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

Short Answer

Expert verified
The radius of curvature for each lens surface should be 22 cm.

Step by step solution

01

Understand the Lensmaker's Equation

The Lensmaker's equation relates the focal length ( f ), the radii of curvature of the two surfaces ( R_1 and R_2 ), and the index of refraction ( n ) of the lens material. The equation is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where R_1 and R_2 are the radii of curvature of the lens surfaces. For the given problem, R_1 = R_2 = R, because the surfaces have equal radii.
02

Simplify the Equation for Equal Radii

Since both radii are equal, R_1 = R_2 = R , we can simplify the lensmaker's equation: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{(-R)} \right) = (n - 1) \left( \frac{2}{R} \right) \] Since \( R_2 \) is on the opposite side of the lens, it is negative.
03

Substitute Known Values

We know that f = 20.0 cm and n = 1.55. Plugging these values into the simplified equation, we have: \[ \frac{1}{20} = (1.55 - 1) \left( \frac{2}{R} \right) \] \[ \frac{1}{20} = 0.55 \left( \frac{2}{R} \right) \]
04

Solve for Radius R

Rearranging the equation from Step 3 to solve for R : \[ R = \frac{2 \times 0.55}{\frac{1}{20}} \] Simplifying, we find: \[ R = \frac{1.1}{0.05} = 22 \text{ cm} \]

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Key Concepts

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

Index of Refraction
The index of refraction, often symbolized as \( n \), is a fundamental property of materials, defining how much light is bent, or refracted, when entering a material. It's a dimensionless number that compares the speed of light in a vacuum to the speed of light in the material. For example, a glass with an index of refraction of 1.55 means light travels 1.55 times slower in the glass than in a vacuum.
  • A higher index signifies that light slows down more and bends more upon entering the material.
  • In the context of lenses, a higher index of refraction can make lenses thinner, as they can bend light more efficiently.
Understanding this concept is crucial in lens design, as it directly influences focal length and curvature requirements.
Focal Length
The focal length of a lens is the distance over which initially collimated rays are brought to a focus. It is usually denoted as \( f \) and measured in centimeters or meters.
  • A shorter focal length means the lens is 'stronger' and converges light rays more quickly.
  • It determines how magnified the image will appear and is crucial for applications like magnifying glasses.
In the context of your exercise, the lens must be designed so its focal length matches the required 20.0 cm, balancing with the index of refraction to ensure correct curvature.
Radii of Curvature
Radii of curvature, represented by \( R_1 \) and \( R_2 \) for the two lens surfaces, are essentially the "curve" or "bend" of the lens surface. In many lenses, especially symmetrical ones, these can be identical, simplifying calculations.
  • The two radii determine how much the lens will bend light, which in turn affects the focal length.
  • For lenses with equal radii as in the exercise, \( R_1 \) equals \( R_2 \), allowing simplification.
The curvature must be accurately calculated using the lensmaker's equation to achieve the desired focal length, particularly important in applications like creating a magnifying glass.
Magnifying Glass
A magnifying glass is a convex lens used to produce a magnified image of an object. The lens parameters, such as focal length and curvature, determine the effectiveness of magnification.
  • The focal length is usually on the order of tens of centimeters, which determines how large and clear the magnified image will appear.
  • With a lower focal length and higher index of refraction, the magnifying glass can be smaller yet still provide significant magnification.
Crafting a magnifying glass involves balancing these properties to ensure it enlarges objects efficiently while remaining lightweight and portable. Understanding these principles allows lensmakers to tailor the glass to the desired specifications, achieving clarity and zoom capabilities.

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Most popular questions from this chapter

A camera lens has a focal length of 200 mm. How far from the lens should the subject for the photo be if the lens is 20.4 cm from the sensor?

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

For a convex spherical mirror that has focal length \(f\) = -12.0 cm, what is the distance of an object from the mirror's vertex if the height of the image is half the height of the object?

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns (\(\mu\)m) is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 \(\mu\)m in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1\(^\circ\) = 60 min), and compare it with the typical experimental value of about 1.0 min. (\(Note\): There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

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