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Ch 34: Geometric Optics
Chapter 34, Problem 34

A lens forms an of an object. The object is 16.0 cm from the lens. The is 12.0 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?

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1
Identify the type of image formed. Since the image is on the same side as the object, it is a virtual image, which is typically formed by a diverging lens.
Use the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Note that for virtual images, \(d_i\) is considered negative.
Substitute the given values into the lens formula. Here, \(d_o = 16.0\,\text{cm}\) and \(d_i = -12.0\,\text{cm}\) (negative because the image is virtual).
Solve the lens formula for \(f\) to find the focal length of the lens.
Determine the type of lens based on the sign of the focal length. If \(f\) is negative, the lens is diverging; if positive, it is converging.

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Key Concepts

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

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens. It is expressed as 1/f = 1/v + 1/u. This formula is essential for determining the focal length of the lens when the object and image distances are known.
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Converging and Diverging Lenses

Lenses can be classified as converging (convex) or diverging (concave). Converging lenses focus parallel rays of light to a point, resulting in a positive focal length, while diverging lenses spread light rays outward, leading to a negative focal length. Identifying the type of lens is crucial for understanding its behavior with light.
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Sign Convention in Optics

In optics, a sign convention is used to determine the signs of object distance, image distance, and focal length. Typically, distances measured in the direction of incoming light are positive, while those measured against it are negative. This convention is vital for correctly applying the lens formula and interpreting the results.
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Related Practice
Textbook Question
An object is 16.0 cm to the left of a lens. The lens forms an 36.0 cm to the right of the lens. (c) Draw a principal-ray diagram.
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Textbook Question
A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored.
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Textbook Question
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 of the arrow formed by paraxial rays incident on the convex surface. Is the erect or inverted?
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Textbook Question
A converging lens with a focal length of 70.0 cm forms an of a 3.20-cm-tall real object that is to the left of the lens. The is 4.50 cm tall and inverted. Where are the object and located in relation to the lens? Is the real or virtual?
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Textbook Question
A converging lens with a focal length of 9.00 cm forms an of a 4.00-mm-tall real object that is to the left of the lens. The is 1.30 cm tall and erect. Where are the object and located? Is the real or virtual?
335
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Textbook Question
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?
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