A photographer takes a photograph of a Boeing 747 airliner...

A photographer takes a photograph of a Boeing 747 airliner (length 70.7 $\mathrm{m} )$ when it is flying directly overbead an altitude of 9.50 $\mathrm{km}$ . The lens has a focal length of 5.00 $\mathrm{m}$ . How long is the image of the airliner on the film?

Key Concepts

Focal Length

Focal length is a primary characteristic of a lens, defining the distance over which initially collimated rays of light are brought to a focus. It is an essential parameter in determining how strongly the lens converges or diverges light, influencing both the image distance and the magnification in an imaging system.

Object Distance

Object distance refers to the distance between the object being imaged and the lens. It plays a critical role in the application of the thin lens equation and in calculating the magnification, thereby affecting both the location and the size of the image formed by the lens.

Thin Lens Equation

The thin lens equation, expressed as 1/f = 1/d? + 1/d?, is a fundamental relation in geometrical optics that connects the focal length (f) of a lens with the distances from the lens to the object (d?) and to the image (d?). It is used to determine where an image will be formed when an object is placed at a given distance from the lens.

Magnification

Magnification is the ratio of the image size to the object size, typically given by -d?/d? in the context of thin lens systems. This concept helps in determining how the dimensions of the object are scaled in the produced image, and the negative sign indicates image inversion in simple lens setups.

Transcript

00:01 Consider the case of a photographer taking a picture of a plane flying overhead.

00:07 The plane is 70 .7 meters long.

00:16 We know that the focal length of the camera that the photographer is using is 5 meters, and we know that the plane is flying at an altitude of 9 .5 kilometers, or equivalently 9 .5 times 10 to the 3 meters.

00:44 We are interested in determining how long the plane will appear on the film of the photographer's camera.

00:53 To begin, consider this equation where s is the object distance, s prime is the image distance, and f is the focal length of the camera.

01:04 We know how far away the object is because we know the altitude at which the plane is flying.

01:10 So s is our 9 .5 kilometers.

01:13 We know the focal length of the camera is 5 meters, so we need to solve for s prime the distance to the image.

01:46 Rearranging our equation, we find that the image distance is equal to the inverse of the difference between 1 over the focal length minus 1 over the object distance.

01:58 We can substitute these values in.

02:16 I accidentally put the object distance where the focus.

02:21 Length should go.

02:26 So this should be five to the focal length minus one over 9 .50 times 10 to the 3 meters inverted.

02:42 So solving this we find that s prime is actually pretty close to 5 .0 meters.

02:51 So we know that the image is basically right at the focal length of the camera...

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