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A printed page is kept pressed by a transparent cube of edge $t$. The refractive index of the cube varies as $\mu(z)=1+\frac{z}{t}$, where $z$ is the vertical distance from bottom of the cube. If viewed from top, then the printed letters appear to be shifted by an amount
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The correct answer is:
$(1-\ln 2) t$
Consider an elemental strip at a height $z$ of thickness $d z$.
$$
\begin{aligned}
\text { Apparent height } & =\frac{\text { Real height }}{\text { RI }}=\frac{d z}{\mu(z)} \\
d h & =\frac{d z}{1+\frac{z}{t}} \\
d h & =\left(\frac{t}{t+z}\right) d z
\end{aligned}
$$
$$
\begin{aligned}
h & =\int_0^z \frac{t}{t+z} \cdot d z \\
h=[t \ln (t+z)]_0^z & =t[\ln (1+t)-\operatorname{In}(t+0)] \\
h & =t \operatorname{In} \frac{2 t}{t}=t \operatorname{In} 2 \\
\text { Shift, } \Delta x & =t-t \ln 2 \\
\Delta x & =(1-\ln 2) t
\end{aligned}
$$
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