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Chapter 23: Problem 52

A concave mirror has a radius of curvature of \(5.0 \mathrm{m} .\) An object, initially \(2.0 \mathrm{m}\) in front of the mirror, is moved back until it is \(6.0 \mathrm{m}\) from the mirror. Describe how the image location changes.

Short Answer

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Answer: As the object moves from 2.0m to 6.0m in front of the concave mirror, the image location moves closer to the mirror, from 5.0m to 3.75m.

Step by step solution

01

Since we are given the radius of curvature \(R\), we can find the focal length \(f\) using the following formula: \(f = \frac{R}{2}\) where: - \(R=5.0\,m\) (radius of curvature) Now let's calculate the focal length: \(f = \frac{5.0}{2} = 2.5\,m\) #Step 2: Calculate the initial image distance#

To find the initial image distance \(d_{i1}\), we use the mirror equation with the given initial object distance \(d_\text{o1}=2.0\,m\). Plugging the values into the equation: \(\frac{1}{2.5}=\frac{1}{2.0}+\frac{1}{d_{i1}}\) Solving for \(d_{i1}\), we get: \(d_{i1} = \frac{1}{\frac{1}{2.5}-\frac{1}{2}} = 5.0\,m\) #Step 3: Calculate the final image distance#
02

Similarly, for the final image distance \(d_{i2}\), we use the mirror equation with the given final object distance \(d_\text{o2}=6.0\,m\): \(\frac{1}{2.5}=\frac{1}{6.0}+\frac{1}{d_{i2}}\) Solving for \(d_{i2}\), we get: \(d_{i2} = \frac{1}{\frac{1}{2.5}-\frac{1}{6}} = 3.75\,m\) #Step 4: Describe how the image location changes#

The initial image distance was \(5.0\,m\), and the final image distance is \(3.75\,m\). Thus, as the object moves from \(2.0\,m\) to \(6.0\,m\) in front of the concave mirror, the image location moves closer to the mirror, from \(5.0\,m\) to \(3.75\,m\).

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