An object is placed 200 cm from converging lens with focal...

An object is placed 20.0 cm from converging lens with focal length 15.0 cm ( A concave mirror with focal length 10.0 cm is located 77.5 cm to the right of the lens. What light goes through the lens, reflects from the mirror, and passes through the lens again, forming a final image. What is the location of the final image? ____ cm to the left of the lens

Transcript

00:01 In this question we have given an object that is placed 20 semi from the converging lens with a focal length of 15 semi a concave mirror with focal length that is 10 semis located 77 .5 semi to the right of the lens now we have to find the location of the final image to the left of the lens so we have given data for this that is object distance that is u1 is equal to 20 semi and we have given the focal length of converging lens that is f1 is equal to this is 15 semi so now let v1 be the position of image so we have a mirror equation that is 1 upon f1 that is equal to 1 upon u1 plus 1 upon v1 now put the data in this equation so this is 1 upon 15 is equal to this is 1 upon 20 plus 1 upon v1 so from here we get the value of v1 this will come out to be 60 semi now this image as object for concave mirror so we can say that now for concave mirror we have the data that is focal length of the mirror is that is f2 is equal to 10 semi and we have given the distance between lens and mirror which is 77 .5 semi so we can say that object distance is given by that is u2 is equal to this is 77 .5 minus 60 so this will comes out to be 17 .5 semi now let v2 be the position of final image so we can that sorry this is the not final image this is image made by the concave mirror so position of image so we can say that if we use the formula of the mirrors of 1 upon f2 is equal to this is 1 upon u2 plus 1 upon v2 so from here we can find the v2 this is 1 upon 10 is 1 upon 17 .5 plus 1 upon v2 so from here we can say that v2 is equal to 23 .3 semi now this image again act as a object for converging lens so we can say that for converging lens we have the focal length that is f1 is 15 semi and object distance that is u3 and this is equal to 77 .5 minus 20 3 .33 so this will come out to be 54 .17 semi so let v3 be the position of final image so we can say that 1 upon f1 is equal to this is 1 upon u3 plus 1 upon v3 so this is 1 upon 15 is 1 upon 54 .17 plus 1 upon v3 so from here we can find the v3 that is 20 .7 semi so this is this is the answer of our given question...

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