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(A) Consider an elemental strip of thickness $dz$ at a height $z$ from the bottom of the cube.The apparent thickness $dh$ of this elemental strip is g

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$(1 - \ln 2) t$

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(A) Consider an elemental strip of thickness $dz$ at a height $z$ from the bottom of the cube.The apparent thickness $dh$ of this elemental strip is given by:$dh = \frac{dz}{\mu(z)} = \frac{dz}{1 + \frac{z}{t}} = \frac{t}{t + z} dz$The total apparent depth $h'$ of the cube as seen from the top is the integral of $dh$ from $z = 0$ to $z = t$:$h' = \int_0^t \frac{t}{t + z} dz = t [\ln(t + z)]_0^t$$h' = t [\ln(2t) - \ln(t)] = t \ln\left(\frac{2t}{t}ight) = t \ln 2$The shift in the position of the printed letters is the difference between the real thickness $t$ and the apparent thickness $h'$:$	ext{Shift} = t - h' = t - t \ln 2 = (1 - \ln 2) t$

$A$ 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 the bottom of the cube. If viewed from the top, then the printed letters appear to be shifted by an amount

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