A converging lens with a focal length of 70.0 cm forms an of a 3.20-cm-tall real object that is to the left of the lens. The is 4.50 cm tall and inverted. Where are the object and located in relation to the lens? Is the real or virtual?

Question

A converging lens with a focal length of 70.0 cm forms an  of a 3.20-cm-tall real object that is to the left of the lens. The  is 4.50 cm tall and inverted. Where are the object and  located in relation to the lens? Is the  real or virtual?

Accepted Answer

Hi everyone. In this practice problem, we are being asked to determine the image and object sessions with respect to a lens and determine whether the image will be real or virtual. We'll have a plano convex lens forming an image of an action figure 12.4 centimeters tall and is placed to its left. The image of the action figure is inverted and has a height of 16. centimeter. The lens is flat on one side and has a focal length of 85 centimeter. We're being asked to determine the image and object positions with respect to the lens and determine whether the image will be real or virtual. The options given are A S O equals 11 27 centimeter si equals 1 18 centimeter and image is real BS O equals negative 1 55 centimeter, SI equals 1 79 centimeter and image is virtual CSO equals 1 50 centimeter, si equals 1 96 centimeter and image is real and D SS O equals 1 88 centimeter si equals negative 1 99 centimeter and image is virtual. So in the problem statement, it is given that the focal length or F is 85 centimeter. So F will be equals to 85 centimeter. The image height is given by Y I and given the problem statement to be 16.2 centimeter or in this case because the image is inverted, then Y I will be equals to negative 16. centimeter Y O. In the problem statement is given to be 12.4 centimeter and image distance from lens si is unknown and S O is also unknown, which are the two things being asked. We know that the object image relationship is given by one divided by S O plus one, divided by SI equals one divided by F. But since we do not know S O and SI, we have to create a relationship of S O and SI by utilizing the map application which can be calculated by taking Y I divided by Y O equals to negative si divided by S O rearranging this, we can then get SI to be equal to Y I divided by Y O multiplied by negative S O just like. So, so we want to plug in the values to get the um relationship between SI and S O si will then be equal to negative of negative 16.2 centimeter four Y I divided by A Y O which is 12.4 centimeter multiply it by S O. In calculating this, we will get SI to then be equal to 1.3 oh six, the two negatives will cancel itself out. So si will be 1.306 S O. So we can then substitute this into our object image distance relationship and also substitute the focal length value which is centimeter into this relationship here. So let's do that. We have one divided by S O plus one divided by SI which is 1.3 oh six S O. So one divided by 1.306 S O equals to one divided by F which is one divided by 85 centimeter. So now that we have a substitute out of this values, we can then actually take the one divided by S O out of this because it, it uh the one divided by S O as the common factor between these two. So one divided by S O multiplied by, in parentheses, one plus one divided by 1.3 oh six, close equals to one divided by 85 centimeter. And then calculating this, we will get one divided by S O multiplied by 1.766. So the in parentheses will be 1.766 equals to one divided by 85 centimeter. And then calculating this for S O, we will get, we will get S O to then be equal to 1. multiplied by 85 centimeter. And that will give us S O of 1 50 cent meter. So there we have the object distance but we can use this and plug it into the relationship that we just created to obtain the image distance, which is 1.3 oh six multiplied by S O which is going to be 1 centimeter and calculating this si will then be equals to 96 centimeter. If we notice the sign of SI is big is positive or SI is bigger than zero. So the image will be real and there we have it S O will then be equal to 1 50 centimeter si equals to 1 96 centimeter and image is going to be real since SI is positive and all of those um answers will correspond to option C in our answers. Choices and option C will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

A converging lens with a focal length of 70.0 cm forms an of a 3.20-cm-tall real object that is to the left of the lens. The is 4.50 cm tall and inverted. Where are the object and located in relation to the lens? Is the real or virtual?

Verified Solution

Key Concepts

Converging Lens

Magnification

Real and Virtual Images