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 \(\frac{4}{3}\)]
A
45 m
B
80 m
C
15 m
D
60 m
Verified Correct Answer: C.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #13898)