Suppose your \(50.0 \mathrm{mm}\) -focal length camera lens is \(51.0 \mathrm{mm}\) away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is \(2.00 \mathrm{cm}\) high?

Optical magnification is a crucial concept in understanding how lenses create images of objects. It refers to the ratio of the size of the image produced by the lens to the actual size of the object. In terms of equations, it is expressed as:

\[\begin{equation}M = \frac{h_i}{h_o} = \frac{d_i}{d_o}\end{equation}\]
where:

• M is the magnification,
• \(h_i\) is the height of the image,
• \(h_o\) is the height of the object,
• \(d_i\) is the image distance from the lens,
• and \(d_o\) is the object distance from the lens.

If the magnification is greater than one, the image is larger than the object; if it is less than one, the image is smaller. A magnification of negative value indicates that the image is inverted relative to the object. Understanding this relationship can help one determine how large an object will appear when viewed through a lens, which is exactly what photographers consider when focusing their cameras.

In the example provided, by applying the magnification formula, we conclude that a 2.00 cm high image corresponds to a 1.00 m high object when projected by a lens 51.0 mm away from the film. Thus, the magnification in this case would be \[\begin{equation}M = \frac{2.00\text{cm}}{100\text{cm}} = 0.02,\end{equation}\]indicating a much smaller, demagnified image.

The focal length of a lens is a key determinant of its optical power. It is the distance from the lens to the focal point, where parallel rays of light converge after passing through the lens. In terms of visualization, consider a lens focusing sunlight onto a piece of paper; the distance at which the light forms the sharpest point of light on the paper is the focal length.

In the step-by-step solution provided, the focal length (f) for the camera lens is 50.0 mm. This is a relatively short focal length, indicating that the lens can focus on objects that are close to it, which is typical for standard photography. Lenses with shorter focal lengths have a wider field of view and a greater depth of field, hence are useful for capturing more of a scene.

• A shorter focal length leads to wider shots,
• while a longer focal length provides a narrower but zoomed-in view.

Focal length does not only determine the depth of field but also affects the magnification and the size of the image. In the thin lens formula\[\begin{equation}\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i},\end{equation}\]a shorter focal length can result in a larger image for a given object distance, which could be leveraged creatively in various fields such as photography and cinematography.

The object-image distance relationship is fundamental in optics and describes how the distances of the object and its image from a lens are interrelated. This relationship is mathematically defined by the thin lens equation:

\[\begin{equation}\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i},\end{equation}\]
where:

• \(f\) represents the focal length of the lens,
• \(d_o\) is the object distance from the lens, and
• \(d_i\) is the image distance from the lens.

From the equation, we see that if we know any two of these variables, we can solve for the third. This equation is essential for predicting where an image formed by a lens will appear relative to the lens and the object.
In the exercise, we're given that the image distance (\(d_i\)) is 51.0 mm from the lens, which, along with the focal length (\(f\)), allows us to calculate that the object (\(d_o\)) must be located 2.55 m away for it to be in focus. This relationship is at the heart of lens design and is pivotal in applications involving cameras, glasses, microscopes, and telescopes. It helps in calculating the necessary distances to obtain clear and sharp images.