Problem 9 - University Physics with Modern Physics

Question

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 \(\mathrm{cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\) cm-tall coin that is 6.00 \(\mathrm{cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

Step-by-Step Solution

Answer

The coin is located \(18.0\, \mathrm{cm}\) from the mirror, the image is \(0.5\, \mathrm{cm}\) tall, upright, and virtual.

Step 1: Analyze the Given Information

We are provided the following details: the radius of curvature \(R = 18.0\, \mathrm{cm}\) and the image distance \(v = -6.0\, \mathrm{cm}\). The negative sign indicates that the image is formed behind the mirror. The height of the object (coin) is \(1.5\, \mathrm{cm}\). We need to find the object distance \(u\) and analyze the image.

Step 2: Use Mirror Formula

The mirror formula is given by \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\). Using the mirror formula for spherical mirrors, we find the focal length \(f\) as \(f = \frac{R}{2} = \frac{18.0}{2} = 9.0\, \mathrm{cm}\).

Step 3: Solve for Object Distance

Rearrange the mirror formula to solve for \(u\): \[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]Substitute the known values: \[ \frac{1}{u} = \frac{1}{9.0} - \frac{1}{-6.0} \]Calculate:\[ \frac{1}{u} = \frac{2 - 3}{18} = \frac{-1}{18} \]Thus, \(u = -18.0\, \mathrm{cm}\).

Step 4: Determine Image Size and Orientation

Use the magnification formula, \(m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}}\).Calculate the magnification:\\[ m = \frac{-6.0}{-18.0} = \frac{1}{3} \]Thus, the image height is:\[ \text{Image height} = m \times \text{Object height} = \frac{1}{3} \times 1.5 = 0.5\, \mathrm{cm} \].The positive sign indicates the image is upright.

Step 5: Determine the Nature of the Image

The image is virtual because it forms behind the mirror, as indicated by the negative image distance \(v = -6.0\, \mathrm{cm}\). Virtual images cannot be projected onto a screen and are always upright when formed by a convex surface.

Key Concepts

Mirror Formula

The mirror formula is an essential tool in understanding how spherical mirrors create images. It relates the focal length (\( f \)), object distance (\( u \)), and image distance (\( v \)) through the equation: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]. This formula allows us to calculate unknown distances by rearranging and substituting known values.

• The focal length (\( f \)) is half the radius of curvature (\( R \)) for spherical mirrors: \( f = \frac{R}{2} \).
• In convex mirrors, the focal length and image distance are considered negative due to their geometric nature. Images form behind the mirror, indicating a virtual image location.

By rearranging the formula to solve for \( u \), we can determine the object distance when focal length and image distance are known. This helps in locating where an object must be placed relative to a spherical mirror to achieve a specific image formation.

Virtual Image

A virtual image in optics is what you see in a mirror—it appears to stand behind the mirror surface. Unlike real images, virtual images cannot be captured on a screen because they don't actually converge in real space.

• In convex mirrors, images are always virtual because they form where light rays appear to diverge from.
• These images retain the same orientation as the object, hence they are upright.
• Convex mirrors, with a computed negative image distance, always give rise to virtual images—suggesting the image is on the same side of the mirror as the object.

Virtual images provide important benefits in practical applications too. For example, in car side mirrors, virtual images allow drivers to view a wider field of vision, enhancing safety.

Magnification

Magnification describes the degree to which an image is enlarged or reduced compared to the actual object. The magnification (\( m \)) formula is given by: \[ m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}} \].

• A magnification value (\( m \)) greater than 1 suggests that the image is larger than the object.
• A value less than 1 indicates reduction; in our case, \( m = \frac{1}{3} \), showing the image is one-third the object's height.
• The sign of magnification indicates image orientation—positive for upright, negative for inverted.

In convex mirrors, the magnification is often less than 1, denoting a minimized view. This property makes convex mirrors invaluable for situations needing wide perspective views, such as in security or rear-view mirrors.