Ch 34: Geometric Optics

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 34: Geometric Optics

Problem 28a

Chapter 34, Problem 28a

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. What is the focal length of the lens? Is the lens converging or diverging?

Verified step by step guidance

Step 1: Begin by identifying the lens equation, which relates the object distance (d_o), image distance (d_i), and focal length (f). The equation is:

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

Step 2: Substitute the given values into the lens equation. The object distance (d_o) is 16.0 cm, and the image distance (d_i) is -12.0 cm (negative because the image is on the same side as the object, indicating a virtual image). The equation becomes:

\frac{1}{f} = \frac{1}{16.0} + \frac{1}{-12.0}

Step 3: Simplify the fractions to calculate the reciprocal of the focal length. Perform the addition of the two terms:

\frac{1}{f} = \frac{1}{16.0} + \frac{-1}{12.0}

. Combine the fractions by finding a common denominator.

Step 4: Once the reciprocal of the focal length is calculated, invert the result to find the focal length (f). Remember that the sign of the focal length will indicate the type of lens: a positive focal length corresponds to a converging lens, while a negative focal length corresponds to a diverging lens.

Step 5: Analyze the sign of the focal length obtained. If the focal length is positive, the lens is converging. If the focal length is negative, the lens is diverging. This conclusion is based on the properties of lenses and their ability to form real or virtual images.

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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.

Converging and Diverging Lenses

Lenses can be classified as converging (convex) or diverging (concave). A converging lens brings parallel rays of light to a focus, resulting in a positive focal length, while a diverging lens causes parallel rays to spread out, resulting in a negative focal length. Identifying the type of lens is crucial for understanding its behavior with light.

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. Understanding this convention is vital for correctly applying the lens formula and interpreting results.

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Related Practice

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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Textbook Question

The glass rod of Exercise 34.22 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod's axis is to be d 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image?

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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 image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

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Textbook Question

A converging lens with a focal length of 9.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?

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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. 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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