Problem 34 - Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\).

Step 2: 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 \(f = - \frac{425}{8} = -53.125\ cm\).

Step 3: 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 \(d_i\).

Step 4: 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 \(-14.36\ cm\).

Step 5: 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.

Key Concepts

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.