Problem 46 - College Physics

Question

(a) Design the lens in a single-lens slide projector that will form a sharp image on a screen \(4.0 \mathrm{~m}\) away with the transparent slides \(6.0 \mathrm{~cm}\) from the lens. (b) If the object on a slide is \(1.0 \mathrm{~cm}\) tall, how tall will the image on the screen be?

Step-by-Step Solution

Answer

(a) Focal length is approximately 5.91 cm. (b) Image height is approximately -66.67 cm (inverted).

Step 1: Use the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance (distance from lens to screen), and \( u \) is the object distance (distance from lens to slide). Substitute \( v = 4.0 \text{ m} \) or \(400 \text{ cm} \) and \( u = 6.0 \text{ cm} \).

Step 2: Calculate Focal Length

Rearranging the lens formula gives \( \frac{1}{f} = \frac{1}{400} + \frac{1}{6} \). Solve for \( f \): Calculate \( \frac{1}{400} = 0.0025 \) and \( \frac{1}{6} \approx 0.1667 \). Then, \( \frac{1}{f} = 0.0025 + 0.1667 \approx 0.1692 \). Thus, \( f = \frac{1}{0.1692} \approx 5.91 \text{ cm} \).

Step 3: Use Magnification Formula for Image Height

The magnification \( m \) is given by \( m = -\frac{v}{u} \). Using \( v = 400 \text{ cm} \) and \( u = 6.0 \text{ cm} \), we get \( m = -\frac{400}{6} = -66.67 \). The height of the image \( h' \) is then given by \( h' = m \times h \), where \( h = 1.0 \text{ cm} \) is the object height.

Step 4: Calculate Image Height

Plug the values into the magnification equation: \( h' = -66.67 \times 1.0 \text{ cm} = -66.67 \text{ cm} \). The negative sign indicates that the image is inverted relative to the object.

Key Concepts

Focal Length Calculation

The focal length of a lens is pivotal in determining how light is focused and an image is formed. To find this, we use the **lens formula**:

• \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

Where:

• \( f \) is the focal length of the lens
• \( v \) is the image distance (distance from lens to screen)
• \( u \) is the object distance (distance from lens to slide)

In the exercise, the image distance \( v \) is 400 cm, and \( u \), the object distance, is 6.0 cm.
By substituting these values into the formula, we solve for the focal length \( f \).
After calculating, the focal length \( f \) is found to be approximately 5.91 cm. This tells us how curved the lens should be to focus the image correctly on the screen. Understanding this helps in designing the optics to achieve clear projections.

Magnification Formula

Magnification is a measure of how much larger or smaller the image is compared to the object. It's calculated using the formula:

• \( m = -\frac{v}{u} \)

Here, \( m \) is the magnification factor, \( v \) is the image distance, and \( u \) is the object distance. In our example, substituting \( v = 400 \) cm and \( u = 6.0 \) cm, gives us a magnification of -66.67.
What does this mean? The image is 66.67 times larger than the object and inverted. The negative sign is significant as it denotes inversion of the image. When you look through a projector lens, understanding this helps in predicting how large your projection will be.

Single-Lens Slide Projector

A slide projector uses a single lens to project an enlarged image of a slide onto a screen. The clarity and size of the image depend on the calculations of focal length and magnification.
Here, we engage a lens that ensures a screen is 4.0 meters away, while slides are held 6.0 cm from the lens. By using the lens formula, the calculated focal length helps in selecting or designing the correct lens.

• This ensures the image is focused sharply on the screen.

Understanding a single-lens system's simplicity helps in effectively setting up and using projectors for presentations.

Image Formation

Image formation is the process wherein light through a lens gets focused to form a visual representation on a screen. This process is governed by the interplay of focal length, object distance, and image distance.
When the lens's focal length is correctly calculated and applied, light rays converge to form a sharp image at a specific location. In the context of this problem:

• The object on the slide is 1.0 cm tall.
• The resulting image on the screen is -66.67 cm tall.

The negative height indicates an inverted image. Understanding this concept allows one to manipulate optical devices efficiently to achieve desired visual outcomes.