Problem 132 - Fundamentals of Physics

Question

A millipede sits \(1.0 \mathrm{~m}\) in front of the nearest part of the surface of a shiny sphere of diameter \(0.70 \mathrm{~m}\). (a) How far from the surface does the millipede's image appear? (b) If the millipede's height is \(2.0 \mathrm{~mm},\) what is the image height? (c) Is the image inverted?

Step-by-Step Solution

Answer

(a) The image appears 0.21 m from the surface. (b) The image height is -0.42 mm. (c) The image is not inverted.

Step 1: Determine the radius of the sphere

The diameter of the sphere is given as \(0.70 \text{ m}\). To find the radius \(R\), we use the formula for the radius: \(R = \frac{d}{2}\). Thus, \(R = \frac{0.70}{2} = 0.35 \text{ m}\).

Step 2: Set up the mirror equation

For mirrors, we use the mirror equation to relate the object distance \(p\), the image distance \(i\), and the radius of curvature \(R\): \(\frac{1}{f} = \frac{1}{p} + \frac{1}{i}\), where \(f\) is the focal length, \(f = \frac{R}{2}\). Thus, \(f = \frac{0.35}{2} = 0.175 \text{ m}\).

Step 3: Solve for the image distance

Rearrange the mirror equation to solve for \(i\): \(\frac{1}{i} = \frac{1}{f} - \frac{1}{p}\). Using \(f = 0.175 \text{ m}\) and \(p = 1.0 \text{ m}\), we have \(\frac{1}{i} = \frac{1}{0.175} - \frac{1}{1.0}\). Solve this to find \(i\).

Step 4: Calculate the image distance

Calculate \(\frac{1}{f} = \frac{1}{0.175} \approx 5.714\) and then \(\frac{1}{i} = 5.714 - 1 = 4.714\). Thus, \(i \approx \frac{1}{4.714} \approx 0.212 \text{ m}\). The image distance \(i\) is measured from the mirror. Since the mirror is a convex mirror, the image distance indicates that the image is virtual and behind the mirror.

Step 5: Determine the distance from sphere surface

Since the image distance \(i\) is calculated relative to the mirror's surface, the distance from the surface where the image appears is simply \(0.212 \text{ m}\).

Step 6: Use magnification to find the image height

The magnification \(m\) is given by the formula \(m = -\frac{i}{p}\). The image height \(h_i\) is given by \(m = \frac{h_i}{h_o}\), where \(h_o = 2.0 \text{ mm}\).

Step 7: Calculate the magnification and image height

Substitute the known values into the magnification formula: \(m = -\frac{0.212}{1.0} = -0.212\). Thus, \(h_i = m \times h_o = -0.212 \times 2.0\ \text{mm} = -0.424\ \text{mm}\). The negative sign indicates the nature of the image being virtual and upright.

Step 8: Determine if the image is inverted

In a convex mirror, the image is virtual, reduced, and upright. Since the magnification calculated is negative, it confirms the image is upright and not inverted.

Key Concepts

Mirror Equation

The mirror equation is a fundamental principle in geometrical optics that helps us understand how images are formed by mirrors. It shows the relationship between the object distance, image distance, and the mirror's focal length. The mirror equation is expressed as:\[\frac{1}{f} = \frac{1}{p} + \frac{1}{i}\]where:

• \( f \) is the focal length of the mirror
• \( p \) is the object distance, or how far the object is from the mirror
• \( i \) is the image distance, indicating where the image forms relative to the mirror

To find the focal length of a spherical mirror, use the formula \( f = \frac{R}{2} \), where \( R \) is the radius of curvature of the mirror. This equation proves to be critical in determining where an image appears and whether it is real or virtual. For convex mirrors, which always produce virtual images, the focal length is considered negative. Thus, using the mirror equation allows you to solve for unknowns, enhancing your understanding of optical systems.

Convex Mirror

Convex mirrors curve outward and are known for their unique ability to reflect light in such a way that spreads it outwards. This property results in a virtual, upright, and smaller image compared to the actual object. These types of mirrors are commonly used in vehicles as side mirrors and for security purposes because they provide a wider field of view.

In a convex mirror, the focal point is located behind the mirror, producing images that are virtual (they cannot be projected onto a screen), upright, and reduced in size. When referring to the mirror equation in the context of a convex mirror, the focal length and image distance will be negative, emphasizing the nature of virtual images. Being aware of these properties helps us understand why images appear smaller and distorted when viewed through a convex mirror.

Image Magnification

Image magnification describes how much larger or smaller the image is compared to the actual object. Magnification (\( m \)) can be calculated using two principles:

• The relationship with image and object distance: \( m = -\frac{i}{p} \)
• The relationship with image and object height: \( m = \frac{h_i}{h_o} \)

where:

• \( h_i \) is the image height
• \( h_o \) is the object height

In this formula, a negative magnification value implies that the image is virtual and upright, typical for images formed by convex mirrors. Conversely, real images have positive magnifications and appear inverted. By calculating the magnification, you gain insight into the nature of the image, helping you visualize essential characteristics such as orientation and size.