A coin is placed at the bottom of a cube of..

Question 1 Report

A coin is placed at the bottom of a cube of glass tern thick. If the refractive index of the glass is $μ$ , how high does the coin appear to be raised to an observer looking perpendicularly into the glass?

t−μ

\frac{t (μ - I)}{μ}

t (1 +

\frac{1}{μ}

)

μ

μ

Answer Details

Let apparent depth = X

$μ$ = $\frac{t}{x}$ $\Rightarrow$ X = $\frac{t}{u}$

raised of coin = t - x

= t - $\frac{t}{μ}$

= $\frac{t (μ - I)}{μ}$

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A coin is placed at the bottom of a cube of glass tern thick. If the refractive index of the glass is ? ? , how high does the coin appear to be raised to an...