JAMB - Physics (2006 - No. 35)
A submarine is observed to rise from a real depth of 80 m to 60 m in water. Calculate the change in apparent depth.
[Refraction index of water \(\frac{4}{3}\)]
[Refraction index of water \(\frac{4}{3}\)]
45 m
80 m
15 m
60 m
Explanation
Refraction index (n) = \(\frac{\text{Real depth}}{\text{App. depth}}\)
For 80 m Real depth: \(\frac{4}{3}\) = \(\frac{80}{\text{App.depth}}\)
App. dept = \(\frac{80 \times 3}{4}\) = 60 m
For 60 m real depth; \(\frac{4}{3}\) = \(\frac{60}{\text{App.depth}}\)
App. depth = \(\frac{60 \times 3}{4}\) = 45 m
∴change in App. depth = 60 - 45 = 15 m
For 80 m Real depth: \(\frac{4}{3}\) = \(\frac{80}{\text{App.depth}}\)
App. dept = \(\frac{80 \times 3}{4}\) = 60 m
For 60 m real depth; \(\frac{4}{3}\) = \(\frac{60}{\text{App.depth}}\)
App. depth = \(\frac{60 \times 3}{4}\) = 45 m
∴change in App. depth = 60 - 45 = 15 m
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