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Rconczve mirror has . focal length 0f 43.0 cm_ (a) What is Its radius Of curvature?(b) Locate the Image when the object distance is 100 cm. (Indicate the side of th...

Question

Rconczve mirror has . focal length 0f 43.0 cm_ (a) What is Its radius Of curvature?(b) Locate the Image when the object distance is 100 cm. (Indicate the side of the mirror with the sign of your answer )Describe the properties of the Image when the object distance 100 cm. (Select all that apply.) Orealvirualupright inverted(c) Locate the Image when the object distance answer. ) cmcm: (Indicate the Slde of the mlrror with the sign of yourDescribe the properties of the Image when the object distan

Rconczve mirror has . focal length 0f 43.0 cm_ (a) What is Its radius Of curvature? (b) Locate the Image when the object distance is 100 cm. (Indicate the side of the mirror with the sign of your answer ) Describe the properties of the Image when the object distance 100 cm. (Select all that apply.) Oreal virual upright inverted (c) Locate the Image when the object distance answer. ) cm cm: (Indicate the Slde of the mlrror with the sign of your Describe the properties of the Image when the object distance 10,0 cm. (Select all that apply-) real virtual upright Inverted



Answers

(a) A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is 0.600 $\mathrm{m}$ . (b) A convex mirror forms a virtual image half the size of the object. Assuming the distance between image and object is $20.0 \mathrm{cm},$ determine the radius of curvature of the mirror.

For this problem On the topic off image formation, we're told that a concave mirror forms an inverted image four times larger than an object. We want to find the focal length of the mirror, assuming we know the distance between the object and the image. Secondly, we want to look at a convex mirror that from the virtual image half the size of the object, and we want to find the radius of coverage off that mirror, assuming again, we know the distance between the image and the object. So, firstly, for concave mirror, we know the magnification is minus four, and this is negative since the images inverted. We also know the magnification is defined as minus cube over P, where accused the image, distance and P is the object distance. So from here, we can see that Q is equal to fall times P, so the image distance is four times the object distance. Now we are also given the fact that the difference between the two Q minus P. 0.6 m and from above this can be written as four p, which is Q minus P, so we can immediately solve for P and Q. So we get P to be 0.2 m and cue to be 0.8 m. If we solve that equation now, we confined the focal length since we have P and Q. We know that one over F one of the focal and F is equal to one over p plus one off cube, and this is 1/0 0.2 m plus 1/0 0.8 m. Okay. And so yeah, if we rearrange this equation, we invert both sides. We confined the focal and F, which is 16 centimeters. So with mathematics, we get the focal length off our concave mirror to be 16 centimeters. Now. We need to do is similar calculation for a convex mirror were given the convex mirror with a magnification off positive or half for a convex mirror. A positive magnification means the images inverted, and we know this is defined as minus que over p as above. We also know that Q The absolute value of Q Plus P is equal to zero 0.2 meters, which is given the distance between the image and the object, and this is equal to minus que plus p and jump above. We can see that this is simply minus Q minus too cute. So he substitute out the P and replaced with Q. So from here refined that the image distance que is equal to minus 6.67 centimeters and then we consult a P and P is 13.3 centimeters. Now to find the radius off radius of coverage off the convex mirror, we use the equation. One over P plus one over Q is equal to two of our where r is the radius of curvature, and so this is equal to one divided by zero point 133 m plus one over que, which is one divided by minus zero point 066 7 m. And if we rearrange and we invert and sold for the ladies of curvature, we get the radius of curvature. Convex mirror are two B minus 26.7 centimeters

So, for part, a seance producing a smaller image located behind the surface of a mirror is what we are told happens in this situation that's going to require a convex Mir. Okay, so again, because the image is, uh, because the image is smaller and located behind the surface of the mirror, the mirror is convex, so we could just go ahead for part A and right, come next. The box I didn't is your solution. Now for Part B, we're going to use the information we're given about the location of the distance from the object to the mirror, the height of the image produced in the height of the original object. We're gonna use that information to find the distance to the image, since the magnification is equal to the height of the image divided by the height of the object, which is equal to negative the distance to the image divided by the distance to the object we can find the distance to the image is equal to negative distance to the object multiplied by the height of the image divided by the height of the object. So playing in all those values into this expression the distance. The image is approximately negative. 25 centimeters. Okay, since it's, uh, negative 25 centimeters, that means it's 25 centimeters behind the surface. So we just write that out because that's really what it's asking us here and asked us for. The locations will say 25 centimeters behind surface. We can go ahead and box that in as our solution for Part B. We'll start a new page here to do part. See. So for part C, it asked us to find the focal length. We can do that by simply using our equation one over the distance to the object which we were given, plus one over the distance to the image we just found that is equal to one over the focal length. So plugging in the values for distance to the object and distance to the image of actually wrote distance the object twice Let me change that. It's supposed to be distance to the image we find that that's equal to won't negative won over 100 and 12 centimetres. So we just invert that to find the focal length when we find the focal length is equal to 112 centimeters. We can go ahead and box. That is your solution for part C. The party asked us to find the radius. Will The Radius as simple radius of curvature is simply two times the Focal Inc So two times 112 centimetres is 224 centimeters and we can go ahead and box. It is their solution for part D.

