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When light rays hit a plane mirror, they reflect off in a way that allows us to see an image of the object in front of the mirror. A plane mirror reflection has several key characteristics:

• The image is the same size as the object.
• It appears to be the same distance behind the mirror as the object is in front of it.
• The image is virtual, meaning it cannot be captured on a screen, as opposed to a real image.
• The image is laterally inverted, or flipped from left to right.

In our photographer's scenario, she is standing 0.750 meters in front of the mirror, so her image appears 0.750 meters behind it. The total distance her camera 'sees' is therefore 0.750 meters + 0.750 meters = 1.5 meters away. Describing this path helps us understand how light interacts with mirrors to create images, setting the stage for applying other concepts like lens formulas.

The lens formula is a critical tool in optics, used to find the relationship between the object distance \(d_o\), the image distance \(d_i\), and the lens's focal length \(f\). The formula is expressed as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This relationship holds true for thin lenses and helps us determine unknown distances when two are given. In the exercise, our focus is on determining the object distance \(d_o\), knowing that the image distance \(d_i\) is 1.5 meters (accounting for the mirror reflection), and the focal length is 19.5 mm.

• First, convert the focal length to meters: \ f = 0.0195 \, \text{m} \.
• Plug the values into the lens formula and rearrange to solve for \(\frac{1}{d_o}\).

Solving gives us an object distance \(d_o\) of approximately 0.0198 m. This provides consistency in determining positions relative to the lens and the light sensors.

Magnification is how much larger or smaller an image appears compared to the object itself. It relates the height of the image to the object and the distances calculated via the lens formula. The magnification \(m\) is given by: \[ m = \frac{d_i}{d_o} \] Where \(d_i\) is the image distance and \(d_o\) is the object distance. In this exercise, we substitute the values: \(m = \frac{1.5}{0.0198}\).

• Calculate to find \(m \approx 75.76\).
• This means the image appears about 75.76 times larger than the object.

Think of magnification as the lens's ability to enlarge the view, which is particularly crucial in photography to capture details clearly and correctly.

The process of forming an image using a lens involves various optical principles, all of which combine to create the final image our eye or sensor perceives. Image formation depends on factors like lens type (convex or concave), distances involved, and bends light, converging or diverging it to form images.

• Real images can be projected on a screen, formed by converging light rays.
• Virtual images, like those in mirrors, can't be captured on screens; they appear where light doesn't actually reach.

In the exercise, once we know the image's magnification and understand the lens formula, we can calculate the image's height on the camera's sensor. Using: \[ h_i = m \times h_o \] Where \(h_o = 0.08 \, \text{m}\), calculate the image height \(h_i\): \(h_i = 75.76 \times 0.08 = 6.0608 \, \text{m}\).

• This result tells us the camera captures a much larger representation of the object, due to the magnifying effects of the lens system.

This calculus gives insight into how cameras function, preparing images suitable for our visual consumption.