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A small pin fixed on a table top is viewed from above from a distance of $50 \mathrm{~cm}$. By what distance would the pin appear to be raised if it is viewed from the same point through a $15 \mathrm{~cm}$ thick glass slab held parallel to the table? Refractive index of glass $=1.5$. Does the answer depend on the location of the slab?
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Verified Answer
Real depth $=15 \mathrm{~cm}$,
$$
\begin{aligned}
&\text { Refractive index } \mu=\frac{\text { real depth }}{\text { apparent depth }} \\
&\Rightarrow 1.5=\frac{15}{\text { apparent depth }} \\
&\Rightarrow \text { apparent depth }=\frac{15}{1.5}=10 \mathrm{~cm} .
\end{aligned}
$$
Distance through which the pin appear to be raised $=15$ $-10=5 \mathrm{~cm}$.
The answer is independent of the location of the slab.
$$
\begin{aligned}
&\text { Refractive index } \mu=\frac{\text { real depth }}{\text { apparent depth }} \\
&\Rightarrow 1.5=\frac{15}{\text { apparent depth }} \\
&\Rightarrow \text { apparent depth }=\frac{15}{1.5}=10 \mathrm{~cm} .
\end{aligned}
$$
Distance through which the pin appear to be raised $=15$ $-10=5 \mathrm{~cm}$.
The answer is independent of the location of the slab.
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