In the first part of this problem, we're going to get the object and m in distance when the object lies beyond the center off. Correct. So we need to get that deal, which is the object Distance and de I, which is at a distance Well, the middle equipment as a wonder wanted by f X equals two one divided by deal plus one you wanted by the high where this effort, the vocal and off the mirror we got it equating about one. It is doing that This deal minus D I is equals to 45.0 center. So from here we can white This idea eyes equals true uh deal oneness 45.0 sentiment and we call it equating about two now and setting this weathering to equate number one along with the value off the focal. And we can write a query. Number one is when you wanted by 30 point is your sentiment is equals to one divided by Dior plus one day wanted by do your Wyness 45.0 centimeters So this can be written as it wonder wanted by 30.0 centimeter equals two here will be to Dior minus 45 point zero centimeter. You wanted by deal into a deal minus 45 points it listen to me. So this will give us and a question which is do a square minus 105 deal plus 108. 1350 and it is equal to zero. Now we can school is quiet and using the using the quadratic formula as a do equals two minus into 100 minus 105 plus minus. Squared off minus 105. Old square minus. Fall into one into 1350 on the square root is up to this point you wanted by two into one. So there you are, sir. The very 40 as equals two 1 to 5 plus minus 75.0. So what is 75.0? Do wanted by to we call it equating about three. So from here we can get to values two questions one with plus sign and one with minus No for the case where object lies beyond the center of nature. We can wipe this quiet and as the deal is equals to 105 centimeter blood. So 1 to 5.0 centimeter rewarded by to So this was your The value for the deal, as do is equals to 90.0 centimeters. So no, you leave with number to become white demand for the for the D I as a the eyes equals two, uh, 90.0 centimeter minus 45 point zero sentiment. So we will get the battle for the i. D. I Z equals two 45 points in the sentiment. Now when the object lies between the focal and 10 the mirror the question will be do your prime and it is equals to 105 centimeter minus So 25 centimeter divided by two. So this would you rather for you said your prime as a your brand is equals to 15 points Little sentiment similarly, we can get the value for D. I prime is Do you have prime is it goes to, uh 15 points little centimeter minus 45.7 centimeters. So we get the value for the Saudi. I primary their prime musicals too minus 30 point zero centimeters. So end of the cushion. Thank you

Hi. In the given problem there is a donkey spherical mirror bulls for talent is given as F is equal to plus 30.0 centimetre and the distance between the positions of the image and the object in front of this concave is very cool mirror. Or given us if this is the center of curvature, this will be the focal land. If the object is here we know the image forms will be here. So if this is the object that stands from the vertex of this mirror, D. O. Then here it will become the image distance D. I. So the gap between object and the image formed which may be either it may be D or minus D. I. If the object is beyond the center of curvature then object distance will be more. The image will be formed between the optical center and the principal focus. In that case this is D. O minus D. I. Is equal to 45 10 centimeter. And and in the second case, if the object is lying between central asia and the focus, principal focus, then the image will be formed beyond central characters. So now it will become geo uh sorry D. I minus deal And that is same 45.0 centimetre. And as you know the positions of the object and image in front of the concave mirror are interchangeable. So we will find only one of the answer this given problem. Then another answer will be in ter convertible. So in the first part of the problem yeah the object lies beyond center of curvature. Then definitely image will be lying between central curvature and principal focus. So this will be to your minus D. is equal to 45 cm. So using that we get an expression for D. I. In terms of deal. So this D. I will be Video object stands -45. So using mirror equation, which is a relation among the object distance deal the industry sends li I earned our focal length. So here it will be for video. This will be the same as D. O. Or D. I. This is one by D. All -45 Is equal to focal length, which is 31 x 30 centimetre. Now taking an L. C. M. Here this is do you Braggadocio -45. And here it will be deal -45 plus. The old Is equal to one x 30. Or we can say this is advise of deal -45 divided by D. O. Square, expanding the bracket. Opening this bracket, do you square minus 45 G. O. Which is equal to one x 30. When making a cross multiplication. We get here if you multiply this 30 but this and this. So it will become 60 G. O -1350 is equal to D. O. Square minus 45. The oh so finally rearranging all these terms we get a quadratic equation. Do you square -105 d. O plus 1000 350 is equal to zero? Which is a quadratic equation of the form a X square plus bx plus C is equal to zero. And the solution of such a quadratic equation is given with the help of the screaming and and here the discriminate is B squared minus for a C. So the root of the situation will be deal Is equal to -7. These -105. Let's minus D. Discriminate. Which is actually b squared minus four is easy for me. This is 0 5 -4. A for Amy's the multiple of Geo Square which is one. C means 1350 Divided by two aims to into one again. Saw Yeah it will come out to be This is 105. Bless minus For the square of this. 105. You will come out to be 11025 and then it will come out to be minus 5000 or hunted. Do you fight it by two? Then this is 100 and five plus minus the square root off It will come out to be 5000 625 Divided by two. So this is D. Oh Is equal to 105. Last 75 divided by two. Again. Now taking the positive sign there are two possibilities. So first of all, if we take positive sign, This deal means the object distance will be 105 plus 75, which is 180 divided by two. Yes, this is 90 cm. And if we take negative sign here, This video will be 105 -75. So it will be 30 x two means 15 centimeter. But this Value of D. O. Which is equal to 15 cm, will not be possible because the distance between D. U. And D. I. Is only 45 is 45 centimeter. So the value of D. O less than 45 centimeter is not possible. So the only possible answer for D. O. Is 90.0 centimetre. So D. I means the distance of image there will be because we have used for it As their equals to deal -45. So this D. I 90 minus 45 means this is 45 centimeter. So these are the two answers for the First part of the problem object has been kept at a distance of 90 cm and images being formed at a distance of 45cm in front of the concave mirror. No, in the second part of the problem, as we know as the distances of the positions of object and image its image four, a spherical mirror are interchangeable. Therefore, in the 2nd case, and object will be lying at a distance less than the radius of curvature means which will be lying before the center of gravity. Sure. Then the image will be lying beyond the center of curvature. Means in this case Now video will become 45 cm, So DI will now become 90 centimetre. So these two are the answers for the second part of the same problem. Thank you.


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