A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0-cm lens being used as the objective. Both the object being viewed and the final are at infinity. (b) Find the height of the formed by the objective of a building 60.0 m tall, 3.00 km away.

Video transcript

Hi everyone. In this practice problem, we're being asked to calculate the height of an image produced by the objective lens of a telescope. We have a refracting telescope used for touristic purposes consisting of an objective and an eye piece. The objective is a convergent lens of focal length, 1.1 m and the eye piece is a convergent lens with a focal length of 0.18 m. If an object is placed very far away from the telescope, the virtual image produced is going to be at infinity. We have a tourist looking through the telescope discovering the Statue of Liberty where the statue stands 93 m tall and is located five kilometers away from the telescope. We're being asked to calculate the height of the image produced by the objective and the options given are a one point oh three centimeter B two point oh five centimeter C 22 centimeter and lastly D 42.3 centimeter. So in order for us to uh solve this problem, we want to recall that the image produced by the objective is obtained using the object image lens equation for the convergent lens of L one. In this case, the object distance lens equation gives us one over S or one divided by S one plus one, divided by S one prime equals to one divided by F one S one is the distance of the object S one prime is the distance of the image. And F one is the focal length. It is given in the problem statement that S one or the uh distance of the object is five kilometers which is the distance of the Statue of Liberty from the telescope. So five kilometer, if we convert that into meter, we will then get 5000 m. It is also given in the problem statement that the focal length or F one for the objective is going to be 1.1 m. So now we want to actually substitute this information into our equation here in order for us to determine as one prime or the uh image distance from the telescope. So let's substitute that in. So we have one divided by S one. So one divided by 5000 m plus one, divided by S one prime equals one divided by 1. m. We want to notice in this case that one divided by 5000 is going to be a lot smaller than one divided one by 1.1. And because of this, in this case, we can determine that S one prime will be equal to the focal length or F one which will then be equals to 1. m. Next, we want to uh determine the linear magnification factor or M which is given by M equals to negative S prime divided by S let's substitute this values and we have s one prime earlier decided to be 1.1 m and S to be 5000 m. So calculating this, we will get our uh linear magnification factor or M to be negative 2.2 times 10 to the power of negative four. Finally, we can actually calculate the actual height of the image produced by the objective where the size of the image obtained or absolute value of Y prime can be calculated by multiplying M or the absolute value of M with the absolute value of Y. So M we obtained previously to be negative 2.2 times 10 to the power of negative four and the Y is obtained or is given in the problem statement which is 93 m. So now we can calculate that to get our Y prime or the height of the image produced where in this case, that will come out to a value of 0.2 oh five m or converting that into centimeter. We want to multiply that by 100 so 0. oh five times 100 centimeter. And in that case, that will give us a Y prime value of two point oh five centimeter. So finally, the height of the image produced by the objective is going to be two point oh five centimeter which will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem. And that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics that's available on our website. But other than that, that'll be it for this one. Thank you.