Question

A vessel is half-filled with a liquid of refractive index $\mu$. The other half is filled with an immiscible liquid of refractive index $1.5\mu$. The apparent depth of the vessel is $50\%$ of the actual depth. Then $\mu$ is,
(A) $1.4$
(B) $1.5$
(C) $1.6$
(D) $1.67$

Answer 1

The vessel is half-filled with some liquid and the other half is filled with the immiscible liquid. So, both liquids cannot mix together. And the apparent depth of the vessel is always less than the actual depth of the vessel. And by using the apparent depth formula, the refractive index can be determined.

Formula used:
Apparent depth formula is expressed as,
$\dfrac{{{d_1}}}{{{\mu _1}}} = \dfrac{{{d_2}}}{{{\mu _2}}}$
Where,
${d_1}$ and ${d_2}$ are the heights
${\mu _1}$ and ${\mu _2}$ are the refractive indexes.

Complete step by step answer:
Given, The refractive index of immiscible liquid, $= 1.5\mu$
The apparent depth of the vessel is $50\%$ of the actual depth.
Apparent depth formula is expressed as,
$\dfrac{{{d_1}}}{{{\mu _1}}} = \dfrac{{{d_2}}}{{{\mu _2}}}\,..............\left( 1 \right)$
when, the apparent depth of the vessel is $50\%$ of the actual depth
then,
$A = \dfrac{{50 \times h}}{{100}} = \dfrac{h}{2}$
Where,
$A$ is the apparent depth
Thus the,
$A = \dfrac{h}{2}$
Substituting the known value in the equation (1),
$A = \dfrac{{{d_1}}}{{{\mu _1}}} + \dfrac{{{d_2}}}{{{\mu _2}}}$
Where,
${d_1} = \dfrac{h}{2}$ and ${d_2} = \dfrac{h}{2}$
${\mu _1} = \mu$ and ${\mu _2} = 1.5\mu$
Then,
The apparent depth is,
$\dfrac{h}{2} = \dfrac{{\left( {\dfrac{h}{2}} \right)}}{\mu } + \dfrac{{\left( {\dfrac{h}{2}} \right)}}{{1.5\mu }}$
We can write the above equation as,
$h\left( {\dfrac{1}{2}} \right) = h\left( {\dfrac{{\left( {\dfrac{1}{2}} \right)}}{\mu } + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{1.5\mu }}} \right)$
By cancelling the $h$ on both sides,
$\dfrac{1}{2} = \left( {\dfrac{{\left( {\dfrac{1}{2}} \right)}}{\mu } + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{1.5\mu }}} \right)$
Simplifying the above equation,
$\dfrac{1}{2} = \dfrac{1}{{2\mu }} + \dfrac{1}{{3\mu }}$
$\dfrac{1}{2} = \dfrac{1}{\mu }\left[ {\dfrac{1}{2} + \dfrac{1}{3}} \right]$
By altering the above equation,
$\mu = 2\left[ {\dfrac{1}{2} + \dfrac{1}{3}} \right]$
By solving the values in brackets,
$\mu = 2\left[ {0.83} \right] \\\ \mu = 1.66 \\\ \mu \simeq 1.67 \\\$

Thus, the refractive index $\mu = 1.67$. Hence, the option (D) is correct.

Note:
- The immiscible liquids are which close not mix with other liquids to give a single phase. If the liquids of similar polarities are mixed together, then, they are said to be miscible liquids.
- When the light rays pass from air into a glass, then the direction of the light rays gets changed. Then the path gets bent which is known as the refractive index. The refractive index of a medium is completely dependent on the frequency of light passing through the medium. The refractive index can be measured by using a refractometer.