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Objectdo = RAn object is a distance do = R from a convex spherical mirror; which also has a radius of curvature R Where is the image formed?no image is formedR/?...

Question

Objectdo = RAn object is a distance do = R from a convex spherical mirror; which also has a radius of curvature R Where is the image formed?no image is formedR/?

object do = R An object is a distance do = R from a convex spherical mirror; which also has a radius of curvature R Where is the image formed? no image is formed R/?



Answers

An object is located at a distance $s_{o}$ from a spherical mirror of radius $R$. Show that the resulting image will be magnified by an amount $$M_{T}=\frac{R}{2 s_{o}+R}$$

Convex mirror of radius of garbage are. Let us try this. So this is the convex mirror of radius of curvature are it is the focal and if equals two already right away to I have to find what maximum distance can of the image can be formed to find. This organizes the formula which is on my Jew plus one. The V equals two on my F as it is convex mirrors. So fee and if would be negative because of the sign convention. So from it he put signs correctly. So as we have to find the maximum distance if we that is one by V equals to one review plus one day if we can write one by f s to buy our two divided by our so by eight weekend right R plus to you now on the V, he calls to argue. Plus by the very by our plus to you Now we divide both the numerator and denominator by the by you So it comes like this. Are you first true? So if we maximize to value a view then you If we put then you become infinity and this term be considered then then the image would be in our by two. That is in the focal and and if we minimize you, the term would be zero. Nearly if we take you tends to zero. Then this term of infinity and the image would be finally infinity. So this is the maximum and the minimum distance the image could be formed.

Question Number twenty four can be a little tricky if you're not careful. So the first thing you want to do is set up your problem. So I've drawn the wall here, and some distance are which is equal to the radius of curvature of this spherical mirror. You place an object which is here, and the question wants to find out. What distance should the merely place such that a really image is formed on the wall? Well, if we know that our is already, it's a curvature. We can get the focal length from that because it's just the radius of curvature divided by two. So we have that written down here. One other thing that we know is the object distance. Well, sorry. I should say the image distance is going to be the distance from the wall to the mirror. So that's d I. And then finally, we can say that Dio is simply going to be equal to D I minus R because this distances are and then if you take the eye minus that, you get our deep O which is here. So, like I said, it can be a little tricky. But once you get it set up, It's not too bad. Now, if we want to solve this, we use the standard equation for these types of situations. Oh, far too many equal signs and you can plug in for our and plug in for Dio. And finally, we want to find the eye. But sadly, we can't really solve for that because we simply have something in terms of our So we're gonna have to get a solution. In terms of our rearranging this, we can write that to d I squared minus for T I R plus R squared is equal to zero. We can solve this using the quadratic equation and the eye for our plus or minus blah, blah, blah. I let you do that part. You get r one plus or minus square root two divided by two. Ah, however, we're only gonna want where the image distance is greater than our. So we're only going to look at the plus side of this. Um, I think so, because this could either be less than so this term. Could you two be less than or greater than one? And we want the one that is greater. So it's going to be a plus. So we get that d ay is equal to one point seven one r and it wants to calculate the magnification sort of as a little bonus. The end. And that's not too tricky because we'LL go to a new page in Is Simply minus D I over Dio and we have that r D. I is one point seven one r and the O is just d ay minus R So one point seven one are minus r the ars cancel and the magnification is minus two point for one, so it is bigger, but it is inverted.

Okay, So even chapter 32 probably 30. So says an object is placed at a distance are one of the wall he'll be our object Distance are so where are is exactly equals radius of curvature of a certain concave Mir. So at what distance from the wall should this beer mere be be placed so that I feel in into the optics or on the wall? One is the magnification of the image. So somewhere over here, we're gonna have come here, and this is gonna be our object is business, and we want the image exactly on the wall. So our total image distance is cos of this cool, so we can also relate. Now, our object distance is gonna be the image distance minus are from this. So using this, we can go to our miracle Asian home. So we have won over the vocally just also to over our for this mere. And this is equal to one over the image distance plus one over the object distance. But we can put that in there for let's p I s R. So now let's go ahead and simplify this. I won't buy everything by our pucks are over. The minus are making multiple everything by D I What we'll go and play with my left hand should be left to r T I squared running us to our square d I equals Bar the eye minus r squared plus R D I. Pace of five. This we should see to d I squared minus for the i r plus r squared equals zero. So we want to solve for the image distance D I here and this is a quadratic equations. We use the quadratic formulas. Let's go over here say D I equals the quadratic formula for this equation. Is or are plus or minus square. 16 are square minus eight R squared all over for you. Cool so we can simplify this baby's in the year of us. Four are plus or minus four are square root of one minus one. Before you simplify that, this is our times one plus or minus two over too cool. So it's put bad in. This is will give us two answers a quadratic equation and should be left us 0.292 are and 1.71 R so we know the image or the near has to be placed on this other side of the object, so it's gonna be greater than one are. So it's gotta be this one. So exactly we want d I to be 1.71 off. Cool. Similarity is the magnification equation to find what this is won over d i. D not. This is 1.71 R over 1.71 are Linus are That's negative. Two for one awesome.

In this particulary case, it is given that the object distance ease greater than the radius, which equals two f. So we will use the farm with that one over F. It was one of our dean art. Thus one of a d i. From here we will count to that one of a d. I equals one of us. Be not one over f minus one of our day. Not now. D not is greater than two. If so, one of our day not is less than one of our two f. And if we choose the native sign negative one. What? Jeannot is greater than negative 1/2 years. So if we replace negative one of what they not we are going to write it like this. And then one of the lot is greater than one of our two f. So hence we can put the inequality sign here and this equals one over f. So we have de I Is this then? Yes. If we have a contest meter if we have Sorry, this is one of a to F. So that's why I work too. If so, if we have the object that do if then the image on sir forms as at u f. If we have d I equals infinity then using the same form of the one of her f equals one over. Do you not equals infinity plus one of our d I We will have d I equals infinite d I equals F So since d I since d o is greater than to s But this than infinity we figured out that the location for the image or the image distance he's greater than if but this time two f so the image is formed between f and U F or focus on the


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