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Objectcne rqureMldayat benkeen Tne3noathe Mrror Knichseparated Dy dustanceTne magnilude Dfthe mlmorsNoinscunvarureanaine ensenasengin46 2 cmLcns Object NinerConsid...

Question

Objectcne rqureMldayat benkeen Tne3noathe Mrror Knichseparated Dy dustanceTne magnilude Dfthe mlmorsNoinscunvarureanaine ensenasengin46 2 cmLcns Object NinerConsidering onlythe Ilght that leaves the ahiec and trvcls frst iowaro the mirror; locate the fina Imjqc lomcd Cnecz (ne syntar resdonse right tne Mimcrsystcmthis image realVirtual?virtualupriqhtInvetted?upriqntinyened(d) Whatover magnification?Tour (esponse allters from Ine corecc 3nchamInore (nanDomale cneck Tour c3icuiations

object cne rqure Mldayat benkeen Tne 3noathe Mrror Knich separated Dy dustance Tne magnilude Dfthe mlmorsNoins cunvarure anaine ensenas engin 46 2 cm Lcns Object Niner Considering onlythe Ilght that leaves the ahiec and trvcls frst iowaro the mirror; locate the fina Imjqc lomcd Cnecz (ne syntar resdonse right tne Mimcr systcm this image real Virtual? virtual upriqht Invetted? upriqnt inyened (d) What over magnification? Tour (esponse allters from Ine corecc 3ncham Inore (nan Domale cneck Tour c3icuiations



Answers

The object in Figure $\mathrm{P} 36.56$ is midway between the lens and the mirror. The mirror's radius of curvature is $20.0 \mathrm{cm},$ and the lens has a focal length of $-16.7 \mathrm{cm} .$ Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. Is this image real or virtual? Is it upright or inverted? What is the overall magnification?

High in the gin problem. There is, uh, gone. Give Linz in combination mint a concave spherical mirrors as shown here in the figure the distance between them has been given as to be five centimeter, and an object has been kept midway between them. This is the object in the form off an arrow. This is the lands. This is the surgical mirror. And this is the object. Or as it hasn't kept midway between the school. So most of all, the release off creature off the on cave hysterical mirror has been given us 20 centimeter. So its focal and opulent of the mirror, maybe half off the radius of curvature and using sign convention, it may become minus 20 by two or minus stand by zero centimeter and the focal land off. Confidence has been given us minus 16.7. And the media No, we have to obtain the position off image formed by this combination off mirror and less. First of all, there is will be moving do worse. The meter off, reflecting from the mirror, they will past through the lens and finally reform the image. So four the meter the object distance with this and we say it be one. So the mirror be one is half off. B means 25.0 by two and using sign convention. This is negative. Centimeter hes minus 12.5 Santa Anita The focal length off the meter. If I m it's called a minus 10 10 centimeter. So using nearer Formula One, bite me even plus one by you. You represent the position off image for this mirror. So this is Cuban. You were the one by f m. Opulent of the mirror being the values, the known values. This is one by minus well, 0.5 Last one buy you one. It's going toe one by minus then so from here, one by q one comes out to be one by 12.5, minus one by then. Hence taking an elysium one by q one comes out to be Elson will be 50 four minus five means minus one by 50. So the position off in age formed by mirror will be minus 50 centimeter and negative sign shows that it is. Yeah. So the left hand side off this combination off this mirror at a distance of 50 centimeter so as the lens itself is at a distance off 25 centimeter from the middle and image is somewhere at this point at a distance of fifties and danger. So no for lens. This really image formed by the mirror will work as a virtual object for the lens or lens image. Real image for one by this medical meter will serve. S worked your object for the lens? Yes, for the lens, the object distance will become. B two is equal to in positive because the subject will be in their direction off, but incident please. So here it would be positive. 50 minus G means we'll say 15. Minus 25 means 25 centimetre. This is the object of distance for the lens, the focal length of the lens fl is doing as minus 16.7 centimeter. So using Len situation one by U two minus one by p two is it will do one by f l. Opening the values one by few to this is missing minus one by P, to which is 25 centimeter is equal to one by 16 point, setting off minus. So rearranging bottoms, you get an expression What do you do? One bike you do is equal to one by 25 minus one by 16.7. Taking the exam again. This is 25 multiplied by 16.7 cm. So in the new military, between 16.7 minus. Then the fight so finally won by you too is equal to minus 8.3, divided by 25 time 16.7. So you do means the distance off image from this unbalance is minus 25 9, 16.7, divided by it 1 ft centimeter. So the skewed comes out to B minus 50 people on zero centimeter approximately no for party. This is party for part B. We have to find the nature of this image as it is being formed by a phone cave lance, and on producing back. So naturally, the nature off the image is a virtual and in part C. We have to find out whether it is direct or inverted with energies. In were big. Thank you

