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 30

Chapter 34, Problem 30

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?

Verified step by step guidance

Step 1: Understand the problem. A converging lens is being used, and we are given the focal length \( f = 70.0 \; \text{cm} \), the height of the object \( h_o = 3.20 \; \text{cm} \), and the height of the image \( h_i = -4.50 \; \text{cm} \) (negative because the image is inverted). We need to find the object distance \( d_o \), the image distance \( d_i \), and determine whether the image is real or virtual.

Step 2: Use the magnification formula to relate the object and image distances. The magnification \( M \) is given by \( M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \). Substitute the known values for \( h_i \) and \( h_o \) to calculate the ratio \( \frac{d_i}{d_o} \).

Step 3: Use the lens equation to relate the focal length, object distance, and image distance. The lens equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substitute the known value of \( f \) and the relationship between \( d_i \) and \( d_o \) from Step 2 into this equation to solve for \( d_o \).

Step 4: Once \( d_o \) is determined, use the magnification formula \( M = -\frac{d_i}{d_o} \) to calculate \( d_i \).

Step 5: Determine the nature of the image. Since the image is inverted and the lens is converging, the image is real. A real image is formed when light rays actually converge at the image location.

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Key Concepts

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

Converging Lens

A converging lens, or convex lens, is a transparent optical device that bends light rays inward to a focal point. The focal length is the distance from the lens to this point, where parallel rays of light converge. In this case, the focal length is 70.0 cm, which is crucial for determining the image location and characteristics when an object is placed in front of the lens.

Magnification

Magnification is the ratio of the height of the image to the height of the object, indicating how much larger or smaller the image appears compared to the object. It can be calculated using the formula: magnification (M) = height of image (h') / height of object (h). In this scenario, the image is 4.50 cm tall, and the object is 3.20 cm tall, which helps determine the nature of the image formed by the lens.

Real vs. Virtual Images

Real images are formed when light rays converge and can be projected onto a screen, while virtual images occur when light rays diverge and cannot be projected. In the context of a converging lens, if the object is placed outside the focal length, a real and inverted image is produced. Conversely, if the object is within the focal length, a virtual and upright image is formed. Understanding this distinction is essential for analyzing the image characteristics in the given problem.

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Related Practice

Textbook Question

For each thin lens shown in Fig. E34.37, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.

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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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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 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?

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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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