Problem 85 - Physics for Scientists and Engineers with Modern Physics

Question

For a camera equipped with a 58 -mm-focal-length lens, what is the object distance if the image height equals the object height? How far is the object from the image on the film?

Step-by-Step Solution

Answer

The object distance is 116 mm, and the distance from the object to the image is 232 mm.

Step 1: Understanding the Magnification Concept

In optics, when the image height equals the object height, it implies that the magnification (M) is 1. Magnification is given by the formula: \( M = \frac{-q}{p} \), where \( q \) is the image distance and \( p \) is the object distance. When \( M = 1 \), it indicates that \( q = p \).

Step 2: Using the Lens Formula

The lens formula relates the focal length \( f \), object distance \( p \), and image distance \( q \) as \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \). Given that the focal length \( f = 58 \text{ mm} \) and \( q = p \), we can substitute \( q \) by \( p \) to simplify the formula as \( \frac{1}{f} = \frac{2}{p} \).

Step 3: Solving for Object Distance \( p \)

Rearrange the equation from Step 2 to solve for \( p \): \( p = 2f \). Substituting \( f = 58 \text{ mm} \), we get \( p = 2 \times 58 = 116 \text{ mm} \).

Step 4: Calculating the Distance from Object to Image

Since the object and image distances are the same \( (p = q) \), the distance from the object to the image is \( q + p = 2p \). Therefore, this distance is \( 2 \times 116 = 232 \text{ mm} \).

Key Concepts

Magnification

Magnification is a measure of how much larger or smaller the image is compared to the object itself. It plays a crucial role in optics to understand the relationship between image size and object size.
In mathematical terms, magnification (\( M \)) is defined as the ratio of the image height to the object height. It can also be expressed using the image distance (\( q \)) and the object distance (\( p \)) according to the formula:

• \( M = \frac{-q}{p}\)

When the image height equals the object height, the magnification is 1, represented as \( M = 1 \).
This specific condition tells us that the image is the same size as the object, leading to the fact that the image distance and object distance are equal, simplifying calculations in various optical settings.

Lens Formula

The lens formula is a key equation in optics that connects the focal length, object distance, and image distance. It helps in figuring out these distances when either the object or the image is positioned through a lens.
The formula 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.
This formula can be re-arranged depending on what you are solving for, and it is vital when working with images formed by lenses.
For example, if two quantities are known, say \( f \) and \( p \), \( q \) can be found easily allowing us to solve many practical optics problems.

Image Distance

The image distance is the distance between the lens and the image formed by the lens. It is usually denoted by the symbol \( q \), and in many optics problems, knowing \( q \) helps to determine other unknown parameters.
For either convex or concave lenses, the image distance could be real or virtual, which affects how images are observed.

• Real images, formed when light converges, have positive image distances.
• Virtual images, appearing where light does not actually meet or coming from behind the lens, have negative image distances.

Knowing about image distance is essential for applications ranging from photography to corrective lenses.

Object Distance

In optics, the object distance is the distance between the object and the lens and is usually represented by \( p \). Determining the object distance is crucial for finding out other aspects of image formation through lenses.
It can be calculated using the lens formula when the focal length and image distance are known, typically by rearranging the formula as needed.
For our exercise, with the focal length (\( f = 58 \text{ mm} \)) and the special condition that the magnification is 1 (meaning \( p = q \)), the object distance was found to be:

• \( p = 2f \)
• Resulting in \( p = 116 \text{ mm} \).

Understanding object distance helps not only in theoretical problems but practical lens settings and adjustments.