Put this question. We're told to look at figure p 23.5 to to see that the object is midway between a lens with a focal focal length. F of negative 16.7 centimeters in a mirror with the radius of curvature are 20 centimeters and they're separated by a distance D equal to 25 centimeters. And for part, A wants us to consider only the light that leaves the object and travels first towards the mirror and locate the final image formed by this system. Okay, so to do this, we're gonna use the lens equation which says that one over F which is equal to two over the radius of curvature since the mirror is the first thing it interacts with here, this is going to be equal to one over p one. The distance to the object from the mere plus one over kyu won the distance to the image formed by this. Well, P one is 12.5 centimeters, So we can rearrange this to solve for Kyu won. We see that q one is equal to ah p one times are divided by two times p one minus R so Q one comes out to equal that d centimeters. It's positive. Therefore, it's to the left of the mirror so we can use Kyu won the image produced by this as the object for the second part here. So for part B to find P to the object for the lens, P two is equal to the distance between the two minus Q one. So this comes out to equal negative 25 centimeters. Now, if you go back to this page, we can use this equation here, which I'm gonna underline. One over f is equal to one over p one plus one over kyu won. We can rearrange that then to solve for Q except for this time instead of one. Were using some script too. Right? So que tu is equal to p two times the focal length of the lens divided by P two minus the focal length of the lens playing those values and we find that que tu is equally negative. 50 0.3 centimeters. Weekend box said in his air solution for part A Oh, this wasn't part. Be sorry about that. This was all part of a here. So to race this okay, part be asked if this image that is produced is riel or virtual well, To do that, we just have to look at the sign of Q two. And since que Tu is negative, it's virtual. So if you two would have been positive, it would have been real making box in it. And this is their solution for part B. Parsi asked. If, uh, if the image that is produced is upright or if it's inverted To do that, we need to know the sign of the magnification. So magnification of thing Entire system is equal to the magnification produced by the mirror, which is Kyu won over P one, multiplied by the magnification produced by the lips so minus Q two over Key, too. And we know all those values, so we complex them in. We find that this is equal to positive 8.5 So since the magnification is positive, that means it's upright, so we can say, uh, in is positive. So it's up right? Making box. It is their solution for C in the part D asked for the value of this magnification. We already found that in part, see, so the value of the magnification M is equal to 8.5 making box set in as our solution for D in the final part of our question.

