Jim plans to take a picture of MeGraw Tower with a 35-mm...

Jim plans to take a picture of MeGraw Tower with a 35-mm camera that has a $50.0$ -mm focal length lens. A roll of $35-\mathrm{mm}$ film is $35 \mathrm{~mm}$ wide; each frame is $24 \mathrm{~mm}$ by $36 \mathrm{~mm}$. The tower has a height of $52 \mathrm{~m}$ and Jim wants a detailed close-up picture. How close to the tower should Jim be to capture the largest possible image of the entire tower on his film?

Key Concepts

Geometric Optics and Similar Triangles

Geometric relationships, particularly similar triangles, are used to relate the actual size of an object to the size of its image produced by the camera lens. This reasoning helps in scaling the object’s dimensions to the film dimensions based on the distances from the lens, which is essential for framing the subject correctly.

Film Format and Frame Constraint

The physical dimensions of the film frame set a limit on the size of the image that can be captured. This concept involves matching the projected size of the subject (after magnification) with the dimensions of the film to maximize the use of the available area, ensuring the subject is fully contained in the image.

Magnification

Magnification is defined as the ratio of the image size to the object size, and can also be expressed as the ratio of the image distance to the object distance. It is critical in determining how large the object appears on the film and is used to ensure that the entire subject fits within the frame.

Thin Lens Formula

This concept relates the focal length of a lens to the distances of the object and the image from the lens using the equation 1/f = 1/d_o + 1/d_i. It is fundamental in determining where the image forms on the film plane relative to the object, and it underpins the basic operation of cameras and optical systems.

Transcript

00:01 So first of all, i'm going to draw a little picture.

00:05 So i have the lens of my camera here.

00:11 I have, here's the midline.

00:15 I have an object here with a height of h.

00:19 In this case, h is 52 meters, which is 5 ,200 millimeters.

00:31 And we have two focal lengths for our camera here.

00:36 F and f is 50 .0 millimeters.

00:45 So i'm going to draw some light rays here.

00:56 And i'm going to end up with an image with a height of h prime here.

01:02 Now, what i want to know is what distance p do i have to be from my object in order to be able to fit all of it into my h prime, which is limited by the film of my camera? i'm going to label q here.

01:20 The film of the camera is a 35 millimeter camera.

01:25 So we have film that is 36 millimeters in width and 24 millimeters in height.

01:33 Now, the answer to this question depends heavily on whether we're allowed to turn our camera vertically or we have to use it horizontally.

01:40 So we're going to solve it both ways.

01:42 The last thing i'm going to label here is i'm going to label this side.

01:50 L.

01:52 And that's just so that i can form these two equivalent triangles here.

01:56 These angles are the same and these right angles are the same.

02:00 So that i can write the proportion f over l is equal to h over h prime, where l is just the difference between f and q and f...