Video: Focussing Light Using a Thin Lens

Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. How far away is an object that is in focus? What is the height of the object if its image is 2.00 cm high?

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Video Transcript

Suppose your 50.0-millimeter focal length camera lens is 51.0 millimeters away from the film in the camera. How far away is an object that is in focus? What is the height of the object if it’s image is 2.00 centimeters high?

Let’s start by highlighting some of the important information we’ve been given. The focal length of the camera lens we’re considering is 50.0 millimeters. We’ll call that value 𝑓. The camera lens is 51.0 millimeters away from the camera film. That’s the image distance which we’ll call 𝑑 sub 𝑖. We wanna find out how far away an object, that is in focus, is from the lens. We’ll call that distance 𝑑 sub 𝑜. In part two, we’re told that this system forms an image with a height of 2.00 centimeters. We’ll call that value ℎ sub 𝑖. Given that image height, we want to know the object height which we’ll call ℎ sub 𝑜.

Let’s begin our solution by drawing a diagram of the situation. We have a lens, a distance 𝑑 sub 𝑖 away from a screen on which an image forms. The image is of an object some distance away from the lens, 𝑑 sub 𝑜, that we want to solve for. The lens forms an image of the object on the screen, and the vertical height of that image, ℎ sub 𝑖, is 2.00 centimeters. We want to solve for the height of the object, ℎ sub 𝑜.

To start solving for 𝑑 sub 𝑜, the object distance, let’s recall a relationship sometimes called the thin lens equation or the lens maker’s equation. This relationship says that one over the focal length of a thin lens is equal to one over the object distance, 𝑑 sub 𝑜, plus one over the image distance, 𝑑 sub 𝑖. Applying this to our situation, we want to mathematically rearrange this expression to solve for object distance 𝑑 sub 𝑜. When we do that, we find that 𝑑 sub 𝑜 equals 𝑑 sub 𝑖 times 𝑓 divided by 𝑑 sub 𝑖 minus 𝑓. The focal length 𝑓 and the image distance 𝑑 sub 𝑖 are given in the problem statement, so we can plug those values in now.

When we enter these values on our calculator, we find that, to three significant figures, 𝑑 sub 𝑜 is 2.55 meters. That’s the distance the object is away from the lens.

Now we want to solve for the height of the object, ℎ sub 𝑜. To do that, we can recall the mathematical relationship for image magnification. Magnification 𝑀 equals negative 𝑑 sub 𝑖 over 𝑑 sub 𝑜 which also equals ℎ sub 𝑖 over ℎ sub 𝑜. Applying this relationship to our scenario, if we treat all values as positive, then the magnitude of negative 𝑑 sub 𝑖 over 𝑑 sub 𝑜 equals ℎ sub 𝑖 over ℎ sub 𝑜. We can rearrange this to solve for object height ℎ sub 𝑜. It equals the magnitude of negative 𝑑 sub 𝑜 over 𝑑 sub 𝑖 times ℎ sub 𝑖. Each of these three values is either given to us in the problem statement or we solved for it in part one. We can plug in these values now.

When we enter these values on our calculator, we find that the object height, ℎ sub 𝑜, is equal to 1.00 meters. That’s how tall the object in this exercise is.

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