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 41a

Chapter 34, Problem 41a

A 1.20 cm tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm.

Verified step by step guidance

Step 1: Use the lens formula to find the image distance (\(d_i\)) for the first lens. The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length of the lens, \(d_o\) is the object distance, and \(d_i\) is the image distance. Substitute \(f = 40.0\ \text{cm}\) and \(d_o = 50.0\ \text{cm}\) into the formula.

Step 2: Rearrange the lens formula to solve for \(d_i\): \(\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\). Perform the subtraction of the reciprocals of \(f\) and \(d_o\) to find \(\frac{1}{d_i}\), then take the reciprocal to find \(d_i\).

Step 3: Determine the magnification (\(M\)) of the first lens using the formula \(M = -\frac{d_i}{d_o}\). Substitute the values of \(d_i\) and \(d_o\) to calculate \(M\).

Step 4: Calculate the height of the image (\(h_i\)) formed by the first lens using the magnification formula \(h_i = M \cdot h_o\), where \(h_o = 1.20\ \text{cm}\) is the height of the object. Substitute the values of \(M\) and \(h_o\) to find \(h_i\).

Step 5: Summarize the results for the first image (\(I_1\)): its location (\(d_i\)) relative to the first lens and its height (\(h_i\)). These values will be used as inputs for further calculations involving the second lens in subsequent parts of the problem.

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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. In this scenario, it will help calculate the image location (I1) created by the first converging lens.

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. This concept is crucial for understanding how the size of the image relates to the size of the object. In this problem, it will be used to find the height of the image formed by the first lens.

Converging Lens

A converging lens, or convex lens, is thicker at the center than at the edges and causes parallel rays of light to converge at a focal point. The focal length is the distance from the lens to this point. Understanding the behavior of converging lenses is vital for analyzing how they form images, especially in multi-lens systems like the one described in the question.

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Thin Lens Equation

Related Practice

Textbook Question

Repeat Exercise 34.41 using the same lenses except for the following changes: The second lens is a diverging lens having a focal length of magnitude 60.0 cm.

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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

BIO The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. Find the radii of curvature of this lens.

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Textbook Question

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. Draw a principal-ray diagram.

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Textbook Question

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted?

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Textbook Question

You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. If the dimensions of the picture on a 35 mm color slide are 24 mm ✖ 36 mm, what is the minimum size of the projector screen required to accommodate the image?

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