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?
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 = \(\frac{\mu t - t}{\mu}\) = \(\frac{t(\mu - 1)}{\mu}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}