show that the focal length of a lance became 4 time than it's original time when it immerged in the water. plzzzzz answer fast it's urgent
and get 10 points....
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HEY MATE_______________________________
1/(object distance) + 1/(image distance) = 1/(focal length),
or simply 1/o + 1/i = 1/f, where i = 5f.
Solving for o, first gives 1/o = 1/f - 1/i = 1/f - 1/(5f) = 5/(5f) - 1/(5f) = (5–1)/(5f) = 4/(5f), which gives o = (5/4)f.
Then to calculate magnification *, M = i/o = (5f)/[(5/4)f] = 5 x (4/5) = 4 .
HOPE THIS HELP YOU..
1/(object distance) + 1/(image distance) = 1/(focal length),
or simply 1/o + 1/i = 1/f, where i = 5f.
Solving for o, first gives 1/o = 1/f - 1/i = 1/f - 1/(5f) = 5/(5f) - 1/(5f) = (5–1)/(5f) = 4/(5f), which gives o = (5/4)f.
Then to calculate magnification *, M = i/o = (5f)/[(5/4)f] = 5 x (4/5) = 4 .
HOPE THIS HELP YOU..
RittijaPal:
why i=5f???
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