An object is 16.0 cm to the left of a lens. The lens forms an 36.0 cm to the right of the lens. (b) If the object is 8.00 mm tall, how tall is the ? Is it erect or inverted?

Video transcript

Hi everyone. In this practice problem, we are being asked to determine the height of an image and whether the image will be inverted or non inverted relative to the fly. We have a one centimeter height fly placed 10 centimeter away to the left of a lens. And on the other side of the lens, the image of the fly is going to be observed 15 centimeter away from the lens. We're being asked to determine the height of the observed image and whether the image is inverted or non inverted relative to the fly. The options given are a height being 0.7 centimeter and image is non inverted B height being 1.5 centimeter and image is inverted C height being 1.7 centimeter and the image is not inverted. And D the height of the image being 2.5 centimeter and the image being inverted. So we want to recall that the object image relationship for a lens can be calculated by the formula of one divided by S plus one, divided by S prime equals to one divided by F S represent the, represents the object distance which in this case, is given to be 10 centimeters S prime represents the um image distance which in this case is 15 centimeter and F is going to be representing the focal length of the lens. So in this case, F is unknown. So let's actually um calculate that first. So rearranging the object image relationship for a lens, we will then get the F to then be equal to S multiplied by S prime divided by S plus S prime. So let's substitute the values in we have 10 centimeter for S multiplied by 15 centimeter for S prime divided by 10 centimeter plus 15 centimeter. And then doing this calculation, we will get the focal length value of 6.0 centimeter just like. So, so in this case, we know that F is going to be bigger than zero, which means that our lens is going to be converging. So the next thing that we want to actually look for is the magnification factor which will determine the height of the image. So the magnification factor we want to recall can be calculated by taking the negative of S prime divided by S. So let's substitute those values. We will have negative 15 centimeter divided by 10 centimeter which will give us the magnification of negative 1.5 centimeter. So the size of the image will then be equals to uh Y prime, absolute value equals to the absolute value of M multiplied by Y. So that will give us uh absolute value of 1.5. Sorry, the magnification should not have centimeter here and it should just be a number of negative 1.5 because the centimeters will be crossed out. And then from there, we will have 1.5 uh the absolute value of one negative 1. multiplied by a one which is A Y which is 1. centimeter. And that will give us a value of 1.5 centimeter for the height of the image. So the size of the flight image is going to be 1.5 centimeter. And since the magnification factor here is negative, so the image is going to then be inverted. So we will then have the answer to then be equal to option B where the height of the image is 1.5 centimeter and the image is infer. 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 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.