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Chapter 33: Problem 107

In a science-fiction novel, an intelligent ocean-dwelling creature's eye functions underwater with a near point of 25 \(\mathrm{cm} .\) This creature would like to create an underwater magnifier out of a thin plastic container filled with air. What shape should the air-filled plastic container have (i.e., determine radii of curvature of its surfaces) in order for it to be used by the creature as a \(3.0 \times\) magnifier? Assume the eye is focused at its near point.

Short Answer

Expert verified
The magnifier should be a planoconvex lens with one surface radius of 2.08 cm.

Step by step solution

01

Understanding the magnification requirement

The creature wants a magnifier that provides a magnification of 3.0. The magnification (\(M\)) for simple magnifiers is defined as \(M = \dfrac{25}{F}\), where \(F\) is the focal length in centimeters. Therefore, to achieve a 3.0 magnification, we can set up the equation: \(3.0 = \dfrac{25}{F}\).
02

Solve for the focal length

To determine the focal length \(F\) that provides the desired magnification, solve the equation \(3.0 = \dfrac{25}{F}\):\[F = \dfrac{25}{3.0} = 8.33\, \text{cm}\]
03

Determine the lens maker's equation

We use the lens maker's formula to find the radii of curvature of the lens needed for this focal length. The lens maker's equation is:\[\dfrac{1}{F} = (n - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)\]where \(n\) is the refractive index of the lens material, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
04

Analyze refraction conditions

The magnifier is underwater, so the refractive index of the lens \(n\) in water becomes: \(n_{air/water} = \dfrac{n_{air}}{n_{water}} \approx \dfrac{1.0}{1.33} \approx 0.75\). Substitute \(F = 8.33\, \text{cm}\) and \(n_{air/water} = 0.75\) into the lens maker's equation: \[\dfrac{1}{8.33} = (0.75 - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)\] This simplifies to:\[-0.12 = -0.25 \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)\] which further simplifies to: \[0.48 = \dfrac{1}{R_1} - \dfrac{1}{R_2}\]
05

Decide radii of curvature for the magnifier

To achieve a positive focal length (convex lens), we can choose one surface with a positive radius and the other with a negative radius or zero if simpler. Possible options:1. Planoconvex lens: \(R_1 = 2.08\, \text{cm}, R_2 = \infty\) gives \(1/2.08\) fulfilling \(0.48\).2. Symmetrical Convex lens: take symmetrical radii like \(R_1 = 4.16\, \text{cm}, R_2 = -4.16\, \text{cm}\).Either configuration would work based on practicality in creating the lens.

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Key Concepts

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

Magnification
Magnification is a key concept in optics and is used to describe how much larger or smaller an object appears in a given optical instrument, like a magnifying glass or a lens. It is often represented by the symbol
  • M = \( \dfrac{25}{F} \)
Where \( M \) stands for magnification, and \( F \) represents the focal length of the lens in centimeters. In the context of simple magnifiers, they are typically defined to have a magnification based on a standard observation distance, often taken as 25 cm. This distance is roughly the closest point at which the average human eye can clearly focus on an object without strain. For the intelligent ocean-dwelling creature in the exercise, this means altering the focal length of their magnifying lens to achieve a specific magnification level. For their desired 3x magnification, one would set \( 3.0 = \dfrac{25}{F} \) and solve to find the focal length that pairs with this magnification, resulting in \( F = 8.33 \, \text{cm} \).

By understanding magnification, one effectively alters the perceived size of objects under observation through careful adjustment of the focal length.
Lens Maker's Equation
The lens maker's equation is an essential piece of optics that assists in determining the design of lenses. It's essentially a bridge between the lens material's properties and the construction parameters required for a particular optical performance.
  • \[ \dfrac{1}{F} = (n - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right) \]
In the equation:
  • \(F\) is the focal length,
  • \(n\) is the refractive index of the lens material,
  • \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
For this underwater magnifier, since the lens was filled with air and submerged in water, the adjusted refractive index became \( n_{air/water} \approx 0.75 \).

Substituting values into the lens maker's equation helps derive the necessary radii of curvature to fulfill the optical design of the magnifier for the ocean creature. This meticulous adjustment is pivotal for the performance of optical instruments in varying environments, such as underwater conditions.
Focal Length
Focal length is an important characteristic of lenses, describing the distance between the lens and the image sensor when the subject is in focus. It determines not only magnification but also the field of view.A shorter focal length translates into a wider field of view, while a longer focal length gives a more narrow field of view. In the context of a magnifying lens
  • The focal length \(F\) influences how much the image will be magnified.
As derived in the solution, we determined \(F = 8.33\, \text{cm}\) to achieve a magnification of 3.0. This was calculated using the dimension of the near point as a reference for practical reading or observation distance.The entire optical system's design will pivot around ensuring the indicated focal length aligns with the desired usage environment, in this case, underwater, braced for the creature鈥檚 unique vision characteristics.Focusing on focal length is crucial for formulating lenses for specific applications, much like tailoring a tool for precise tasks.

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Most popular questions from this chapter

The proper functioning of certain optical devices (e.g., optical fibers and spectrometers) requires that the input light be a collection of diverging rays within a cone of halfangle \(\theta\) (Fig. \(33-50\) ). If the light initially exists as a collimated beam (i.e., parallel rays), show that a single lens of focal length \(f\) and diameter \(D\) can be used to create the required input light if \(D / f=2 \tan \theta .\) If \(\theta=3.5^{\circ}\) for a certain spectrometer, what focal length lens should be used if the lens diameter is \(5.0 \mathrm{~cm} ?\)

(II) A planoconvex lens (Fig. 33-2a) has one flat surface and the other has \(R=15.3 \mathrm{~cm} .\) This lens is used to view a red and yellow object which is \(66.0 \mathrm{~cm}\) away from the lens. The index of refraction of the glass is 1.5106 for red light and 1.5226 for yellow light. What are the locations of the red and yellow images formed by the lens?

A converging lens with focal length of \(13.0 \mathrm{~cm}\) is placed in contact with a diverging lens with a focal length of \(-20.0 \mathrm{~cm} .\) What is the focal length of the combination, and is the combination converging or diverging?

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?

What is the magnifying power of a \(+4.0-\mathrm{D}\) lens used as a magnifier? Assume a relaxed normal eve.
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