How tall is his image on the detector

Join Study Groups. Create your own study plan. Join live cram sessions. Live student success coach. Cornell University. Farnaz M. Simon Fraser University. Zachary M. Hope College. Meghan M. McMaster University. Physics Mechanics Bootcamp Lectures Math Review - Intro In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true.

Algebra - Example 1 In mathematics, algebra is one of the broad parts of mathematics, together with number theory, geometry and analysis. Recommended Videos Suppose your A classic 35 -mm film came…. A simple single-lens camer…. The problem wants to know how high is the image of the men in this camera in this detector.

So how we going to describe this? Let's remember that exists that equation to calculate they'd actually correlate the distance with the height often object. So this is the magnification and the magnification is the negative distance off the image divided by the distance off the object, which is also equal the height off the image the size of the image divided by the size of the object. So that's the equation we go into use to find the size off the image.

So let's isolate the size of the image here. Well, the size of the image is just negative. D I the oh de multiplies the size the height off the man. Uh, I forgot one information. We have the height of the man. He has 2 m of height. Okay, eso let's look to the equation.

We have, uh, h zero. We have t zero, but we do not have the distance off the image. And that's probably the most physical thing you need to know to solve this problem because think about that. The distance off the distance between the man and the camera is 10 m, and this is much larger than the focal length off the camera. Therefore, we can presume we can assume that the distance off the image is being formed in the focal point off this camera. So we can assume that because because the distance off the object is much larger, then the focal length we can assume that the distance off the image is going to be exactly the focal land for the camera.

Therefore, we can write this equation here minus negative. Because this is millimeters divided by 10 de multiplies to So we can cross this here. Therefore the height of the image The size of the image is negative one times 10 to the minus two meters.

Well, we can simplify this and say that the size of the image is negative. One ST Emitter. The negative sign in here Onley represents that the image is inverted. So this is an inverted image.

So the final answer is actually the absolute value off the height. His image appears on a detector that is 50 mm behind the lens. How tall is his image on the detector? An object is The lens forms an image Part A What is the focal length of the lens?

Part B Is the lens converging or diverging? Part C If the object is 8. Part D Is it erect or inverted? The image formed by a convex lens is 5. The image is If the object is 2. A camera creates a real image of a tree 40 m away. The image is formed 3. Find the magnification. Ruff, the 50 cm tall Labrador Retriever stands 3 m from a plane mirror and looks at his image. What is Ruffs image position and height? Part AWhat is the focal length of the lens? Part BIs the lens converging or diverging?

Part CIf the object is 8. Part DIs it erect or inverted? An object is placed A converging lens of focal length 20 cm is used to form a real image 1. How far from the lens is the object?

A camera with a The focal length of a converging lens is 0. An object is place 1. The distance between the lens and the image is? Suppose you place an object 8 cm in front of a converging lens and the image appears 16 cm on the other side of the lens.

What is the focal length of the lens? A candle burns 1.