A coin is placed at the bottom of a beaker containing water (refractive index =4/3) to a depth of 12cm. By what height the coin appears to be raised when seen from vertically above?
9cm
16cm
3cm
4cm
A
9cm
B
16cm
C
3cm
D
4cm
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Solution
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RI=real depthapparent depth
Substituting, 43=12apparentdepth
Therefore, apparentdepth=12×34=9
The height by which it appears to be raised is 12−9=3cm
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