Problem 10 - College Physics

Question

\(\bullet\) Examining your image in a convex mirror whose radius of curvature is \(25.0 \mathrm{cm},\) you stand with the tip of your nose 10.0 \(\mathrm{cm}\) from the surface of the mirror. (a) Where is the image of your nose located? What is its magnification? (b) Your ear is 10.0 \(\mathrm{cm}\) behind the tip of your nose; where is the image of your ear located, and what is its magnification? Do your answers suggest reasons for your strange appearance in a convex mirror?

Step-by-Step Solution

Answer

The nose image is at \(-5.56\) cm with magnification 0.556; the ear image is at \(-7.69\) cm with magnification 0.3845. This causes distorted views in convex mirrors.

Step 1: Identify the Given Values and Formulas

We have a convex mirror with a radius of curvature \( R = 25.0 \text{ cm} \) and the distance from object to mirror \( d_o = 10.0 \text{ cm} \). The focal length \( f \) for a convex mirror is \( f = \frac{R}{2} \), but negative, so \( f = -12.5 \text{ cm} \). We'll use the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) to find the image distance \( d_i \), and the magnification formula \( m = -\frac{d_i}{d_o} \).

Step 2: Calculate the Image Distance for the Nose

Substitute \( f = -12.5 \text{ cm} \) and \( d_o = 10.0 \text{ cm} \) into the mirror equation:\[\frac{1}{-12.5} = \frac{1}{10} + \frac{1}{d_i}\]Solving for \( d_i \) gives:\[\frac{1}{d_i} = \frac{1}{-12.5} - \frac{1}{10} = -0.08 - 0.1 = -0.18\]Therefore, \( d_i = -5.56 \text{ cm} \) (approx). The negative sign indicates the virtual image.

Step 3: Calculate the Magnification for the Nose Image

Using the magnification formula for the image of the nose:\[m = -\frac{d_i}{d_o} = -\frac{-5.56}{10} \approx 0.556\]This indicates the image is smaller than the object.

Step 4: Calculate the Position and Magnification for the Ear

The distance of the ear from the mirror is \( d_{oe} = 10.0 + 10.0 = 20.0 \text{ cm} \). Use the mirror formula:\[\frac{1}{-12.5} = \frac{1}{20} + \frac{1}{d_{ie}}\]Solving for \( d_{ie} \), we have:\[\frac{1}{d_{ie}} = \frac{1}{-12.5} - \frac{1}{20} = -0.08 - 0.05 = -0.13\]Thus, \( d_{ie} = -7.69 \text{ cm} \) (approx). For magnification:\[m = -\frac{d_{ie}}{d_{oe}} = -\frac{-7.69}{20} \approx 0.3845\]This indicates the image is even smaller.

Step 5: Explain the Distortion in Appearance

The images are formed closer to the mirror and are smaller than the actual objects. This results in the strange appearance typical of convex mirrors, where closer objects appear larger compared to those farther away, all appearing reduced in size.

Key Concepts

Mirror Equation

The mirror equation is a fundamental concept in optics that helps us understand how images are formed by mirrors. It is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where:

• \( f \) is the focal length of the mirror.
• \( d_o \) is the object distance, the distance from the object to the mirror.
• \( d_i \) is the image distance, the distance from the image to the mirror.

For a convex mirror, the focal length \( f \) is always considered negative because it is a diverging mirror. Convex mirrors spread out light rays that strike them, which is why the focal point is virtual and located behind the mirror. By using this equation, you can solve for any unknown distance if the other two are known. It's a powerful tool for determining the position and nature of an image.

Image Distance

Image distance \( (d_i) \) is crucial to understanding where an image will form when using a mirror. For convex mirrors, the image distance is always negative, indicating that the images are virtual and formed behind the mirror. In the exercise, we used the mirror equation to find the image distance for the nose and the ear.
For the nose, substituting the given values in the mirror equation resulted in \( d_i = -5.56 \text{ cm} \), indicating that the image is formed 5.56 cm behind the mirror.
For the ear, we calculated \( d_i = -7.69 \text{ cm} \), showing the image is even further behind the mirror. These calculations reveal how images appear reduced and virtual on a convex mirror.

Magnification

Magnification in mirror optics refers to how much larger or smaller an image is compared to the actual object. It is represented by the formula \( m = -\frac{d_i}{d_o} \).
- Here, \( m \) is the magnification.- \( d_i \) is the image distance, which is negative for virtual images.- \( d_o \) is the object distance.
In the example, the magnification for the nose was calculated to be approximately 0.556, meaning the image is smaller than the actual nose. For the ear, the magnification was about 0.3845, indicating an even smaller image. This is why objects in convex mirrors appear smaller and why nearby objects appear relatively larger than those farther away.

Radius of Curvature

The radius of curvature \( (R) \) refers to the radius of the sphere from which a mirror segment is taken. It is twice the focal length \( (f) \) of the mirror, expressed as \( R = 2f \).
In the context of convex mirrors, the radius of curvature is positive, but the focal length is considered negative. This is essential for solving the mirror equation and understanding image formation. In the exercise, the convex mirror had a radius of curvature of 25 cm, which allowed us to find the focal length as \( f = -\frac{R}{2} = -12.5 \text{ cm} \).
Understanding the radius of curvature helps in knowing how the mirror focuses light, giving insight into the image size and distortion properties associated with convex mirrors.