Problem 32 - College Physics

Question

A spherical mirror at an amusement park has a radius of \(10 \mathrm{~m}\). If it forms an image that has a lateral magnification of \(+2.0,\) what are the object and image distances?

Step-by-Step Solution

Answer

Object distance is 2.5 m, image distance is -5 m.

Step 1: Understand the Given

We are provided with a spherical mirror that has a radius of curvature, \( R = 10 \text{ m} \), and a magnification, \( m = +2.0 \). The sign \( + \) in magnification indicates that the image is upright and same side as the object. We need to find the object distance \( p \) and image distance \( q \).

Step 2: Use the Mirror Formula

The mirror formula is \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \), where \( f \) is the focal length. The focal length is half the radius of curvature for spherical mirrors, \( f = \frac{R}{2} \). Substitute \( R = 10 \text{ m} \) so \( f = 5 \text{ m} \).

Step 3: Use the Magnification Formula

Magnification \( m \) is also given by the formula \( m = -\frac{q}{p} \). We know \( m = +2.0 \), so \( -\frac{q}{p} = 2.0 \). Hence, \( q = -2p \).

Step 4: Substitute in the Mirror Formula

Substitute \( q = -2p \) into the mirror formula \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \):\[\frac{1}{5} = \frac{1}{p} + \frac{1}{-2p}\]

Step 5: Simplify and Solve for Object Distance

Combine the terms:\[\frac{1}{5} = \frac{1}{p} - \frac{1}{2p}\]This simplifies to:\[\frac{1}{5} = \frac{2 - 1}{2p} = \frac{1}{2p}\]Therefore, \( 2p = 5 \), so \( p = 2.5 \text{ m} \).

Step 6: Calculate Image Distance

Using \( q = -2p \), substitute \( p = 2.5 \text{ m} \):\[q = -2 \times 2.5 = -5 \text{ m}\]

Key Concepts

Mirror Formula

The mirror formula is fundamental in understanding how spherical mirrors work. It connects the object distance, image distance, and focal length. The formula is expressed as:

• \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)

where:

• \( f \) represents the focal length of the mirror.
• \( p \) is the distance from the object to the mirror.
• \( q \) is the distance from the image to the mirror.

For spherical mirrors, especially concave or convex ones, the focal length \( f \) is half the radius of curvature \( R \). Hence, \( f = \frac{R}{2} \). Understanding this relationship is key because it tells us about how light converges or diverges after reflecting off a mirror. The mirror formula helps us determine where an image will form and whether the image will be real or virtual.

Magnification

Magnification in mirrors refers to how much bigger or smaller the image is compared to the object. The formula for magnification \( m \) in the context of mirrors is:

• \( m = -\frac{q}{p} \)

where:

• \( m \) is the magnification.
• \( q \) is the image distance.
• \( p \) is the object distance.

A positive magnification value means the image is upright relative to the object, while a negative value indicates an inverted image. In the problem example, a magnification of \(+2.0\) tells us that the image is twice the height of the object and upright. This helps in predicting not only the size but also the orientation of images formed by mirrors.

Image Distance

The image distance \( q \) is critical in finding the position of the image formed by a mirror. It is the distance from the mirror surface to where the image is located. Using the mirror and magnification formulas together allows us to solve for \( q \). We often rearrange the magnification formula to find:

• \( q = -mp \)

In situations like our example, where magnification \( m \) is given as \(+2\), the image distance \( q \) can be calculated by substituting this value into the formula. It helps to know whether the image is formed on the same side as the object or on the opposite side.

Object Distance

The object distance \( p \) is simply the distance from the object to the mirror. It is a crucial component in the mirror formula. Knowing either \( q \) or \( p \), we can find the other using the relationship:

• \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)

Object distance and image distance are directly linked through the focal length. By substituting known values into this formula, we can solve for object distance. In problems involving spherical mirrors, this approach lets us understand how varying the object's position affects the characteristics of the image produced.