A coin is placed in the bottom of a beaker containing water (refractive
index = 4/3) at a depth of 12 cm. By what height the coin appears to be
raised when seen vertically above?
Clue: Acual depth = Refractive index x Apparent depth
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Explanation:
Answer
RI=
apparent depth
real depth
Substituting,
3
4
=
apparentdepth
12
Therefore, apparent depth=
4
12×3
=9
The height by which it appears to be raised is 12−9=3cm
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