Problem 16 - College Physics

Question

\(\cdot\) The stainless steel rear end of a tanker truck is convex, shiny, and has a radius of curvature of 2.0 \(\mathrm{m} .\) You're tailgating the truck, with the front end of your car only 5.0 \(\mathrm{m}\) behind it. Making the not very realistic assumption that your car is on the axis of the mirror formed by the tank, (a) determine the position, orientation, magnification, and type (real or virtual) of the image of your car's front end that forms in this mirror; (b) draw a principal-ray diagram of the situation to check your answer.

Step-by-Step Solution

Answer

Image is virtual, upright, smaller, located 0.833 m behind the mirror with magnification 0.166.

Step 1: Identify the Given Values

From the problem, we know that the radius of curvature, \( R \), is 2.0 m, and the object distance \( d_o \) is 5.0 m. Since the mirror is convex, the radius of curvature is negative (\( R = -2.0 \) m).

Step 2: Calculate the Focal Length

The focal length \( f \) is related to the radius of curvature \( R \) by the formula \( f = \frac{R}{2} \). Since \( R = -2.0 \) m, we calculate \( f = \frac{-2.0}{2} = -1.0 \) m.

Step 3: Use the Mirror Equation

The mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substituting the known values, we have \( \frac{1}{-1.0} = \frac{1}{5.0} + \frac{1}{d_i} \). Solving for \( d_i \):\[\frac{1}{d_i} = \frac{1}{-1.0} - \frac{1}{5.0} = -1 - 0.2 = -1.2\]So, \( d_i = -\frac{1}{1.2} \approx -0.833 \) m.

Step 4: Determine Image Orientation and Type

Since the image distance \( d_i \) is negative, the image is virtual and located 0.833 m behind the mirror. Convex mirrors always produce virtual images.

Step 5: Calculate Magnification

The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Substituting the known values, we find \( m = -\frac{-0.833}{5.0} = 0.1666 \). Thus, the image is upright and smaller than the object.

Step 6: Draw Principal-Ray Diagram

Drawing a principal-ray diagram involves three key rays: 1) A ray parallel to the axis reflects through the focal point, 2) A ray through the center of curvature reflects back along its path, and 3) A ray to the focal point reflects parallel to the axis. For a convex mirror, these rays appear to diverge but can be traced back to a point behind the mirror, confirming the image's location.

Key Concepts

Ray Diagrams

Ray diagrams are essential tools in optics used to graphically determine the properties of images formed by mirrors or lenses. For convex mirrors, these diagrams can help visualize why images are always virtual and upright. In our scenario, you're using a convex mirror on the tanker truck. Here's how to draw the ray diagram for this case:

• First, draw the mirror as a slightly curved line that bows towards the object (your car).
• Identify the principal axis – a straight horizontal line passing through the center of the mirror.
• Represent three major rays:
  • Ray 1 goes parallel to the principal axis, and upon reflection, it appears to diverge from the focal point behind the mirror.
  • Ray 2 heads towards the mirror's focal point but emerges parallel in reflection.
  • Ray 3 aims at the mirror's center of curvature and reflects back upon itself.

All these rays, when extended backward behind the mirror, converge at a point called the virtual image position. The image appears smaller and upright, confirming the nature of a convex mirror.

Mirror Equation

The mirror equation is a fundamental relation used to predict how mirrors focus light and form images. It relates the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) in this equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]For our convex mirror scenario, it's crucial to remember:

• The focal length (\(f\)) is negative, which indicates the mirror is convex.
• You need to plug in the radius of curvature (\(R\)) to find the focal length using \(f = \frac{R}{2}\).
• Substitute known values and solve for the unknown, typically \(d_i\) when \(d_o\) and \(f\) are known.
• A negative \(d_i\) indicates a virtual image.

In the given problem, substituting \(d_o = 5.0 \text{ m}\) and \(f = -1.0 \text{ m}\), we essay that: \[ \frac{1}{d_i} = \frac{1}{-1.0} - \frac{1}{5.0} = -1.2\] Thus, \(d_i\) calculated gives us the position of a virtual image.

Virtual Images

A virtual image is a perception of an object obtained from diverging lens or mirror systems where the light rays do not physically converge. For convex mirrors, virtual images have specific properties:

• Located behind the mirror – they cannot be projected onto a screen because they do not actually "exist" where the rays seem to converge.
• Always appear upright and smaller – that's why convex mirrors are often used for rear-view surfaces to provide a wider field of view.
• These images are perceived when the brain traces back the diverging rays to apparent points of origin.

In the truck example, the virtual image of the car appears at 0.833 m behind the mirror, confirming the mirror's inherent properties.

Magnification

Magnification helps you understand the size relationship between the image and the object viewed through a mirror. For mirrors, the magnification (\(m\)) is defined by:\[ m = -\frac{d_i}{d_o} \]Here's how it applies in the context of convex mirrors:

• A positive magnification value implies that the image is upright.
• If the magnitude is less than one, the image is smaller than the object.
• Conversely, if greater than one, the image is larger.

In the example with the tanker truck, since \(d_i = -0.833 \text{ m}\) and \(d_o = 5.0 \text{ m}\), substituting into the formula results in:\[ m = 0.1666 \]This confirms that the virtual image is smaller and upright, typical of convex mirrors.