A converging lens with a focal length of 9.00 cm forms an of a 4...

Question

A converging lens with a focal length of 9.00 cm forms an  of a 4.00-mm-tall real object that is to the left of the lens. The  is 1.30 cm tall and erect. Where are the object and  located? Is the  real or virtual?

Accepted Answer

Hi everyone in this practice problem, we're being asked to determine how far away from the lens an image and an object is located and whether the image is erect or inverted, we'll have a highly polished double convex lens with 100 30 millimeter focal length and a 44 millimeter diameter forming an image of an object placed in front of it. The image formed by the convex lens is 2.1 centimeter tall and erect and the height of the object is 15 millimeter. We're being asked to find out how far away from the lens is the image and the object located and whether the image is erect or inverted. The options given are a S O equals 2.7 centimeter. I equals negative 7.9 centimeter and image is inverted. BS O equals 3.7 centimeter as I equals negative 5. centimeter and image is erect cso equals negative 1.6 centimeter as I equals 5.4 centimeter. And image is inverted and lastly, DS O equals negative 8.4 centimeter as I equals 8. centimeter and image is erect. So given in the problem statement, the height of the object which is going to be equals to Y O is equals to 15 millimeter which in this case or 15 millimeter will be equals to 1.5 centimeter. The height of the image is Y I which is given to be 2. centimeter. The focal length of the lens is given to be F equals to 100 and millimeter which in centimeter that will correspond to 13 centimeter. As we know the object image relationship can be obtained or given by the formula of one divided by S O plus one divided by SI equals to one divided by F. And the magnification or M will be equals to Y I divided by Y O or at the same time will be equal to negative SI divided by S O. So in this case, Y I divided by Y O will be equal to negative SI divided by S O and to obtain SI, we can rearrange this formula to then get negative S O multiplied by Y I divided by Y O just like. So we want to plug in the values. So we get a relationship between SI and S O. So SI will then be equals to negative Y I divided Y by A Y O which is going to be 2.1 centimeter divided by a 1.5 centimeter multiplied by S O. Calculating this si will be equals to negative 1.4 of S O. So applying the thin lens formula and plugging in this given values, we can then substitute si into our object image relationship here. So we will get one divided by S O plus one, divided by si will be substituted with negative 1.4 S O equals to one point F. And in this case, the F is 13 centimeters. So let's actually substitute that end. So you'll get one divided by S O plus one, divided by negative 1.4 S O equals to one divided by 13 centimeter. So I'm gonna pluck the one uh divided by S O out of this formula. So we get one divided by S O multiplied by, in parentheses, one divided by negative 1.4 plus one equals to one divided by 13 centimeter. Calculating this uh expression in parentheses, we will then get one divided by S O multiplied by 0.286 equals to one divided by 13 centimeter. Doing further calculation, we will then get S O to be equals to 3.7 C meter. So now we want to uh pluck this back into our uh relationship between SI and S O to get SI so that si will then be equal to negative 1.4 multiplied by 3.7 cent meter. And that will give us SI to then be equal to negative 5. cent the meter. So the sign of the SI B negative here actually correspond or indicate that the image is going to be erect. So there we have it S O will be equal to 3.7 centimeter. Si will be equal to negative 5.2 centimeter or the distance is essentially 5.2 centimeter and the image is going to be erect because SI is uh smaller than zero or negative. So in this case, the option uh B will actually correspond to the answer to this practice problem and 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.

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A converging lens with a focal length of 9.00 cm forms an of a 4.00-mm-tall real object that is to the left of the lens. The is 1.30 cm tall and erect. Where are the object and located? Is the real or virtual?

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