A coin is placed at the bottom of a cube of glass tern 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
Correct Answer: B
Explanation
Let apparent depth = X \(\mu\) = \(\frac{t}{x}\) \(\Rightarrow\) X = \(\frac{t}{u}\) raised of coin = t - x = t - \(\frac{t}{\mu}\) = \(\frac{t(\mu - I)}{\mu}\)
