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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.

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 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.

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.