fluorescent lamp of length 1 m is placed horizontally at a depth of 1.2 m below a ceiling. a plane mirror of length .6 m placed below the lamp parallel to and symmetric to the lamp at a distance 2.4 m from it.find the length in meters of the reflected patch of light on the ceiling.
plz answer quickly. who answer correctly. I will mark him or her brainliest.
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Answered by
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Hello
Let CD be the fluorescent lamp placed below the ceiling at a depth of 1.2 m.
Let AB be the plane mirror placed below the lamp and PQ be the reflected patch on the ceiling.
By symmetry of triangles, MN=0.2m
PE=1m and QF=1m
Therefore, PQ=1+1+1=3m.
Hope this helps u:))
Plz mark me as a brainliest
Let CD be the fluorescent lamp placed below the ceiling at a depth of 1.2 m.
Let AB be the plane mirror placed below the lamp and PQ be the reflected patch on the ceiling.
By symmetry of triangles, MN=0.2m
PE=1m and QF=1m
Therefore, PQ=1+1+1=3m.
Hope this helps u:))
Plz mark me as a brainliest
Tushakar:
what is MN, pq, pf?
you copied it from net
I also see but I don't understand on it.
so I asked here
OK sorry
but you also copied it
ok
if you able to draw diagram
then I mark you brainliest.
Answered by
10
Understand the figure.....1st
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