A trough 12.0cm deep is filled with water of refractive index \(\frac{4}{3}\). By how much...
PHYSICS
JAMB 1991
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
Correct Answer: A. 3.0cm
Explanation
\(\frac{4}{3}\) = \(\frac{12}{\text {Apparent depth}}\)
Apparent depth = 9cm
Height by which coin is raised = 12 - 9
= 3cm
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