A trough 12.0cm deep is filled with water of refractive index \(\frac{4}{3}\). By how much would a coin at the bottom of the trough appear to be displaced when viewed vertically from above the water surface?
A
3.0cm
B
6.0cm
C
9.0cm
D
16.0cm
Verified Correct Answer: A.
Physical Principles & Numerical Solution
\(\frac{4}{3}\) = \(\frac{12}{\text {Apparent depth}}\)
Apparent depth = 9cm
Height by which coin is raised = 12 - 9
= 3cm
Source: Myschool Past Questions Archive (Record #28188)