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Chapter 25: Problem 34

An object that is \(25 \mathrm{~cm}\) in front of a convex mirror has an image located \(17 \mathrm{~cm}\) behind the mirror. How far behind the mirror is the image located when the object is \(19 \mathrm{~cm}\) in front of the mirror?

Short Answer

Expert verified
The image is located approximately 14.36 cm behind the mirror when the object is 19 cm in front of it.

Step by step solution

01

Identifying Given Variables and Formula

We know that the object distance, denoted by \(d_o\), is 25 cm and the image distance, denoted by \(d_i\), is -17 cm (negative because it is behind a convex mirror). We must find the focal length \(f\) using the mirror equation \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\).
02

Calculating the Focal Length

Substitute the known values into the mirror equation: \[\frac{1}{f} = \frac{1}{25} + \frac{1}{-17}\] Calculate:\[\frac{1}{f} = \frac{1}{25} - \frac{1}{17} = \frac{17 - 25}{425} = \frac{-8}{425}\]Thus, the focal length \(f = - \frac{425}{8} = -53.125\ cm\).
03

Using the Focal Length to Find New Image Distance

Now, we need to find where the image forms when the object is 19 cm in front of the mirror. Set \(d_o = 19\) cm in the mirror equation, and use the focal length found before:\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)Substitute and solve for \(d_i\):\[\frac{1}{-53.125} = \frac{1}{19} + \frac{1}{d_i}\] \[-\frac{1}{53.125} - \frac{1}{19} = \frac{1}{d_i}\] Calculate with common denominators and solve for \(d_i\).
04

Solving for New Image Distance

By solving the equation:\[-\frac{1}{53.125} - \frac{1}{19} = \frac{1}{d_i}\]Convert to common denominators and calculate:\[-\frac{19 + 53.125}{53.125*19} = \frac{1}{d_i}\]Find the least common multiple and solve for \(d_i\):\[d_i = \frac{53.125 \times 19}{72.125} \approx -14.36\ cm\] Therefore, the new image distance \(d_i\) is approximately \(-14.36\ cm\).
05

Conclusion

The result shows that the image forms approximately 14.36 cm behind the mirror, which is consistent with how images are formed in convex mirrors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mirror equation
The mirror equation is essential for understanding how images form in mirrors. It combines the object distance \(d_o\), image distance \(d_i\), and focal length \(f\). The equation is given by:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
This formula relates the three quantities, enabling us to find one if the other two are known. For convex mirrors, it is important to remember:
  • Image distances \(d_i\) are negative because images form behind the mirror.
  • The focal length \(f\) is also negative, reflecting the mirror's diverging nature.
By substituting the known values into the mirror equation, you can solve for any given variable. This is particularly helpful when dealing with differing object distances, allowing us to predict where the image will be located.
focal length calculation
Understanding how to calculate the focal length of a mirror is crucial. The focal length is the point where parallel rays either converge or appear to diverge after reflecting off the surface of the mirror.
In our exercise, we needed the focal length of a convex mirror, calculated using the mirror equation:
  • Substitute the object distance \(d_o = 25\ cm\) and image distance \(d_i = -17\ cm\).
  • Use the equation: \( \frac{1}{f} = \frac{1}{25} + \frac{1}{-17} \).
Simplifying this, we found \(f \approx -53.125\ cm\).
The negative sign indicates the focal point is virtual. In practical terms:
  • Convex mirrors always have a negative focal length.
  • A consistent negative focal length across calculations indicates the reliability of the mirror equation.
Accurate focal length calculation is significant because it is a fundamental property of the mirror that affects all image formation scenarios.
object distance in mirrors
The distance of an object from a mirror, known as the object distance \(d_o\), is a pivotal factor influencing image formation. For convex mirrors, here's what happens:
  • Images form behind the mirror, defined by a negative image distance \(d_i\).
  • The object distance \(d_o\) is always positive since it is measured in front of the mirror.
In our problem, we first had an object standing at 25 cm from the mirror. The image was 17 cm behind. Then, the distance was reduced to 19 cm, affecting where the image appeared.
By refocusing on the mirror equation with the updated \(d_o\), we can determine the new \(d_i\).
  • Adjusting the object distance directly leads to changes in image distance.
  • It highlights the inverse relationship between object and image distances in the equation.
Understanding this relationship is key. It allows students to predict how moving an object affects its reflected image position, an essential skill in optics.

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Most popular questions from this chapter

A concave mirror has a focal length of \(12 \mathrm{~cm}\). This mirror forms an image located \(36 \mathrm{~cm}\) in front of the mirror. What is the magnification of the mirror?

A plane mirror and a concave mirror \((f=8.0 \mathrm{~cm})\) are facing each other and are separated by a distance of \(20.0 \mathrm{~cm}\). An object is placed \(10.0 \mathrm{~cm}\) in front of the plane mirror. Consider the light from the object that reflects first from the plane mirror and then from the concave mirror. Using a ray diagram drawn to scale, find the location of the image that this light produces in the concave mirror. Specify this distance relative to the concave mirror.

The drawing shows a top view of a square room. One wall is missing, and the other three are each mirrors. From point \(P\) in the center of the open side, a laser is fired, with the intent of hitting a small target located at the center of one wall. Identify six directions in which the laser can be fired and score a hit, assuming that the light does not strike any mirror more than once. Draw the rays to confirm your choices.

Consult Interactive Solution \(\underline{25.25}\) at for insight into this problem. An object is placed in front of a convex mirror, and the size of the image is one-fourth that of the object. What is the ratio \(d_{0} / f\) of the object distance to the focal length of the mirror?

Using the mirror equation and the magnification equation, show that for a convex mirror the image is always (a) virtual (i.e., \(d_{\mathrm{i}}\) is always negative) and (b) upright and smaller, relative to the object (i.e., \(m\) is positive and less than one).
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