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

• Correct answer
  A

  $(1-\ln 2) t$
• B

  $(2 \ln 2-1) t$
• C

  $\frac{t}{2 \ln 2}$
• D

  $\frac{2 t}{3 \ln 2}$

Solution

Consider an elemental strip at a height $z$ of thickness $d z$. $$ \begin{aligned} \text { Apparent height } & =\frac{\text { Real height }}{\text { RI }}=\frac{d z}{\mu(z)} \\ d h & =\frac{d z}{1+\frac{z}{t}} \\ d h & =\left(\frac{t}{t+z}\right) d z \end{aligned} $$ $$ \begin{aligned} h & =\int_0^z \frac{t}{t+z} \cdot d z \\ h=[t \ln (t+z)]_0^z & =t[\ln (1+t)-\operatorname{In}(t+0)] \\ h & =t \operatorname{In} \frac{2 t}{t}=t \operatorname{In} 2 \\ \text { Shift, } \Delta x & =t-t \ln 2 \\ \Delta x & =(1-\ln 2) t \end{aligned} $$

Hence, the correct answer is option: 'A'.