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A square loop of wire of side 4 cm is placed 20 cm away from a concave mirror of focal length 8 cm. What is the area enclosed by the image of the wire (assume that the center of the loop is on the axis of the mirror and the two sides are normal to the axis)?

 

 

Asked by modi72879 21st February 2018, 9:54 PM
Answered by Expert
Answer:
mirror equation that involves focal lenth f, object-to-mirror distance u and image-to-mirror distance v is given by
begin mathsize 12px style 1 over f equals 1 over u plus 1 over v end style
begin mathsize 12px style 1 over 8 equals 1 over v plus 1 over 20 end style
we get v = 40/3 cm.  magnitude of magnification begin mathsize 12px style equals space v over u equals fraction numerator begin display style bevelled 40 over 3 end style over denominator 20 end fraction space equals space 2 over 3 end style
area of image = 4×4×(2/3)×(2/3) = 7.11 sq.cm
Answered by Expert 22nd February 2018, 12:30 PM
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