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Ch 35: Optical Instruments
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 35: Optical InstrumentsProblem 31
Chapter 35, Problem 31

A 1.0-cm-tall object is 110 cm from a screen. A diverging lens with focal length -20 cm is 20 cm in front of the object. What are the focal length and distance from the screen of a second lens that will produce a well-focused, 2.0-cm-tall on the screen?

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Step 1: Start by analyzing the diverging lens. Use the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For the diverging lens, \( f = -20 \ \text{cm} \) and \( u = -20 \ \text{cm} \) (negative because the object is on the same side as the incoming light). Solve for \( v \), the image distance.
Step 2: Determine the position of the image formed by the diverging lens. The image distance \( v \) calculated in Step 1 will serve as the object distance for the second lens. Note that the image formed by the diverging lens is virtual, so \( v \) will be negative.
Step 3: For the second lens, the total distance between the object (formed by the diverging lens) and the screen is 110 cm. Use this information to calculate the required focal length \( f_2 \) of the second lens. The lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) will be applied again, where \( u \) is the distance from the second lens to the virtual image formed by the first lens, and \( v \) is the distance from the second lens to the screen.
Step 4: Use the magnification formula \( M = \frac{h'}{h} = -\frac{v}{u} \) to verify the height of the final image. Here, \( h' \) is the height of the final image (2.0 cm), and \( h \) is the height of the object (1.0 cm). Ensure that the calculated magnification matches the given height of the final image.
Step 5: Combine the results from Steps 3 and 4 to determine the focal length \( f_2 \) and the position of the second lens relative to the screen. Ensure that the calculations satisfy both the lens formula and the magnification condition.

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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, expressed as 1/f = 1/v + 1/u. This formula is essential for determining the position of the image formed by a lens and is crucial for solving problems involving multiple lenses.
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Lens Maker Equation

Magnification

Magnification (M) is the ratio of the height of the image (h') to the height of the object (h), given by M = h'/h = -v/u. Understanding magnification is vital for determining how the size of the image changes relative to the object, which is necessary for achieving the desired image height on the screen.
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Mirror Equation

Diverging Lens Characteristics

A diverging lens, characterized by a negative focal length, causes parallel rays of light to spread out as if they originated from a focal point on the same side as the object. This property affects how images are formed and is important for calculating the necessary parameters for a second lens to achieve a specific image size.
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Thin Lens Equation
Related Practice
Textbook Question

A scientist needs to focus a helium-neon laser beam (⋋ = 633 nm) to a 10-μm-diameter spot 8.0 cm behind a lens. What minimum diameter must the lens have?

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Textbook Question

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Textbook Question

The cornea, a boundary between the air and the aqueous humor, has a 3.0 cm focal length when acting alone. What is its radius of curvature?

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Textbook Question

A 15-cm-focal-length converging lens is 20 cm to the right of a 7.0-cm-focal-length converging lens. A 1.0-cm-tall object is distance L to the left of the 7.0-cm-focal-length lens. What are the height and orientation of the final image?

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Textbook Question

Infrared telescopes, which use special infrared detectors, are able to peer farther into star-forming regions of the galaxy because infrared light is not scattered as strongly as is visible light by the tenuous clouds of hydrogen gas from which new stars are created. For what wavelength of light is the scattering only 1% that of light with a visible wavelength of 500 nm?

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Textbook Question

A common optical instrument in a laser laboratory is a beam expander. One type of beam expander is shown in FIGURE P35.28. The parallel rays of a laser beam of width w₁ enter from the left. What is the width w2 of the exiting laser beam?

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