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The focal length of a lens is a crucial concept in optics. It indicates how strongly the lens converges or diverges light. Mathematically, it's the distance between the lens and the sensor (or film) when the lens is focused on a distant object. A shorter focal length means the lens bends light more sharply, creating a wider field of view. Conversely, a longer focal length provides a narrower view.
In a practical scenario, like with a movie camera, the focal length determines the scope of view and the magnification of the image. In our problem, the camera's lens has a focal length of 75 mm (or 0.075 m when converted). This relatively short focal length suggests the camera's ability to capture a wide scene, making it perfect for close-to-medium range filming. Understanding this parameter is key to mastering how lenses influence image formation.

Object distance refers to the distance between the lens and the object being photographed or filmed. In optical systems, this is an essential determinant of the image's properties. The object distance (\(d_o\)) along with the focal length, helps in determining where the image will be formed.
In the given problem, the person stands 27 meters away from the camera. This large distance indicates that the person is relatively far from the camera's lens. Before applying the lens formula, it's vital to ensure consistent units for both focal length and object distance. Proper conversion from meters to say, meters, aligns with the metric system used in the focal length. Knowing this distance allows us to utilize formulas to find where the image will form behind the lens.

Image distance is the length at where the image of the object is formed from the lens. It is denoted as \(d_i\) in the lens formula.
The relationship between image distance and object distance is governed by the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). This equation helps us calculate the image distance when the object distance and focal length are known.
In our exercise, substituting the values for \(f = 0.075\) m and \( d_o = 27 \) m into the lens formula helps us compute \(d_i\). Solving it gives us an image distance \(d_i \) of about 0.0753 m. This tells us how far behind the lens the film would need to be positioned to capture a focused image of the person.

The magnification of an image is a key metric and is described by the magnification formula \(m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\). Here, \(h_i\) is the height of the image, while \(h_o\) is the height of the object. On the distance side, it similarly compares the image distance (\(d_i\)) to the object distance (\(d_o\)).
Using this formula, we can find out how much larger or smaller the imaged object will appear. For this problem, with a calculated image distance of 0.0753 meters and an object distance of 27 meters, we find a tiny magnification value of approximately 0.0027889.
When this magnification is applied to the object height (1.8 meters), it results in a small image height of about 5.02 mm. This calculation illustrates how objects that are farther away are represented quite smaller in size on the film, an essential concept in visualizing scale and detail in photography.