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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: 1f=(n1)(1R11R2) 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: 1f=(n1)(1R1(R))=(n1)(2R) Since R2 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: 120=(1.551)(2R) 120=0.55(2R)
04

Solve for Radius R

Rearranging the equation from Step 3 to solve for R : R=2×0.55120 Simplifying, we find: R=1.10.05=22 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 R1 and R2 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, R1 equals R2, 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

The radii of curvature of the surfaces of a thin converging meniscus lens are R1 = +12.0 cm and R2 = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

You want to view through a magnifier an insect that is 2.00 mm long. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.032 radian?

An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

What is the size of the smallest vertical plane mirror in which a woman of height h can see her full-length image?

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