A camera lens has a fixed focal length of magnitude \(50.0 \mathrm{mm} .\) The camera is focused on a 1.0 -m-tall child who is standing \(3.0 \mathrm{m}\) from the lens. (a) Should the image formed be real or virtual? Why? (b) Is the lens converging or diverging? Why? (c) What is the distance from the lens to the film? (d) How tall is the image on the film? (e) To focus the camera, the lens is moved away from or closer to the film. What is the total distance the lens must be able to move if the camera can take clear pictures of objects at distances anywhere from \(1.00 \mathrm{m}\) to infinity?

To determine if the image is real or virtual, we will use the lens formula, which is given by: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Where f is the focal length, \(d_o\) is the distance of the object from the lens, and \(d_i\) is the distance of the image from the lens. We are given \(d_o = 3.0 \mathrm{m}\) and f = 50.0 mm = 0.05 m. Plugging in the values, we can solve for \(d_i\). \(\frac{1}{0.05} = \frac{1}{3.0} + \frac{1}{d_i}\) Now, solve for \(d_i\): \(d_i = \frac{1}{\frac{1}{0.05} - \frac{1}{3}} \approx 0.0476 \mathrm{m}\) Since \(d_i\) is positive, the image is real (a virtual image would have a negative \(d_i\)).

A converging lens (convex) forms a real image, while a diverging lens (concave) forms a virtual image. Since we found that the image is real in part (a), the lens must be converging (convex).

The distance from the lens to the film is the same as the image distance, which we calculated in part (a) as \(d_i \approx 0.0476 \mathrm{m}\).

To find the height of the image, we'll use the magnification formula: \(M = \frac{h_i}{h_o} = \frac{d_i}{d_o}\) Where \(h_i\) is the height of the image, \(h_o\) is the height of the object, \(d_i\) is the image distance, and \(d_o\) is the object distance. We already know \(d_i \approx 0.0476 \mathrm{m}\), \(d_o = 3.0 \mathrm{m}\), and \(h_o = 1.0 \mathrm{m}\). Plugging in these values, we can solve for \(h_i\): \(h_i = M \cdot h_o = \frac{0.0476}{3.0} \cdot 1.0 \approx 0.0159 \mathrm{m}\) The height of the image on the film is approximately 0.0159 m.

To calculate the total distance the lens must move, consider the minimum and maximum focused distances: When the camera can take clear pictures of objects at the minimum distance (1.00 m), \(d_o = 1.00 \mathrm{m}\), and we can calculate the minimum image distance \(d_{i \min}\) using the lens formula. Plug in the values and solve for \(d_{i \min}\): \(d_{i \min} = \frac{1}{\frac{1}{0.05} - \frac{1}{1}} \approx 0.0667 \mathrm{m}\) For objects at an infinite distance, the image distance will be equal to the focal length: \(d_{i \max} = f = 0.0500 \mathrm{m}\) Now, calculate the total distance the lens must move: Total Distance = \(d_{i \max} - d_{i \min} = 0.0667 - 0.0500 \approx 0.0167 \mathrm{m}\) The lens must move approximately 0.0167 m in total to be able to take clear pictures of objects at distances from 1.00 m to infinity.