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Chapter 11: Problem 16

An object initially at the focus of a concave mirror is brought closer to the mirror. As the object is moved toward the mirror's surface, (A) the image gets larger until it is of infinite size when the object reaches the mirror surface. (B) the image gets smaller until it is twice the object size when the object reaches the mirror surface. (C) the image gets smaller until it becomes equal to the object size when the object reaches the mirror surface. (D) the image gets larger until it is twice the object size when the object reaches the mirror surface. (E) the image stays the same size.

Short Answer

Expert verified
Answer: (A) As the object approaches the mirror's surface, the image gets infinitely large.

Step by step solution

01

Recall the Mirror Formula and Magnification Formula for concave mirrors

The Mirror Formula is given by: \[\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}}\] Where \(f\) is the focal length of the mirror, \(d_{o}\) is the object distance, and \(d_{i}\) is the image distance. The Magnification Formula is given by: \[M = \frac{h_{i}}{h_{o}} = -\frac{d_{i}}{d_{o}}\] Where \(M\) is magnification, \(h_{i}\) is the image height, and \(h_{o}\) is the object height.
02

Set up the problem

Initially, the object is at the mirror's focus. Hence, the object distance is equal to the focal length, i.e., \(d_{o} = f\). As the object moves towards the mirror's surface, the object distance \(d_{o}\) will decrease. We need to find the correct answer choice based on the change in image size as the object distance decreases.
03

Analyze the answer choices and eliminate incorrect ones

(A): As the object reaches the mirror surface, \(d_{o}\) approaches zero. The Mirror Formula implies that when \(d_{o} \to 0\), the image distance (\(d_{i}\)) approaches negative infinity, which implies the image gets infinitely large. Therefore, (A) is correct. (B), (C), (D), and (E) do not correctly describe the image size behavior.

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Most popular questions from this chapter

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