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 \( m' = m^2 \).

Step 1: Understand the Image Formation

To solve this problem, we need to consider the image formation in spherical mirrors using the mirror equation and magnification. The relevant equations are the mirror equation \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \) and the formula for lateral magnification \( m = -\frac{q}{p} \).

Step 2: Locate the Two Ends of the Object

The object of length \( L \) lies along the central axis. Consider its two ends: one end at a distance \( p \) (let's say the nearer end to the mirror) and the other at \( p + L \) from the mirror. Use the mirror equation to find the image location \( q_1 \) and \( q_2 \) of these two points.

Step 3: Apply the Mirror Equation to Both Ends

For the first end at distance \( p \):\[ \frac{1}{p} + \frac{1}{q_1} = \frac{1}{f} \implies q_1 = \frac{pf}{p - f} \]For the second end at distance \( p + L \):\[ \frac{1}{p+L} + \frac{1}{q_2} = \frac{1}{f} \implies q_2 = \frac{(p+L)f}{p+L-f} \]

Step 4: Calculate the Length of the Image

The length of the image \( L' \) is the difference between the image positions: \[ L' = |q_2 - q_1| \]Substitute the expressions for \( q_1 \) and \( q_2 \) from Step 3: \[ L' = \left| \frac{(p+L)f}{p+L-f} - \frac{pf}{p-f} \right| \]

Step 5: Simplify the Expression for \( L' \)

Simplify the expression obtained in Step 4:\[ L' = L \left( \frac{f}{p-f} \right)^2 \]This involves algebraically manipulating the fractions to achieve the desired result.

Step 6: Calculate Longitudinal Magnification \( m' \)

Longitudinal magnification \( m' \) is defined as \( m' = \frac{L'}{L} \). From Step 5, we have:\[ m' = \left( \frac{f}{p-f} \right)^2 \]Also, the lateral magnification \( m = \left( \frac{f}{p-f} \right) \). Hence, \( m'^2 = m^2 \), as required.

Key Concepts

Longitudinal Magnification

Longitudinal magnification is an important concept when dealing with spherical mirrors, as it measures how much the length of an object along the optical axis is magnified by the mirror. This type of magnification is different from lateral magnification, which pertains to magnification perpendicular to the axis. In the exercise, we are given an object of length \(L\) on the central axis of a spherical mirror.

• The longitudinal magnification, represented as \(m'\), is calculated using the formula \(m' = \frac{L'}{L}\).
• Here, \(L'\) is the length of the image formed, which is derived using the specific formula \(L' = L\left(\frac{f}{p-f}\right)^2\), where \(f\) is the focal length and \(p\) is the object distance from the mirror.

This concept helps in understanding how the mirror affects the perceived size of an object along its axis. Longitudinal magnification in this context turns out to be related to the square of the lateral magnification: \((\frac{f}{p-f})^2\), showcasing a unique relationship based on the geometry of the mirror.

Mirror Equation

The mirror equation is foundational in understanding how images are formed by spherical mirrors. It is a mathematical relationship that ties together the focal length of the mirror, the object distance, and the image distance. The standard form of the mirror equation is:

• \[ \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \]

In this equation:

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

The mirror equation is crucial since it lets us determine the position of the image formed by the mirror. This understanding was applied in the exercise to find the image distances \(q_1\) and \(q_2\) for the two ends of the object. By solving this equation for different values of \(p\) (object's ends), we find corresponding \(q\)'s that help in determining the length \(L'\) of the image.

Lateral Magnification

Lateral magnification describes the change in size of an image formed by a spherical mirror relative to its object, measured perpendicular to the principal axis. It assesses how a mirror scales an image in width and height but not depth.

• The formula for lateral magnification \(m\) is \(m = -\frac{q}{p}\).
• This equation shows that lateral magnification depends on both the object distance \(p\) and the image distance \(q\).
• In simpler terms, it indicates how much larger or smaller the image is compared to the actual object.

In the exercise, it was shown that \(m = \frac{f}{p-f}\) directly relates to the optical properties of the mirror. A fascinating result derived from this is that longitudinal magnification \(m'\) was found to be the square of lateral magnification \(m\). This highlights the interplay between different types of magnification and the geometry of mirrors, underscoring the importance of comprehending both dimensions of magnification.