The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.

In this problem, we know the following: - Frame dimensions: 24mm x 36mm - Focal length (f): 50mm - Height of person (h): 182cm = 1820mm (converted to mm)

In this case, the height of the frame is 36mm and we need to fit the person in that height. We can use the magnification formula: \(M = \frac{h'}{h} = \frac{q}{p}\) We know that height of person (h) is 1820mm and the maximum height in the frame (h') is 36mm. Then, \(M = \frac{36}{1820} = \frac{q}{p}\) Now, we will use the lens formula: \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\) Substitute the magnification formula into the lens formula and solve for p: \(\frac{1}{50} = \frac{1}{p} + \frac{p}{36p}\) First we will calculate the value of q: \(q = Mp = \frac{36}{1820}p\) Now substitute this value in the lens formula: \(\frac{1}{50} = \frac{1}{p} + \frac{1}{\frac{36}{1820}p}\) Solving for p, we get: \(p_{portrait} = 1949.5mm\)

In this case, the height of the frame is 24mm, and we need to fit the person in that height. We can use the magnification formula again: \(M = \frac{h'}{h} = \frac{q}{p}\) We know that height of person (h) is 1820mm and the maximum height in the frame (h') is 24mm. Then, \(M = \frac{24}{1820} = \frac{q}{p}\) Using the same method as in Step 2, calculate the value of q: \(q = Mp = \frac{24}{1820}p\) Now substitute this value in the lens formula: \(\frac{1}{50} = \frac{1}{p} + \frac{1}{\frac{24}{1820}p}\) Solving for p, we get: \(p_{landscape} = 2924.2mm\)

The minimum distance for the person to stand from the camera so that their image fits on the film in portrait orientation is 1949.5mm, and in landscape orientation is 2924.2mm.