A coin is placed at the bottom of a cube of glass t cm thick. If the refractive index of the glass is \(\mu\), how high does the coin appear to be raised to an observer looking perpendicularly into the glass?
A
\(\frac{1}{t - \mu}\)
B
\(\frac{t(\mu - I)}{\mu}\)
C
t (1 + \(\frac{1}{\mu}\))
D
-\(\mu\)
E
\(\mu\)t
Verified Correct Answer: B.
Physical Principles & Numerical Solution
Let apparent depth = x
\(\mu\) = \(\frac{t}{x}\) ⇒ x = \(\frac{t}{u}\)
Apparent displacement of coin: s = real depth - apparent depth, s = t - x = t - \(\frac{t}{u}\)
s = \(\frac{\mu t - t}{\mu}\) = \(\frac{t(\mu - 1)}{\mu}\)
Source: Myschool Past Questions Archive (Record #28721)