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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Ch 34: Geometric Optics
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 33, Problem 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.
Verified step by step guidance1
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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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 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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Related Practice
Textbook Question
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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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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