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Ch 34: Geometric Optics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 34, Problem 57b

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

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Identify the given values: The focal length of the lens \( f = 8.00 \; \text{cm} \) and the object height \( h_o = 1.00 \; \text{mm} \). The goal is to find the height of the image \( h_i \).
Use the lens formula to find the image distance \( v \): \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where \( f \) is the focal length, \( u \) is the object distance, and \( v \) is the image distance. Rearrange to solve for \( v \): \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \).
Determine the magnification \( M \) of the lens using the formula \( M = \frac{h_i}{h_o} = \frac{v}{u} \). Rearrange to solve for \( h_i \): \( h_i = M \cdot h_o \).
Substitute the values of \( v \), \( u \), and \( h_o \) into the magnification formula to calculate the image height \( h_i \). Ensure that all units are consistent (e.g., convert \( h_o \) to cm if necessary).
Interpret the result: The sign of \( h_i \) will indicate whether the image is upright or inverted, and the magnitude will give the size of the image formed by the magnifier.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Focal Length

The focal length of a lens is the distance from the lens to the focal point, where parallel rays of light converge. For a simple magnifier, a positive focal length indicates that it is a converging lens. The focal length is crucial for determining how the lens will magnify an object and affects the size of the image formed.
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Magnification

Magnification is the ratio of the height of the image to the height of the object. It can be calculated using the formula: magnification (M) = image height (h') / object height (h). In the case of a magnifier, the magnification also depends on the focal length and the distance of the object from the lens, allowing us to determine how much larger the image appears compared to the actual object.
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Thin Lens Formula

The thin lens formula relates the object distance (d_o), image distance (d_i), and focal length (f) of a lens: 1/f = 1/d_o + 1/d_i. This formula is essential for understanding how lenses form images and can be used to find the image distance when the object distance and focal length are known. It helps in calculating the position and size of the image produced by the magnifier.
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Related Practice
Textbook Question

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. What angular magnification is obtainable with the lens if the object is at the focal point?

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Textbook Question

BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. Is she nearsighted or farsighted?

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Textbook Question

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. What is the magnitude of the linear magnification produced by the objective?

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Textbook Question

A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0 cm lens being used as the objective. Both the object being viewed and the final image are at infinity. Find the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away.

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Textbook Question

BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. What focal length contact lens is needed, and what is its power in diopters?

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Textbook Question

Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. What is the angular magnification of the microscope?

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