JAMB - Physics (1991 - No. 24)
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?
3.0cm
6.0cm
9.0cm
16.0cm
Тайлбар
\(\frac{4}{3}\) = \(\frac{12}{\text {Apparent depth}}\)
Apparent depth = 9cm
Height by which coin is raised = 12 - 9
= 3cm
Apparent depth = 9cm
Height by which coin is raised = 12 - 9
= 3cm