Hi there suffer this problem. We have an object that stands on the central axis of an s vertical mirror. Now the information for this problem is provided in this table. And since we are working with the Problem nine, we will have that. The object distance is equal to 1818, positive, finally 18 cm and we are told that the mirror is concave and The focal length it is also given and that is 12. Now we know that for a concave mirror and the focal length should be positive. So we will have that focal point in this case is equal to a positive value of 12 12 centimeters. Now, for historical mirrors, the focal length is related to the radius of Corbett er art by the following, we know that that is equal to the radius of corporate work over to. Now we also know that the object distance be the image distance I and the focal length are related by the following equation. One over B is equal to well plus one overt. The image distance is equal to the inverse of the focal length. Now, with that set and we know that for if the value of I is positive then we will attain a real image. And if it is negative, we will obtain a virtual image. Now we also know that the correspondent lateral magnification is defined as minus the ratio between the image and the image distance and the object distance. Um Now we also know that the value of M is positive for an upright images that is non inverted images and negative for inverted images. Now, real images are formed on the same side as the object. Well, Burton images are formed on the opposite side of the mirror. So we need to take into account that information in order to solve this problem. Now, for the part A of this problem, what we need to determine is the radius of Corvette or Now as I said from the beginning we have that the radius of Corbett Sure, it's simply that we need to solve from this. So we will have that the radius of curvature sure is equal to two times the focal length. So since we have the focal length We will have that the radius of curvature is equal to two times that value. That is positive and it is 12cm. So we will obtain that the radius of Corbett sure is positive 24 centimetres. Now for part B of this problem, we are asked um about the image distance. I now we know that that is related by the other distances from this equation. Now, solving for the image distance, I, we will obtain that that is the product between the distances P and F the focal point and the object distance over the difference between these two quantities. Now in here, we just need to simply substitute these values that we know. So we know that for The image distance and sorry the object distance we have 18th and d matures The focal length is 12 cm and in the denominator we have 18 cm -12 cm. So the image distance it is a positive value and that is 30 citizen diameters. Now for party of this problem we need to obtain the lateral magnification and as I said from the beginning this is defined as main minus the ratio between the image distance over the object distance. So in here we just need to simply substitute those two values. That is 30 citizen diameters and okay The object distance that it is 18 cm. So from this we obtain a negative value and that is -2. Now for part D of these problems, what we need to obtain is the is determined and whether the image that is formed is real or virtual. Now to answer that question we can we need to see the sign that we obtain for the image distance as you can see that that value that we obtain it is positive. So the image is real, the image is real, we can demonstrate that by capital are now for party of this problem, we are asked whether the image that is formed, it is inverted or not inverted in relation to the object. Now for that we need to see the sign of the value that we obtain for the magnification as you can see that the value is negative, so the image is inverted. Image inverted. So we can identify this by capital I and finally for the F question for this problem, we are asked whether and the image that is formed is on the same side of the mirror or on the opposite side, and a real image is formed on the same side of the object. So we can conclude that is formed on the same side as the object. Yeah, so this is a solution for this problem. Thank you.

Hi there. So for this problem we have an object that stands on the central axis of a spherical mirror. And the information for this problem is provided in this table. Now, since we are working with the problem problem 10, The object distance it is given and that is 15. The type of mirror it is a concave mirror and we are also given the value for the focal length. Now we know that for and For a concave mirror, the focal length is positive, so we will have the difficult length is positive and that it has a value of 10 cm. Now the object Distance is also positive and that is 15 cm. Now the first question for this problem is to obtain the radius of Corvette or now we know that the radius of Corbett or is equal to two times the focal length. So we just in here need to substitute that value for the focal length. We know that the focal length is positive and that is tencent amateurs. So from here we obtain a positive value of 20 centimeters Now for part B of this problem, we are asked about the image distance now to obtain that, we know that the image distance is defined as the product between the object distance times the focal length um and the difference between these two quantities now and here we just need to simply substitute the values for it. We have that both of these are positive. One of them is 15 cm for B and tents and damages for the focal length. Now this over T -10 and Sorry P -F. So that is 15 centimeters minus tents and imagers. So using our calculator, we obtain about you off positive 30 m Now for parsi of this problem we are asked for the lateral um magnification M so we know that this is defined as minus the ratio between the image distance and the object distance. So in here we again need to substitute the values for this to the value that we just have obtained of 30 centimeters. Also in here it isn't emitters, we call that and This over the object distance which is positive and that is 15cm. So from here we obtain a value of -2. Now for party of this problem, we are asked whether the image studies form is real or bare tool. To answer that question, we need to see the sign of the value that we obtain from and the image distance and we can see that that value is positive, then the image that we obtain it is real. Image is real, so we can um label that as capital art. Now for party of this problem, we are asked whether the image is inverted or not inverted. Now to answer that question we need to see the sign of for the value of the magnification. Now as you can see we obtain a negative value so the image is inverted and we can represent it this by capital I the letter capital I. So for the F. Question for this problem, we need to answer the question of whether the image that is formed is on the same side of the mirror or on the opposite side. Now we know that for real images that they form on the same side as the object, so they form on the same side as the object. So this is a solution for this problem.


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