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Ch 34: Geometric Optics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 34: Geometric OpticsProblema 61b
Capitolo 34, Problema 61b

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.

Guida verificata passo dopo passo
1
Step 1: Understand the problem. The objective lens forms a real image of the building, which is 60.0 m tall and located 3.00 km away. The height of the image formed by the objective lens can be calculated using the magnification formula for lenses.
Step 2: Recall the magnification formula for a lens: \( M = -\frac{h_i}{h_o} = \frac{f}{d_o} \), where \( h_i \) is the height of the image, \( h_o \) is the height of the object, \( f \) is the focal length of the lens, and \( d_o \) is the object distance.
Step 3: Convert the given distances into consistent units. The height of the building \( h_o \) is 60.0 m, and the object distance \( d_o \) is 3.00 km, which should be converted to meters: \( d_o = 3000 \, \text{m} \). The focal length of the objective lens \( f \) is given as 95.0 cm, which should be converted to meters: \( f = 0.95 \, \text{m} \).
Step 4: Substitute the known values into the magnification formula to find \( h_i \). Rearrange the formula to solve for \( h_i \): \( h_i = -h_o \cdot \frac{f}{d_o} \). Plug in \( h_o = 60.0 \, \text{m} \), \( f = 0.95 \, \text{m} \), and \( d_o = 3000 \, \text{m} \).
Step 5: Simplify the expression to calculate the height of the image \( h_i \). Note that the negative sign indicates the image is inverted. The final numerical calculation can be performed to find the exact value of \( h_i \).

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Focal Length

The focal length of a lens is the distance from the lens to the point where parallel rays of light converge or appear to diverge. It is a critical parameter in lens design, affecting the magnification and image formation. In telescopes, the focal lengths of the objective and eyepiece lenses determine the overall magnification and clarity of the viewed image.
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Magnification

Magnification in optics refers to the ratio of the height of the image produced by a lens to the height of the object being viewed. It is calculated using the formula M = - (f_objective / f_eyepiece), where f represents the focal lengths. Understanding magnification is essential for determining how large an object appears through a telescope, especially when observing distant objects.
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Similar Triangles

The concept of similar triangles is fundamental in optics for calculating heights and distances in image formation. When light rays from an object pass through a lens, they create a triangle with the object and another triangle with the image. By using the properties of similar triangles, one can derive relationships between the heights of the object and image, as well as their respective distances from the lens.
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Domanda del libro di testo

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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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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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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