Esperanza uses a 35 -mm camera with a standard lens of focal...

Esperanza uses a 35 -mm camera with a standard lens of focal length $50.0 \mathrm{~mm}$ to take a photo of her son Carlos, who is $1.2 \mathrm{~m}$ tall and standing $3.0 \mathrm{~m}$ away. (a) What must be the distance between the lens and the film to get a properly focused picture? (b) What is the magnification of the image? (c) What is the height of the image of Carlos on the film?

Esperanza uses a 35 -mm camera with a standard lens of focal length $50.0 \mathrm{~mm}$ to take a photo of her son Carlos, who is $1.2 \mathrm{~m}$ tall and standing $3.0 \mathrm{~m}$ away.
(a) What must be the distance between the lens and the film to get a properly focused picture? (b) What is the magnification of the image? (c) What is the height of the image of Carlos on the film?

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Key Concepts

Thin Lens Equation

The thin lens equation (1/f = 1/d? + 1/d?) is a fundamental principle in optics that relates a lens's focal length, the distance from the lens to the object, and the distance from the lens to the image. It is essential for determining where an image will be focused on a film or sensor.

Focal Length

The focal length of a lens is the distance over which initially collimated rays are brought to a focus. It is a critical lens property that influences both the size of the resulting image and the distance from the lens at which the image is formed.

Magnification

Magnification in an optical system is the ratio of the image size to the object size, often expressed as m = d?/d?. This concept explains how much larger or smaller the image is compared to the actual object, based on the distances involved.

Object and Image Distance

Knowing the distances between the lens, the object, and the image plane is crucial for proper image formation. These distances, linked by the thin lens equation, determine not only the focus but also the scale (via magnification) and clarity of the final image.

Transcript

00:01 We're going to use two equations here.

00:05 We're going to use the thin lens equation, which is 1 over p plus 1 over q is equal to 1 over f.

00:16 In other words, 1 over the distance to the object from the lens, plus 1 over the distance to the image from the lens equals 1 over the focal length of the lens.

00:27 And we're also going to use that the magnification is equal to negative q over p.

00:37 Or q and p are the same.

00:40 So first of all, we're going to write down what information we have.

00:44 We know that focal length is 50 .0 millimeters.

00:52 And i have a height of my object of 1 .2 meters, which is the same.

01:00 Same thing as 1 ,200 millimeters.

01:04 Just going to convert to millimeters, so all of my values are the same units.

01:10 And i have a p, a distance to the object, of 3 .0 meters, which is 3 ,000 millimeters.

01:22 Now to solve for q, the first part of the question is find the distance from the lens to the image that we must have for a clear picture.

01:32 So for a clear picture, we need to use the fin lens equation.

01:36 So i'm going to say that 1 over 3 ,000 millimeters plus 1 over q is equal to 1 over 50 .0 millimeters.

01:57 Rearranging that, i'm going to get that q is equal to 50 .8.

02:10 847 millimeters or approximately 50 .8 millimeters...

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