MCQ
A beaker containing liquid is placed on a table, underneath a microscope which can be moved along a vertical scale. The microscope is focused, through the liquid onto a mark on the table when the reading on the scale is $a$ .

It is next focused on the upper surface of the liquid and the reading is $b$. More liquid is added and the observations are repeated, the corresponding readings are $c$ and $d$. The refractive index of the liquid is

  • $\frac{{b - d}}{{d - c - b + a}}$
  • B
    $\frac{{b - d}}{{d - c - b + a}}$
  • C
    $\frac{{d - c - b + a}}{{b - d}}$
  • D
    $\frac{{b - d}}{{a + b  - c - d}}$

Answer

Correct option: A.
$\frac{{b - d}}{{d - c - b + a}}$
a
$\mathrm{H}_{\mathrm{epp}_{1}}=\frac{\mathrm{H}_{\mathrm{R}_{1}}}{\mu}=\mathrm{b}-\mathrm{a}$

$\mathrm{H}_{\mathrm{R}_{1}}=\mu(\mathrm{b}-\mathrm{a})$           .......$(1)$

$\mathrm{H}_{\mathrm{app}_{2}}=\frac{\mathrm{H}_{\mathrm{R}_{2}}}{\mu}=\mathrm{d}-\mathrm{c}$

$\mathrm{H}_{\mathrm{R}_{2}}=\mu(\mathrm{d}-\mathrm{c})$          .......$(2)$

But $\mathrm{H}_{\mathrm{R}_{2}}-\mathrm{H}_{\mathrm{R}_{1}}=\mathrm{d}-\mathrm{b}$        ........$(3)$

$\mu[\mathrm{b}-\mathrm{a}-\mathrm{d}+\mathrm{c}]=\mathrm{d}-\mathrm{b}$

$\boxed{\mu  = \frac{{{\text{d}} - {\text{b}}}}{{{\text{b}} - {\text{a}} - {\text{d}} + {\text{c}}}}}$ or $\boxed{\mu  = \frac{{{\text{b}} - {\text{a}}}}{{{\text{d}} - {\text{c}} + {\text{a}} - {\text{b}}}}}$

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