Problem 15 - College Physics

Question

.. In a museum devoted to the history of photography, you are setting up a projection system to view some historical 4.0 inch \(\times 5.0\) inch color slides. Your screen is 6.0 \(\mathrm{m}\) from the projector lens, and you want the image to be 4.0 ft \(\times 5.0\) ft on the screen. (a) What focal-length lens do you need? (b) How far from the lens should you put the slide?

Step-by-Step Solution

Answer

Focal length = 455 mm, slide distance = 500 mm from lens.

Step 1: Understand the Problem

We need to project an image from a slide with dimensions 4.0 inches by 5.0 inches onto a screen, where the image needs to be 4.0 ft by 5.0 ft. The distance from the projector to the screen is 6.0 m. We're asked to find the focal length of the lens required and the distance the slide should be placed from the lens.

Step 2: Convert Units

Convert the dimensions of the image on the screen from feet to inches, because the slide dimensions are in inches: 1 ft is 12 inches, so 4.0 ft is 48 inches and 5.0 ft is 60 inches.

Step 3: Calculate Magnification

The magnification, \( M \), is the ratio of the image dimensions to the object dimensions. Calculate \( M \) using either the width or height. Here, using width: \( M = \frac{48}{4} = 12 \).

Step 4: Use the Lens Formula

Use the lens formula, \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance (distance from slide to lens), and \( d_i \) is the image distance (distance from lens to screen, 6.0 m or 6000 mm).

Step 5: Relate Magnification to Distances

The magnification also relates the distances: \( M = \frac{d_i}{d_o} \). Thus, \( d_o = \frac{d_i}{M} = \frac{6000 \text{ mm}}{12} = 500 \text{ mm} \).

Step 6: Solve for Focal Length

Substitute \( d_o = 500 \text{ mm} \) and \( d_i = 6000 \text{ mm} \) into the lens formula: \( \frac{1}{f} = \frac{1}{500} + \frac{1}{6000} \). Calculate \( \frac{1}{f} = \frac{1}{500} + \frac{1}{6000} \approx 0.0022 \), so \( f \approx \frac{1}{0.0022} \approx 455 \text{ mm} \).

Step 7: Conclusion

The focal length needed for the lens is approximately 455 mm, and the slide should be placed 500 mm from the lens.

Key Concepts

Lens Formula

The lens formula is a fundamental equation in optics that relates the focal length of a lens to the object and image distances. The formula is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:

• \(f\) is the focal length of the lens,
• \(d_o\) is the distance from the object (here, the slide) to the lens,
• \(d_i\) is the distance from the lens to the image (distance to the screen in this exercise).

In this exercise, the image distance, \(d_i\), is given as 6000 mm (6.0 m). Finding \(d_o\) requires using the magnification, as both these distances tie together with it.By plugging \(d_o\) and \(d_i\) into the lens formula, you can solve for the focal length \(f\) which ends up governing how focused or sharp the image will be on the screen.The closer the object distance to the required focal point, the clearer the image will appear.

Magnification

Magnification is the process of enlarging the appearance of an object through an optical system. In our scenario, the magnification, \(M\), is the ratio of the image size on the screen to the size of the actual slide. This can be calculated using:\[ M = \frac{h_i}{h_o} = \frac{d_i}{d_o} \]Where:

• \(h_i\) is the image height on the screen,
• \(h_o\) is the original slide height,
• \(d_i\) is the image (or screen) distance,
• \(d_o\) is the object (or slide) distance.

For our projection system problem, the magnification is found by comparing the width (or height) of the projected image and the original slide. Here it is calculated as 12. This magnification factor scales up the dimensions of the slide proportionally to the final image on the screen. Once known, it also informs how to position the slide relative to the lens for the desired outcome.

Focal Length

The focal length of a lens is a critical factor in determining how it will focus light and create an image. It is defined as the distance between the lens and the point where parallel rays of light converge to a single point. In this exercise, the focal length necessary to project the slide clearly onto the screen is approximately 455 mm. This value indicates the strength and precision of the lens needed to ensure the final image is well-focused and reaches the desired size. The choice of focal length depends on:

• The distance between the object and the image,
• The required image size on the screen,
• And the physical space available for the projection setup.

A lens with the right focal length ensures that each part of your slide is accurately magnified and displayed on the screen, creating a crisp and proportional image.

Projection System

A projection system is crucial for enlarging small images or slides onto bigger screens. It consists of three main components: the object (slide), the lens, and the screen. In this setup, the objective is to convert a small picture into a much larger one that maintains fidelity to the original. The process works as follows:

• Light travels through the lens from the slide, bending and altering the path to enhance size and visibility.
• The focal length of the lens determines how light rays are focused onto the screen.
• The distance between each part (slide to lens, and lens to screen) plays a key role in clarity and size.

Achieving the desired image size, as seen in this exercise, involves adjusting the slide's placement and selecting a lens with the correct focal length. A meticulous alignment of these factors results in a well-projected image. The system must also account for potential distortion or focus issues, ensuring the lens selected can accommodate all requirements for the intended display.