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A concave mirror with a radius of curvature of 1 m is illuminated by a candle located on the symmetry axis 3 m from the mirror: Where is the image of the candle?53 ...

Question

A concave mirror with a radius of curvature of 1 m is illuminated by a candle located on the symmetry axis 3 m from the mirror: Where is the image of the candle?53 cm in front of the mirror30 cm in front of the mirror60 cm in front of the mirror60 cm behind the mirror30 cm in front of the mirror

A concave mirror with a radius of curvature of 1 m is illuminated by a candle located on the symmetry axis 3 m from the mirror: Where is the image of the candle? 53 cm in front of the mirror 30 cm in front of the mirror 60 cm in front of the mirror 60 cm behind the mirror 30 cm in front of the mirror



Answers

A candle is placed 15.0 cm in front of a convex mirror. When the convex mirror is replaced with a plane mirror, the image moves 7.0 cm farther away from the mirror. Find the focal length of the convex mirror.

In this question, it is told that there is a concave middle years and distance of objects from the military is given to us. We know that in the stands of Objects they noted by you here and we have to use the sign convention and the distances given 30 semi. So I'm going to use here you request to -30 semi. No the videos of car waiter also given to us the idea so Curvature is denoted by capital R. And it is going to us 47. So we have to use the -40 semi. And now we have to find here the sense of image to find the distance of image. What I'm going to do is that we are going to apply middle regulation, I am going to write it. We know that mirror articulation needs one upon me less one upon you. If it was to do upon our I'm going to put the value of the U. n. r. That is -30 and -40. So I get here one upon the Less one upon -30 it was to go upon -40 after reading one up on 30 regard here one upon were it was too When I'm from 30 -2 upon 40 Now we have to simplify this one upon 30 minus You were born 40 after taking calcium here. The L. C. M. S. 1 20. Okay and we are here 14 -60. After subtracting we got here value of one upon way that is -1 x 60 from here. You can say that We should be equals two minus 60. These minus 60 means we can say that the image is located. Act distance 6 to Tha Me in front of on gail made that. So this is the answer for the given question. Okay thank you.

Gays were in Chapter 32. Problem 12. So this problem we have a small candle here. Kendall was this place 35 centimeters. Those d or D equals 30 five's. The news is from a con cave near. So she's crawl issue here, here, here it says this has a radius of character of one force. And okay, so for a it says, What is the locally of this year? Well, first you're years. We should know that Oakland Zone is just a radius of courage of herbal over too, which in this case, 12 interviews so cool. And then Part B now asked us, Where were the image of the candle being located? So now we can use our lens equation here, which tells us one over the object distance. What's one over the image distance? It was won over the football X. So now we rearrange this equation we can solve for the image distance this is given as the object This news times for Boeing's go over Abner existence minus of uncle. So I believe we have all of these things already. So we conclude this initiative vicious 35 centimeters times 24 7 wounds over 35 minus tongue for Samy years People with calculators evaluated You should get 76 7 years has our image Distance. Okay, so now part C s Will it be inverted are upright So to think about this, we can think about the magnification. And since we have a RV object since that is less than the image distance this means there is a negative magnification and always for negative magnification Sze the image is inverted So this will be an inverted image Cool.

Solving party of this problem. So using the lands equation, I can write one by F. Is equal to one by day, not plus one by D. I. Which can be further internet, one by 30 centimeters plus one by 20 centimeter, solving it further, I can write the value of physical to 13, multiplication 20 by 30 plus 20 which is equal to 12 centimeter since the focal length of the Middle East positive. So it is a I can't give mirror, it is a concave mirror. So the correct answer will be two. Now solving part B. So here I can use the formula R. Is equal to um So just putting the value of F. Here so I can write to multiplication 12 centimeters. So finally I get the value of R is equal to 24 centimeters. So this is the radius of the car waited for this problem.

Okay in this question we have to find out the image distance. It different distances. We have given the objective is 30 centimetre. The radius of curvature is, it will do again, 30 centimetre. And we know that focal length is given us Radius of curvature divide by two, so which is 30 centimetre, Divide by two. So for Carlin We get is 15 centimetre. And the image distances, we have to find out the main distance uh 40 centimeter, 30 centimetre, 15 centimeters and five centimeters. So when When an object is at 40 centim, When dinner is equal to 40 centim, then we have to first find out the image states From the mirror formula, we know that one out of F. He's able to run out of dinner Plus one out of D. I. So one out of D. I. He will do one out of f -1 out of Dinner. So one out of D. I. It went to one out of 15 minus one hour 40, so one out of D. I. Equal to 40 minus 15, divide by 600. So D. I. Get, so one of the D. I. Is equal to $25. 600. So D. I. V. Get so D. I. Get is 24 centimeters. So the images a real image and Magnification of the images given S. M. one equal to minus image distance out of the object distance. So and when we get His -24 Divide by 40. So and when we get is minus 0.6 as the magnification is less than No one. Therefore the images inverted as more of the magnification is greater than well therefore the images magnified. And in the second part we have to find out find out the image distance distance. When the object distances 30 centimeters again from the formula we know that one out of every single to one out of. Do you not plus one Out of the Eye? So by rearranging this formula we get one out of f minus went out of d nutt. So we get one out of the eye equal to It. got one out of 15 minus one out of 30, 30, so one out of b. i. It will do 2 -1 out of 30 so D I get is equal to 30 centimetre. And the magnification of unification is given and to equal to minus the eye out of dinner. So the magnification begun -30 out of 30. So magnification we get is minus one As the magnification is less than one. So the images inverted as more of the magnification Is greater than one. Therefore the images magnified. And In the 3rd part again we have to find out the image distance when the object is placed at Dina is equal to 15 cm. So again we know that one out of a physical to one out of the inert plus one out of the eye. So by rearranging this formula we get one out of two years with one out of every -1 out of dinner. So one out of the eye equal to one out of 15. My next run out of 15. So the image distances zero are there for the magnification. No image, no image is for. Uh huh. But when the objectives when object Dinner displays at five cm then the in a distance become one out of dinner Plus one out of the eye. Rearranging this formula we get one out of every -1 of Gardena. So one out of the eye it went to one out of 15 -1 out of five. So one out of the eye It went to 1 -3 out of 15. So one out of D. I want to -2 out of 15. So this become, so the image distance become minus seven by two. So this image is a virtual image and the magnification becomes minus D. I. Owed off Adina, so this becomes minus into minus seven times two, divide by five. So we get the image magnification as the magnification is greater than one. So therefore it is the images upright, upright and magnified magnified.


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