Chapter 34: Problem 36
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.
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.
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.
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.