JAMB - Physics (1984 - No. 27)
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?
\(\frac{1}{t - \mu}\)
\(\frac{t(\mu - I)}{\mu}\)
t (1 + \(\frac{1}{\mu}\))
-\(\mu\)
\(\mu\)t
Explanation
Let apparent depth = x
\(\mu\) = \(\frac{t}{x}\) \(\Rightarrow\) x = \(\frac{t}{u}\)
Apparent displacement of coin: s = real depth - apparent depth, s = t - x = t - \(\frac{t}{u}\)
s = \(\frac{\mu t - t}{\mu}\) = \(\frac{t(\mu - 1)}{\mu}\)
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