# Video: Image Produced Using a Thin Lens

A camera with a 100 mm-focal length lens is used to photograph the Sun. What is the height of the sun’s image at the camera’s photosensor? Use a value of 1.40 × 10⁶ km for the diameter of the Sun and use a value of 1.50 × 10⁸ km for the distance to the Sun.

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### Video Transcript

A camera with a 100-millimetre focal length lens is used to photograph the Sun. What is the height of the sun’s image at the camera’s photosensor? Use a value of 1.40 times 10 to the sixth kilometres for the diameter of the Sun and use a value of 1.50 times 10 to the eighth kilometres for the distance to the Sun.

We’re told the focal length of the camera lens is 100 millimetres. Let’s call that 𝑓. We’re also told that the diameter of the Sun is 1.40 times 10 to the sixth kilometres, which we’ll call ℎ sub 𝑜 for the height of the object. The distance between the Sun and the camera is 1.50 times 10 to the eighth kilometres. We will call that 𝑑 sub 𝑜, the object distance. We want to solve for the height of the Sun’s image on the photosensor. We will call that ℎ sub 𝑖.

Let’s start our solution by drawing a diagram of this scenario. In this situation, we have the lens of our camera taking in light from the Sun at distance 𝑑 sub 𝑜 away from the lens and image forms on the camera’s photosensor. And we want to know what the height of that image, ℎ sub 𝑖, is.

To figure out this image height, we’ll use two mathematical relationships. The first is sometimes called the lens equation. And it states that one over the lens focal length is equal to the inverse of the object distance plus the inverse of the image distance. The second mathematical relationship we’ll rely on to solve for ℎ sub 𝑖 is the lens magnification equation. This relationship says that magnification 𝑀 equals image height divided by object height, which is also equal to the negative of the image distance divided by the object distance.

If we apply this magnification relationship to our scenario, we see that the image height is equal to the object height times negative the image distance divided by object distance. We’ve been given the object height and distance in the problem statement. Now, let’s use the lens equation to solve for the missing piece 𝑑 sub 𝑖, the image distance.

We can mathematically rearrange the lens equation to solve for image distance 𝑑 sub 𝑖. It equals the product of the focal length and the object distance divided by the object distance minus the focal length. We can plug in for both those values — focal length and object distance — from the information we’ve been given.

When we enter these values, we’re careful to make sure they’re in units of metres. The focal length in metres is 0.100 and the object distance in metres is 1.50 times 10 to the 11th. Altogether, these values give us an image distance of 0.100 metres, the same as the focal length of the lens.

Now that we know 𝑑 sub 𝑖, we’re ready to solve for ℎ sub 𝑖, the height of the image formed. Once again being careful to plug in numbers with units of metres, we find that when we combine these numbers, we find an image height of negative 0.933 millimetres. That’s the spatial extent of the inverted image as it appears on the photosensor.