QQuestion
(a) A coin kept inside water [μ = 4/3] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.
(b) How is the critical angle related to the refractive index of a medium?
✓Answer
(a) Let the real depth of the coin be x.
μ = \frac{\text{Real depth}}{\text{Apparent depth}}=\frac{4}{3}
Therefore, apparent depth = \frac{3x}{4}.
\text{Shift}=x-\frac{3x}{4}=\frac{x}{4}or, \frac{x}{4}=3\,\text{mm}
x = 12 mm
Hence, the depth of the coin in water is 12 mm.
(b) C=\sin^{-1}\left(\frac{1}{\mu}\right)
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