A photographer offers his camera, who lens has a \$50 \mathrmmm\$ focal distance length, to emphasis on an object \$2.0 \mathrmm\$ away. The then desires to take it a photo of things that is \$40 \mathrmcm\$ away. How far, and in i beg your pardon direction, should the lens move to focus on this second object?

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Okay, so this problem, we have actually a photographer making use of his camera to focus on object the is 2 m away. So let's put right here the distance off this object that is true meters far from the camera. Then he turns. He turns his camera to an additional object due to the fact that he desires to take it a snapshot off this 2nd object and also this object iss at a distance that's dubbed the zero Prime. Because there's an additional distance. It's a various distance, which is 40 centimeters. Okay. And also the difficulty wants to know just how far and also in i beg your pardon direction that he must. That must move the lens to emphasis on the 2nd object. Okay, so again, come simplify, we have actually this photographer responding his camera to the first object here. Then he wants to take it a photo off this second object here. Therefore the focal the focus off their lands need to change. And also he wants to know in which direction and how far this adjust must happen. So an initial of all, the focal size off the camera, this is, ah, amount that we understand which is 50 millimeters. Okay, for this reason let's remember the there is a equation for lens in the camera, i m sorry is one separated by F that is same ops. This is no a whole. Over there is equal one split by the street off the image, plus one separated by the distance off the object. So this is, uh, typical equation come the physics that lens, and we desire to know exactly how much the distance off the picture will change. Because think around that once we focus on the very first object, we have a deep Why the street off the very first image? Then once we focus on the second in the second object, we have a different d y d y prime. Okay, so the change between these 2 distance below is what we desire thio or prize in this problem. Therefore, we must discover what is the very first distance turn off the image. And also we're walking to find that using the lens equation. So the first, the I, it's going to be one split by 50 millimeters, which is 50 times 10 to the minus three, which is same one divided by 2 plus one split by D. I. Okay, so as well is the street off the object 2 m. Therefore, we have the right to say that one divided by T I, uh, is 10 come the power of three split by 50 miners, 0.5, i beg your pardon is one fifty percent and this is equal 19 0.5. Therefore, we deserve to say that the ns ISS one split by 19.5. Okay, what about the second distance off the 2nd image? for this reason let's put below again. We have one separated by 50 time 10 to the minor street since it's the focal length lunch. Uh, currently we have actually one separated by 0.4, i m sorry is 40 centimeters add to one divided by the i prime. This after we calculate is going to it is in one divided by D prime. It's basically 17.5. Therefore, we have actually the street off the 2nd image one divided by 17 0.5 meters. Okay, this is meters. This is also meters. Yeah, well and what us want, we desire to recognize in which direction and also how much we need to adjust the focal lens. OK, therefore, this is walking to it is in d the adjust in the distance. Sorry. Let's rewrite this d an altering the street is walk to be the absolute worth off the i prime miners. D I. Therefore, this is one divided by 17.5 minus one, separated by 19.5 in the absolute value. And also calculating this, we discover that this change and because of this in the lens need to be same five suggest 80 6 millimeters. Okay, so that's the readjust off the lens. And also in i m sorry direction? Well, far from the detector. Okay, so imagine that we have a camera here. This is the lens. This is the detector. For this reason the lands should relocate in this direction here, away from the detector. For this reason that's the prize is publication here our way from the detector.

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And also that's the final answer. Many thanks for watching.