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A lensmaker wants to make a magnifying glass from glass with \(n=1.55\) and with a focal length of \(20.0 \mathrm{~cm} .\) If the two surfaces of the lens are to have equal radii, what should that radius be?

Short Answer

Expert verified
The radius should be 22.0 cm.

Step by step solution

01

Understanding the Lensmaker's Equation

To solve the problem, we will use the Lensmaker's Equation which is given by: \[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]where \(f\) is the focal length, \(n\) is the refractive index, \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces of the lens. In this problem, the lens surfaces are equiconvex, so \(R_1 = R_2 = R\).
02

Simplify the Formula

As the lens has equal radii, \(R_1 = R_2 = R\). The equation simplifies to: \[\frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (n - 1) \left( \frac{2}{R} \right)\]Given \(n = 1.55\) and \(f = 20.0\) cm, we can substitute these values into the equation.
03

Substitute Known Values

Substitute the given values into the simplified equation:\[\frac{1}{20.0} = (1.55 - 1) \cdot \frac{2}{R}\]which leads to the equation:\[\frac{1}{20.0} = 0.55 \cdot \frac{2}{R}\]
04

Solve for R

Rearrange the equation to solve for \(R\):\[\frac{1}{20.0} = \frac{1.1}{R}\]Multiply both sides by \(R\) and then by 20 to isolate \(R\):\[R = 1.1 \times 20.0 = 22.0 \text{ cm}\]
05

Final Answer

After solving the equation, the radius of curvature for each lens surface should be \(22.0 \text{ cm}\). Both surfaces should have this radius to achieve the given focal length of 20 cm.

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Key Concepts

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

Focal Length
The focal length of a lens is one of the most critical parameters in lens optics. It refers to the distance from the center of the lens to the focal point, where parallel rays of light either converge to a point or appear to diverge from a point. In the context of lenses, the focal length determines how strongly the lens converges or diverges light. For a converging lens, such as the one in our example, a positive focal length indicates that it brings light rays together. In our exercise, the magnifying glass is designed to have a focal length of 20 cm. This means that parallel rays of light entering the lens will meet 20 cm away from the lens's center, forming a small, inverted image at that point.
Refractive Index
The refractive index of a material is a measure of how fast light travels through it. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. In simpler terms, it indicates how much light bends or slows down when entering a material from air. For our lensmaking problem, the glass has a refractive index ( ) of 1.55. This means light will travel 1.55 times slower in the glass than in air. A higher refractive index typically implies a greater bending of light and, consequently, a stronger lens.
Equiconvex Lens
An equiconvex lens is a symmetrical, convex lens that has the same radius of curvature on both surfaces. This symmetrical design is beneficial for reducing optical aberrations and for simplifying calculations, as seen in the simplified Lensmaker's Equation used in our solution. Since the radii of curvature ( R_1 and R_2) are equal, the equiconvex lens exhibits balanced optical properties. In the problem given, because the lens is equiconvex, we set both radii to be equal, simplifying the equations and subsequently the calculations. This design feature is crucial in achieving the desired focal length by influencing the pathway of light through the lens evenly from both sides.
Radii of Curvature
The radius of curvature in a lens refers to the radius of the sphere from which a lens segment is cut. It determines how curved the lens surfaces are: a smaller radius indicates a more curved lens surface, while a larger radius implies a flatter surface. In the problem we worked on, both surfaces of the lens are equiconvex, necessitating them to have equal radii of curvature. This uniformity ensures that light entering the lens will be converged equally from both sides, maintaining symmetrical optical properties. Through the mathematical derivation using the Lensmaker's Equation, we determined this radius to be 22 cm. Consequently, each surface had to be machined with precision to this specification to achieve the desired performance of the magnifying glass.

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Most popular questions from this chapter

The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{~cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we can ignore it. The depth of a typical human eye is around \(25 \mathrm{~mm}\). (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were \(25 \mathrm{~cm}\) in front of the eye? If not, where would it focus that text, in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{~mm},\) where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

Consider a concave mirror that has a focal length \(f\). In terms of \(f\), determine the object distances that will produce a magnification of (a) \(-1,(b)-2,\) and \((c)-3\).

A concave mirror has a radius of curvature of \(34.0 \mathrm{~cm} .\) (a) What is its focal length? (b) A ladybug \(7.50 \mathrm{~mm}\) tall is located \(22.0 \mathrm{~cm}\) from this mirror along the principal axis. Find the location and height of the image of the insect. (c) If the mirror is immersed in water (of refractive index 1.33 ), what is its focal length?

The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. If you place a concave glass lens into a tank of a liquid that has an index of refraction that is greater than that of the lens, what will happen? A. The lens will no longer be able to create any images. B. The focal length of the lens will become longer. C. The focal length of the lens will become shorter. D. The lens will become a converging lens.

When an object is \(16.0 \mathrm{~cm}\) from a lens, an image is formed \(12.0 \mathrm{~cm}\) from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is \(8.50 \mathrm{~mm}\) tall, how tall is the image? Is it upright or inverted? (c) Draw a principal-ray diagram.

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