Problem 124 - Fundamentals of Physics

Question

A short straight object of length \(L\) lies along the central axis of a spherical mirror, a distance \(p\) from the mirror. (a) Show that its image in the mirror has a length \(L^{\prime},\) where \(L^{\prime}=L\left(\frac{f}{p-f}\right)^{2}\) (Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification \(m^{\prime}\left(=L^{\prime} / L\right)\) is equal to \(m^{2},\) where \(m\) is the lateral magnification.

Step-by-Step Solution

Answer

The image length is \(L' = L\left(\frac{f}{p-f}\right)^2\), and longitudinal magnification \(m' = m^2\).

Step 1: Understand the Problem

We need to find the image length of a straight object placed along the principal axis of a spherical mirror and show it equals \(L' = L\left(\frac{f}{p-f}\right)^2\). This involves finding the positions of the image for each end of the object.

Step 2: Use Mirror Equation for Each End

The mirror equation is given by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(f\) is the focal length, \(p\) is the object distance, and \(q\) is the image distance. For the object's far end at distance \(p+L\), calculate its image distance \(q_1\) using this formula.

Step 3: Calculate Image Distances

For the near end of the object at \(p\), apply the mirror equation to find the image distance \(q_0\). For the far end, at \(p + L\), apply the mirror equation again to find \(q_1\).

Step 4: Derive Expression for Image Length

The length of the image \(L'\) is given by the difference in image distances \(q_0\) and \(q_1\). Substituting these from the mirror equation results, we find \(L' = q_1 - q_0\). Simplify this using the relations from the mirror equation to arrive at \( L' = L\left(\frac{f}{p-f}\right)^2 \).

Step 5: Understand and Calculate Lateral Magnification

Lateral magnification, \(m\), is \(m = -\frac{q_0}{p}\). This represents how much the image of any point at \(p\) is enlarged or contracted laterally.

Step 6: Verify Longitudinal Magnification

The longitudinal magnification \(m' = \frac{L'}{L}\). Substitute \(L'\) from Step 4 to get \(m' = \left(\frac{f}{p-f}\right)^2\). Since we previously found \(m = -\frac{q_0}{p}\), calculate \(m^2\) and demonstrate it equals \(m'\).

Key Concepts

Mirror Equation

The mirror equation is crucial when working with spherical mirrors. It connects the focal length

• The formula is: \[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\]where:
  • \( f \) is the focal length,
  • \( p \) is the object distance from the mirror, and
  • \( q \) is the image distance from the mirror.

To find where an image forms, we input the distance to the object (\( p \)) and solve for the image distance (\( q \)).
This equation helps determine whether the image is real (if \( q \) is positive) or virtual (if \( q \) is negative).
By applying this equation to both ends of an object lying along the axis of a spherical mirror, we can find the image positions for each end, leading to further calculations of image dimensions.
It's essential to grasp this equation as it is the foundation for understanding how images form in spherical mirrors.

Longitudinal Magnification

Longitudinal magnification relates to how the depth or length of an object's image is increased or decreased. It specifically examines how the size of the image along its optical axis changes.

• The relationship can be expressed by:\[m' = \frac{L'}{L}\]where:
  • \( m' \) is the longitudinal magnification,
  • \( L' \) is the image length, and
  • \( L \) is the actual length of the object.

This gives us a ratio representing how much the object's depth or length appears to have been stretched or compressed.
Using the mirror equation, we derive that the longitudinal magnification is equal to the square of the lateral magnification: \[m' = m^2\]This conclusion shows the impact of the spherical mirror on the object's depth perception.
Grasping this concept helps you understand three-dimensional impacts on images, as it considers not just a linear, but a volumetric change.

Lateral Magnification

Lateral magnification explains how much an object's image size changes across the horizontal axis. It measures the ratio of the image height to the object height and helps us see the enlargement or reduction of an image from an upright object.

• Defined by:\[m = -\frac{q_0}{p}\]where:
  • \( m \) is the lateral magnification,
  • \( q_0 \) is the image distance for the near end of the object, and
  • \( p \) is the object distance.

A negative magnification indicates an inverted image, while a positive magnification means the image is upright. This notion reflects how mirrors affect the size appearance of objects as seen from a sideways perspective.
Understanding lateral magnification is crucial because it allows for determining the extent of an image's enlargement or diminution, providing insight into how a mirror can reshape the view of physical objects.