answer

Expert verified
(a) Converging lens. (b) 5.41 diopters.

Step by step solution

01

Determine the type of lens

Since the sunlight is observed to focus at a point behind the lens, the lens must be converging. Converging lenses cause parallel rays of light (like sunlight) to converge to a point known as the focal point.
02

Find the focal length

The focal length of the lens is given as the distance from the lens to the point where sunlight is focused, which is 18.5 cm. Since this is a converging lens, the focal length is positive, so \(f = 18.5\, \text{cm} = 0.185\, \text{m}\).
03

Calculate the power of the lens

The power \(P\) of a lens in diopters (D) is calculated using the formula \(P = \frac{1}{f}\), where \(f\) is the focal length in meters. Substituting the given focal length, we have: \[ P = \frac{1}{0.185} \approx 5.41\, \text{D} \]

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Key Concepts

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

Converging Lens
A converging lens, often referred to as a convex lens, is a type of lens that causes light rays that are initially parallel to come together at a single point. This characteristic is due to the lens's shape, which is thicker at the center than at the edges. Converging lenses are widely used in a variety of optical devices, including:
  • Cameras
  • Magnifying glasses
  • Projectors
  • Eyeglasses for farsightedness
When light, such as sunlight, passes through a converging lens, it bends (or refracts) towards the lens's axis, eventually meeting at a focal point on the other side of the lens. This is why sunlight converges to a point behind the lens, affirming that the lens is indeed a converging lens.
Focal Length
The focal length of a lens is the distance from the lens to its focal point, where light rays converge. In our previous exercise, the focal length was given as 18.5 cm, or 0.185 meters. The focal length is crucial in determining how strong the lens is. A short focal length indicates a powerful lens that bends light sharply, while a longer focal length corresponds to a weaker lens with less bending power.
In the context of converging lenses, the focal length is always considered positive. This positivity indicates the lens's ability to converge light rays efficiently. The focal length is affected by the lens's curvature and the refractive index of the material from which it is made. Understanding the focal length helps in:
  • Calculating the lens's power
  • Designing optical systems for precise focusing
  • Choosing the right lens for correcting vision
Diopters
Diopters, often abbreviated as D, are the units used to express the optical power of a lens. The power of a lens tells us how strongly it can bend light, which is inversely related to the focal length. The formula used to calculate power in diopters is:\[P = \frac{1}{f}\]where \(f\) is the focal length in meters.
In our example, with a focal length of 0.185 meters, the lens power was calculated to be approximately 5.41 diopters. Thus, a lens with a 5.41 diopter power is quite effective in focusing light over a short distance.
Diopters are particularly important in vision correction:
  • Eyeglasses prescriptions are given in diopters, indicating the lens strength needed to correct vision.
  • Positive diopters correct farsightedness (hyperopia), using converging lenses.
  • Negative diopters are used for nearsightedness (myopia), involving diverging lenses.

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Most popular questions from this chapter

(II) A microscope has a 1.8 -cm-focal-length eyepiece and a \(0.80-\mathrm{cm}\) objective. Assuming a relaxed normal eye, calculate (a) the position of the object if the distance between the lenses is \(16.8 \mathrm{~cm},\) and \((b)\) the total magnification.

A telephoto lens system obtains a large magnification in a compact package. A simple such system can be constructed out of two lenses, one converging and one diverging, of focal lengths \(f_{1}\) and \(f_{2}=-\frac{1}{2} f_{1},\) respectively, separated by a distance \(\ell=\frac{3}{4} f_{1}\) as shown in Fig. \(51 .\) (a) For a distant object located at distance \(d_{\mathrm{o}}\) from the first lens, show that the first lens forms an image with magnification \(m_{1} \approx-f_{1} / d_{0}\) located very close to its focal point. Go on to show that the total magnification for the two-lens system is \(m \approx-2 f_{1} / d_{0} .(b)\) For an object located at infinity, show that the two-lens system forms an image that is a distance \(\frac{5}{4} f_{1}\) behind the first lens. (c) A single 250 -mm-focal-length lens would have to be mounted about 250 \(\mathrm{mm}\) from a camera's film in order to produce an image of a distant object at \(d_{\mathrm{o}}\) with magnification \(-(250 \mathrm{mm}) / d_{\mathrm{o}} .\) To produce an image of this object with the same magnification using the two-lens system, what value of \(f_{1}\) should be used and how far in front of the film should the first lens be placed? How much smaller is the "focusing length" (i.e., first lens-to-final image distance) of this two-lens system in comparison with the \(250-\mathrm{mm}\) "focusing length" of the equivalent single lens?

Two converging lenses, one with \(f=4.0 \mathrm{cm}\) and the other with \(f=44 \mathrm{cm},\) are made into a telescope. \((a)\) What are the length and magnification? Which lens should be the eyepiece? \((b)\) Assume these lenses are now combined to make a microscope; if the magnification needs to be \(25 \times\) , how long would the microscope be?

(II) A small insect is placed \(5.85 \mathrm{~cm}\) from a +6.00 -cm-focallength lens. Calculate ( \(a\) ) the position of the image, and (b) the angular magnification.

(II) Reading glasses of what power are needed for a person whose near point is \(105 \mathrm{~cm}\), so that he can read a computer screen at \(55 \mathrm{~cm} ?\) Assume a lens-eye distance of \(1.8 \mathrm{~cm}\).
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