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Chapter 18: Problem 57

An object is placed at \(20 \mathrm{~cm}\) from a convex mirror of focal length \(10 \mathrm{~cm}\). The image formed by the mirror is (A) real and at \(20 \mathrm{~cm}\) from the mirror. (B) virtual and at \(20 \mathrm{~cm}\) from the mirror. (C) virtual and at \((20 / 3) \mathrm{cm}\) from the mirror. (D) real and at ( \(20 / 3) \mathrm{cm}\) from the mirror.

Short Answer

Expert verified
The image formed by the convex mirror is virtual and located at \(\frac{20}{3}\mathrm{cm}\) from the mirror.

Step by step solution

01

Apply sign conventions

According to the sign convention for mirrors: 1. The object distance (u) is negative for objects in front of the mirror. 2. The image distance (v) is positive for virtual images formed behind the mirror (opposite side of the object). 3. The focal length (f) is positive for concave mirrors and negative for convex mirrors. For this problem, as the object is in front of the mirror and it is a convex mirror, we have: Object distance, \(u = -20\mathrm{~cm}\) Focal length, \(f = -10\mathrm{~cm}\)
02

Use the mirror formula to find image distance (v)

Using the mirror formula, \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), we can now find the image distance: \(\frac{1}{-10}= \frac{1}{-20} + \frac{1}{v}\)
03

Solve for image distance (v)

Now, isolate the variable \(v\) on one side of the equation: \(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-20}\) \(\frac{1}{v} = \frac{-1}{-10} + \frac{-1}{-20}\) \(\frac{1}{v} = \frac{2 + 1}{20}\) So, \(v = \frac{20}{3}\) Since the image distance (v) is positive, it means the image is formed behind the mirror, which implies the image is virtual.
04

Choose the correct answer

From our calculation, the image formed by the convex mirror is virtual and located at \(\frac{20}{3}\mathrm{cm}\) from the mirror. Thus, the correct answer is: (C) virtual and at \(\frac{20}{3}\mathrm{cm}\) from the mirror.

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