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Douoly Concave Lens (DCCL) has rdus {magn -udeir_T=r_2=O.m) 0/3 Sr4an object d s.arce (D=O.lo) away What is tne image distance (=2}T0-10+0.06-.06 -U-2 M~0,.24 MFeed...

Question

Douoly Concave Lens (DCCL) has rdus {magn -udeir_T=r_2=O.m) 0/3 Sr4an object d s.arce (D=O.lo) away What is tne image distance (=2}T0-10+0.06-.06 -U-2 M~0,.24 MFeedbackCalculate the focel distance tnen use tne Iena Mumor equerion The sign of the radius you get fro,m [ne Lectwe Notes *13, page

Douoly Concave Lens (DCCL) has rdus {magn -udeir_T=r_2=O.m) 0/3 Sr4an object d s.arce (D=O.lo) away What is tne image distance (=2} T0-10 +0.06 -.06 -U-2 M ~0,.24 M Feedback Calculate the focel distance tnen use tne Iena Mumor equerion The sign of the radius you get fro,m [ne Lectwe Notes *13, page



Answers

(II) An object 3.0 mm high is placed 16 cm from a convex mirror of radius of curvature 16 cm. (a) Show by ray tracing that the image is virtual, and estimate the image distance. (b) Show that the (negative) image distance can be computed from Eq. 23-2 using a focal length of (c) Compute
the image size, using Eq. 23-3.

We have to find the radius of curvature of convex surface solution here given some value. Is there focal length off your first of all focal land of planning, convex lens Is 0.3 m And Reflective Index. That is the value of musical to seven x 4. So we have to find the value of various so we know that the power of the powerful is very cool surface. It's PS is exposed to P. M plus two p.m. So this is one upon F. S. Is equal to burn upon FM pledged to upon F two. Again We can write this cell burn upon a fairly big push to new -1, one Upon our 1 -1 upon our. So you can see that And this is new -1 and this is one upon art C -1 Upon Infinity. So put the values again. So one upon F. L. Is equal to seven by four minus one and this is one upon RC minus zero and you can see three by four and multiply by but upon arts so this is a value of one upon fl. So we know that the value is and F M is equal to RC upon to. So this is the situation second and this is first and used in their situation. So the value of burn upon Fs is important to to upon artsy plus six upon ford galaxy. So this is the value of putting and by solving this we get F. S. is equal to four upon 14 RC. So the value is mm R. c. The call to 1.05. Meet it. So now this is the value of party So we can see here. The value of art says 1.05 m, and the correct option is zero for 1.05 m and at the.

Solving party of this problem. So here I can write a formula one by your physical to one by denote plus one by D. I do not see the object distant. Dhmh distance. Also the distance between the image and object is given by the formula D. Is equal to be not plus D. I. Rewriting it. I can write the value of the note is equal to D minus D. I, solving it for that. I can write the expression at one by F. Is equal to one by D minus D. I. Plus one by D. I. On further simplification, I can write the expression age. D. A. Is equal to d minus under hood. The Squire -4 BF by two. On putting the value in the shape of expression, I can write the value of the I. Is equal to 49 miners under route 49 square minus four multiplication, 49 multiplication minus 2, 33 by two. On solving it further, I finally get the value of the eye is equal to minus 85.1 centimeter. D. I must be projected. The other solution each committed as it has a projected value. Therefore, the image format at a distance of 85.1 centimeter and the image format on the job, on the object side of the lens. So I'm just writing it here, the image formed on the object side, the image from on the object side of the given lands of the given lands, and it is equal to and it is equal to 85.1 cm.

In this question you have uh lands with given values R. One and R. Two. And we are also given the reflective index of the lands we want to find uh image, distance, lateral magnification, determine whether the images real or virtual invited or not invited. And whether is it on the same side as the object or opposite side. Okay. So first we write down what we are given. Okay we are given the objectives tends to be positive 29 cm and it's 1.65 R one is plastic 85 cm. Are too is infinity. So from here we can calculate the line. Mhm. Using Mhm One over F. He goes to in my last one. One of our one. Yes. One of our two. Okay this is equal to one of their P. As one of the I. Yeah, so you can choose to calculate the one of the uh write me from here. Okay this is equal to 1.65 minus one. Okay. Times one over 35 minus zero. Okay, because our two is infinity and then minus. I know what 29. Okay. So from here regards. Okay. Mhm. Being adapted. Mhm. Mhm. Mhm Yeah, negative 323 divide by 20300 And then we get uh be invoked and our eyes is mm negative um 62.8 um cm Okay. So this is the answer for part A then for be calculated the electoral magnification. Okay. And he goes to negative my overpay and this is negative times negative 62.8. Do that by 29. And this is mhm Yeah, positive 2.2. Mhm. Yeah. Cable policy. Since I is negative. So the image is my trouble. Yeah. Then in party. Mhm. Since I m is positive. So the images non inverter. Mhm. Yeah. And then I e the image it's fun on the same site as the object. Mhm. Okay. So this is the answer for party and that's all for this question. Mm hmm.

So for this problem, we are given the focal length of the conclave. Near as 36 centimeters so equals 36 centimeters. We get that the image distance is the distance of the images 1/3 of the object distance. And we want to find out the object and image distances respectively. So first we're gonna use the mirror equation, which is one over f was one over a distance of the image, plus one over distances object of the image. And here we would plug in what we know. Therefore, we have won over 36 equals one over distance of the object, plus one over distance of the object invited by three this therefore, this equals one over 36 equals one over distance of the object, plus three over distance of the object. Therefore, we get that 1/36 centimeters equals four over a distance of the object, then cross multiplying across most. By this, we get that the distance of the object equals four times 36. Therefore, distance of the object equals 144 centimeters from the mirror. And then to find the distance of the image, we were just plug improve with what we know. We know that this is the image equals one third distance of the object. Therefore, we just plug in 1 44 divided by three and therefore we get the distance of the image equals 48.


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