Question

You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are 24 mm \(\times\) 36 mm, what is the minimum size of the projector screen required to accommodate the image?

Short answer

(a) The focal length is approximately 14.75 cm. (b) The minimum screen size is 1.44 m x 2.16 m.

Step-by-step solution

Understanding the Problem

We need to project an image using a slide projector. The projector lens needs to focus an image from a slide onto a screen 9 meters away. We first need to find the lens's focal length using the lens formula. Then, we'll calculate the minimum screen size needed to accommodate the projected image dimensions.

Analyzing Given Values

The distance from the lens to the screen (image distance, \(v\)) is 9.00 m or 900 cm (since consistent units are required). The distance from the slide to the lens (object distance, \(u\)) is 15.0 cm. The slide dimensions are 24 mm \(\times\) 36 mm.

Applying the Lens Formula for Focal Length

The lens formula is \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). Substitute \(v = 900\) cm and \(u = 15\) cm into the equation: \[ \frac{1}{f} = \frac{1}{900} + \frac{1}{15} \] Calculate \(\frac{1}{f}\), then invert to find \(f\).

Calculating \( \frac{1}{f} \)

First, compute the reciprocal values: \( \frac{1}{900} = 0.00111 \) and \( \frac{1}{15} = 0.0667 \). Add them together to get \[ \frac{1}{f} = 0.0667 + 0.00111 = 0.06781 \]

Determining the Focal Length

Invert \( \frac{1}{f} \) to find \(f\): \[ f = \frac{1}{0.06781} \approx 14.75 \text{ cm} \] The required focal length is approximately 14.75 cm.

Calculating Projected Image Size

To find the screen size, calculate the magnification \(M\), where \(M = \frac{v}{u} = \frac{900}{15} = 60\). Then, multiply the slide dimensions by the magnification:\(24 \text{ mm} \times 60 = 1440\text{ mm} = 1.44 \text{ m}\) and \(36 \text{ mm} \times 60 = 2160\text{ mm} = 2.16 \text{ m}\).

Determining Minimum Screen Size

The screen must be able to accommodate an image of size 1.44 meters by 2.16 meters.

Key concepts

Lens Formula

The lens formula is an essential tool in optics that helps us find the relationship between the focal length, the object distance, and the image distance. This formula is expressed as \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where:

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

To solve problems using the lens formula, you substitute the known values of \( v \) and \( u \) into the equation. Then, solve for \( f \) to determine the required lens focal length. This formula is applicable for both convex and concave lenses, with sign conventions determining the nature of the image formed. Remember to keep the units consistent for accurate calculations. For instance, in our example, we converted all distances to centimeters before substitution.

Focal Length Calculation

The focal length calculation is a straightforward process derived from the lens formula. Once you have the values for the image distance \( v \) and the object distance \( u \), calculating the focal length \( f \) requires substituting these values into \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). In our specific problem, based on the given object and image distances of 15.0 cm and 900 cm respectively:

• First, calculate the reciprocals: \( \frac{1}{900} = 0.00111 \) and \( \frac{1}{15} = 0.0667 \).
• Add these reciprocals: \( 0.0667 + 0.00111 = 0.06781 \).
• Finally, invert this sum to find the focal length: \( f = \frac{1}{0.06781} \approx 14.75 \text{ cm} \).

This method provides a practical approach to determining the focal length necessary to project an image precisely on a screen set at a specific distance.

Magnification Calculation

Magnification is a key concept that measures how much larger or smaller the image appears compared to the object. It's calculated using the formula \( M = \frac{v}{u} \), where:

• \( M \) is the magnification factor.
• \( v \) is the image distance from the lens.
• \( u \) is the object distance from the lens.

For our exercise, substitute the image and object distances to get \( M = \frac{900}{15} = 60 \). This means the image is magnified 60 times the size of the original slide. Magnification can be positive (upright image) or negative (inverted image). Here, it translates the small slide dimensions of 24 mm by 36 mm into a much larger image on the screen.

Projected Image Dimensions

The projected image dimensions are determined by applying the calculated magnification to the original object size. This gives an idea of how large the image will appear when projected. Given the magnification factor of 60 from our exercise:

• For the width: \( 24 \text{ mm} \times 60 = 1440 \text{ mm} = 1.44 \text{ m} \).
• For the height: \( 36 \text{ mm} \times 60 = 2160 \text{ mm} = 2.16 \text{ m} \).

Thus, the minimum size of the screen required is 1.44 m by 2.16 m to accommodate the entire projected image. It's crucial to ensure the screen size matches this calculation so that the image can be clearly viewed without any cropping or distortion. The dimensions give insight into how our understanding of magnification directly affects real-world applications like slide projections.