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

Knight Calc 5th Edition

Ch 35: Optical Instruments

Problem 32b

Chapter 35, Problem 32b

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?

Verified step by step guidance

Step 1: Start by analyzing the first lens (7.0-cm focal length). Use the lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Substitute \( f = 7.0 \, \text{cm} \) and \( d_o = L \) (the distance of the object from the first lens). Solve for \( d_i \), the image distance for the first lens.

Step 2: Determine the nature of the image formed by the first lens. If \( d_i > 0 \), the image is real and on the opposite side of the lens. If \( d_i < 0 \), the image is virtual and on the same side as the object. The image formed by the first lens will act as the object for the second lens.

Step 3: Calculate the object distance for the second lens (15-cm focal length). The second lens is 20 cm to the right of the first lens. The object distance for the second lens, \( d_o' \), is given by \( d_o' = 20 - d_i \), where \( d_i \) is the image distance from the first lens. Use this value as the object distance for the second lens.

Step 4: Apply the lens equation again for the second lens: \( \frac{1}{f'} = \frac{1}{d_o'} + \frac{1}{d_i'} \), where \( f' = 15 \, \text{cm} \) is the focal length of the second lens, \( d_o' \) is the object distance for the second lens, and \( d_i' \) is the image distance for the second lens. Solve for \( d_i' \), the final image distance.

Step 5: Determine the height and orientation of the final image. Use the magnification formula \( M = M_1 \times M_2 \), where \( M_1 = -\frac{d_i}{d_o} \) is the magnification of the first lens and \( M_2 = -\frac{d_i'}{d_o'} \) is the magnification of the second lens. The height of the final image is given by \( h' = M \times h \), where \( h = 1.0 \, \text{cm} \) is the height of the object. The orientation is determined by the sign of \( M \): if \( M > 0 \), the image is upright; if \( M < 0 \), the image is inverted.

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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 the lens, which is crucial for solving problems involving multiple lenses.

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. It indicates how much larger or smaller the image is compared to the object and also provides information about the orientation of the image (inverted or upright).

Ray Diagrams

Ray diagrams are graphical representations used to determine the path of light through lenses. By drawing principal rays (such as the parallel ray, the focal ray, and the central ray), one can visually locate the image formed by the lens and assess its characteristics, including height and orientation.

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Ray Diagrams for Converging Lenses

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