Question

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.

Short answer

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

Step-by-step solution

List the known values

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)

Find the minimum object distance in portrait orientation

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: \(p_{portrait} = 1949.5mm\)

Find the minimum object distance in landscape orientation

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: \(p_{landscape} = 2924.2mm\)

Report the minimum object distances in both orientations

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.