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Bxbonru tRaitadiverging E miror has from the mirTor focal lcngth of 0,12 m Showing _ The Ituge fored is located 0.070 m your caleulations, determine where the objec...

Question

Bxbonru tRaitadiverging E miror has from the mirTor focal lcngth of 0,12 m Showing _ The Ituge fored is located 0.070 m your caleulations, determine where the object Is locatcd05.0 cm tall object [S cm away from converging mittor The image is A student places 12 cm from the mirror Showing our calculations , determine the height and orientation ofthe image.A 1,S cm tall object is placed [2 cm awaY from # convcrging mirror The image IS magnified t0 four times its Original hcight and inverted;

Bxbonru tRaita diverging E miror has from the mirTor focal lcngth of 0,12 m Showing _ The Ituge fored is located 0.070 m your caleulations, determine where the object Is locatcd 0 5.0 cm tall object [S cm away from converging mittor The image is A student places 12 cm from the mirror Showing our calculations , determine the height and orientation ofthe image. A 1,S cm tall object is placed [2 cm awaY from # convcrging mirror The image IS magnified t0 four times its Original hcight and inverted; Showing your calculations. detenine the mirror'$ focal length



Answers

An observer to the right of the mirror–lens combination shown in Figure P23.62 sees two real images that are the same size and in the same location. One image is upright, and the other is inverted. Both images are 1.50 times larger than the object. The lens has a focal length of 10.0 cm. The lens and
mirror are separated by 40.0 cm. Determine the focal length of the mirror. (Don’t assume the figure is drawn to scale.)

Hi. In the given problem, we are having a combination off on the plans Means a converging lands men. A concave mirror means up converging near like this. This is a combination. No, an object has been kept here. At some point, the total care between the lens and the mirror has been given us 40 centimeter. So first of all this lens reform are really and inverted image magnified image off this object. Let the object be a B. So the image may be present it life. You dish big fish. This is the distance. Well object for this. Lance, let the distance be X in magnitude. So for the lens a local and has been given us then went zero centimeter. And the linear magnification here has been given us. M one is equal to 1.5 times as it is given that the image is 1.5 times larger than the size off the object and as the images in workers. So we can say this is minus one wine white and we know in age is this union magnification is actually no issue off the distance off image. Let it be you. One means this, this is the distance will finish duda Distance off the object, which will be minus X negative. Sign has the news as sports sign convention. So this stays minus on 15 Israel took you one by minus X. So finally we can say Cuban is nothing but 1.5 times off x. So if we use painless equation one by Cuban minus one by Stephen is equal to one by f. L the opulent of plants putting the known values here. This is one by 1.5 times old ex or Cuban miners. One by minus. X for Beaven is equal to one by 10 the focal length off this lens. So there it becomes one Bible 10.5 times off X plus one by X equals movement by 10 or taking Elsom off. Left inside, this is 1.5 x So one last 1.5 is equal to one by 10. Or we can say this is true. 100.5 times off. One wine Fine, fine. Divided by coupon five divided by 1.5 times off. X equals proven by 10. Making a cross multiplication 1.5 x is equal to 25 or X is equal to DE at 25 by 1.5. So can slim all these things. We get the value of X to be 16.7 centimeter. So this is the position off object for this converging lens. So the distance off image it as Swedish from this lens will come out to be. Q one is equal to 1.5 times off X or we can see. Q one is equal to 1.5 times off. X rituals to 50 by 15 means 250 by 15 sore. It can be given as canceling the zeros. 15 Got cancer. So here it comes out to be 25 centimeter Finally. Now this in age a dash me dash will serve as, ah, work your object for the media And another image is also being formed at the same location which may be treated as a double dash beatable dash. So this, maybe consider as the image formed by the con que Mira So the other image other real image. A double dash beatable dash is being formed by the on gave It's very cool mirror and as this image edible dash mutable dashing is having the same size as that Offiah dash me dash, which is working as a virtual object for it and having the same locations. So the distance this distance off image a dash me dash from the pole off this s radical mirror should be equal toe excellent yourself curvature. So the radius off your preacher off the meter is equal to this distance. 40 the distance between mirror and less plus Q one. So this in magnitude this is 40 less Q one because that's what people us 25 means There's a 65 centimeter the radius of curvature because when the object is kept at the center of provincial off, are hysterical mirror the images forms that just seem position at the same center courage and having the same size. So finally, the focal and off the mirror comes out to be half offered. Serious, of course, major means 65 by two so we can see the focal lamp off. Hysterical. Nearly started do 20.5 centimeter. So this is the answer off the given problem. Thank you

Here. The image distance will come out to be one of the F minus one of our deal. Inwards off that it was one over. Forgot this negative. 40 centimeter minus one over object distance. It is virtual. So that's negative. 50 centimeters in verse. So we have immense distances. Was negative. 207 m native imply Saturdays. Watch your words.

Compute the position in focal length of the converging Cine lens, which will project the image of a lamp Magified four times upon the screen, 10 m from the lamp. Okay, So s. o. plus s. i. s. 10 cm because the screen is 10 cm from the lamp, they told us that the magnification is four times, but this is a converging lens and we know that converging lenses to get it to have a real image, it needs to have a negative magnification. So the magnifier magnification is a negative for is equal to negative S. I. Over S. O. So S. I. is equal to four times s. O. So four times s. O. We can plug this into our first equation, so S. O. Plus four times S. O. is 10 cm, which means S. O. is great. So this gives us the value of, this is five, so is equal to 10 cm, so is so is equal to two centimeters and that's I. Is equal to eight centimeters. So then we have one over F. Is equal to one over espo plus one over S. I. We know S. O. We know as I. So we can find F. & f. s. 1.6 meters

So in this question we learn about the basic concept of optics. So the given question there are two m, one middle is conversing middle and another one is diverse in middle and there's a source which is at a 30 centimeter distance from the first middle. So we can first will find the science convention. Oh so for me that one this direction would be taken as positive and for middle to this direction would be taken as positive. Right? So for the first reflection in Emelin the value of you would be minus 30 centimetre. The focal length is given as minus 20 centimeters. So therefore we can use the middle formula to find the value of. We laid. Uh huh. So then we simplify this. We get the value of three years minus 60 centimeters. Now this would act as a now this image would act as a source for the second middle. So for the second reflection, the value of you would be 60 minus 30 plus X. Which is equals to 30 minus six. Mhm. Right. He had the access this distance. That is the distance between source and middle. To the value of we would be minus X. Because the image is formed at the source and the value of focal land is 20 centimeters. So we can use the middle formula to find the value of X. Uh huh. Yeah. So this would be a quite a thick equation. And the solution would be yeah X equals to 10 plus minus 50 divided by two, which means either 20 centimeters or minor studies centimeters insects cannot be negative. So the value of X would be 20 centimeters. So the total distance between the two m would be 20 plus 30 which is equal to 50 centimeters. Right, So this is a total distance between the two mills. And in the second part we have to find the location of the image formed by the single reflection from M. Two. So that location we have already found, which is the value of X. That is 20 centimeters minus 20 right. Mhm. So the value of access minus 20 centimeter, and the image would be found at source, which is 20 centimeters similar to.


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