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
Explanation
Correct Option
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Classicboymoses
4 years ago
Refractive index = real depth/apparent depth
R. Index = u
Real depth = t
App = ?, Let's call it Y
u = t/y
y= t/u = app. depth
Its distance when viewed by an observer = real depth - app. Depth
= t-y
Recall y = t/u, then we substitute
By observer = t-(t/u)
= (ut-t)/u
= t(u-1)/u
B is correct 😁
