Problem 20 - Physics

Question

The outside mirror on the passenger side of a car is convex and has a focal length of \(-7.0 \mathrm{~m}\). Relative to this mirror, a truck traveling in the rear has an object distance of \(11 \mathrm{~m}\). Find (a) the image distance of the truck and (b) the magnification of the mirror.

Step-by-Step Solution

Answer

(a) The image distance is approximately -4.28 m. (b) The magnification is approximately -0.389.

Step 1: Understand the Mirror Equation

We begin the solution by identifying the relevant formula to use, which is the mirror equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.

Step 2: Plug in Known Values and Solve for Image Distance

Given \( f = -7.0 \) m and \( d_o = 11 \) m, substitute these values into the mirror equation: \( \frac{1}{-7.0} = \frac{1}{11} + \frac{1}{d_i} \). Rearrange to solve for \( \frac{1}{d_i} = \frac{1}{-7.0} - \frac{1}{11} \). Calculating this gives \( \frac{1}{d_i} = -0.1429 - 0.0909 = -0.2338 \). Thus, \( d_i = \frac{1}{-0.2338} \approx -4.28 \) m.

Step 3: Find the Magnification

The magnification \( m \) is found using the magnification formula: \( m = \frac{d_i}{d_o} \). Substituting the values \( d_i = -4.28 \) m and \( d_o = 11 \) m gives \( m = \frac{-4.28}{11} \approx -0.389 \).

Step 4: Interpret the Results

The negative image distance indicates that the image is virtual and located on the same side of the mirror as the object. The negative magnification suggests the image is inverted relative to the object and its magnitude less than one indicates it is smaller than the original object.

Key Concepts

Focal Length

The focal length of a mirror is crucial in determining how an image is formed. For convex mirrors, which are often used in vehicle mirrors to provide a wider field of view, the focal length is always negative. This is because the focal point is behind the mirror. If we consider the convex mirror on a car's passenger side, it has a focal length of \( f = -7.0 \, \text{m} \). This negative value reminds us that the image formed by a convex mirror will always be virtual and smaller.

In the mirror equation, the focal length is used with the object distance to find the image distance. The negative sign indicates that the focus is not a real point since convex mirrors never converge light rays.

Image Distance

When we talk about the image distance in mirrors, we're discussing how far the image is "virtually" located from the mirror. The mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) is a valuable tool for calculating this. Here, \( f \, d_o \, \, \text{and} \, d_i\) stand for focal length, object distance, and image distance, respectively.

In the exercise, substituting \( f = -7.0 \, \text{m} \) and \( d_o = 11 \, \text{m} \) into the equation, and solving for \( d_i \), we find it to be approximately \( -4.28 \, \text{m} \). The negative sign tells us that the image is formed on the same side of the mirror as the object, typical of virtual images in convex mirrors.

Mirror Equation

The mirror equation is the relationship between the focal length, object distance, and image distance. It is written as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). This simple equation lets us solve for one unknown when the other two quantities are known.

Convex mirrors have a distinctive property: they only form virtual images. By using this equation, we can determine how and where these images appear. Convex mirrors make use of their negative focal lengths in the mirror equation, inherently indicating that the rays appear to diverge from a point behind the mirror.

Magnification

Magnification is a measure of how much larger or smaller the image is compared to the object itself. It is determined by the formula \( m = \frac{d_i}{d_o} \). This formula allows us to see by how much a mirror enlarges or shrinks the apparent size of the object's image.

In our problem, the magnification calculated as \( m = \frac{-4.28}{11} \, \text{or} \, -0.389 \), shows two things. First, the negative value indicates the image's orientation is inverted, though this inversion doesn't change the object's context in a convex mirror – the image remains upright but is smaller than the actual object due to the mirror's characteristics. The magnitude less than one confirms the image is indeed reduced in size, matching our expectation for a convex mirror.