
A stationary swimmer S, inside a liquid of refractive index, ${\mu _1}$ is at distance $d$ from a fixed point P inside the liquid. A rectangular block of width $t$ and refractive index ${\mu _2}\left( {{\mu _2} > {\mu _1}} \right)$ is now placed between S and P. S will observe P to be at a distance.
A) $d - t\left( {\dfrac{{2{\mu _1}}}{{{\mu _2}}} - 1} \right)$.
B) $d - t\left( {1 - \dfrac{{{\mu _2}}}{{3{\mu _1}}}} \right)$.
C) $d - t\left( {1 - \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right)$.
D) $d - t\left( {\dfrac{{{\mu _1}}}{{{\mu _2}}} - 1} \right)$.
Answer
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Hint:The formula of an apparent distance can be used to calculate the distance which is observed when seen from a body under a different medium. When there is change of refractive index from one medium to another medium then the object distance is observed at some apparent distance this happens as the two medium has different refractive index.
Formula used: The formula for apparent distance for a body in a medium with ${\mu _1}$ as refractive index and another object is placed in a medium ${\mu _2}$ and when one object is seen from other medium is given by, $t' = t\left( {1 - \dfrac{1}{\mu }} \right)$ where $t$ is the actual distance of the object and $t'$ is the apparent distance of the object and $\mu $ is the ratio of refractive index of the two mediums.
Complete step-by-step answer:
As it is given that a rectangular block of width $t$ can is placed at the medium with refractive index ${\mu _2}$ and the swimmer is at a refractive index of${\mu _1}$. The refractive index of rectangular block with respect to
Swimmer is $\mu = \dfrac{{{\mu _2}}}{{{\mu _1}}}$.
Now apply the formula for apparent distance,
$t' = t\left( {1 - \dfrac{1}{\mu }} \right)$ Where $t$ is the width of the rectangular block.
Replace the value of $\mu = \dfrac{{{\mu _2}}}{{{\mu _1}}}$ in the relation above,
\[
t' = t\left( {1 - \dfrac{1}{\mu }} \right) \\
t' = t\left( {1 - \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) \\
\]
The distance at which S will observe the P to be is,
$
d' = d - t' \\
d' = d - t\left( {1 - \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) \\
$
So the correct answer for this problem is C.
Note: The formula for calculating the apparent image for different refractive index of the medium should be remembered by students for solving such types of problems. We have subtracted the apparent distance from the distance $d$ while solving is because the distance between the observer and point P is given as $d$ and from S the apparent image will be formed before point P.
Formula used: The formula for apparent distance for a body in a medium with ${\mu _1}$ as refractive index and another object is placed in a medium ${\mu _2}$ and when one object is seen from other medium is given by, $t' = t\left( {1 - \dfrac{1}{\mu }} \right)$ where $t$ is the actual distance of the object and $t'$ is the apparent distance of the object and $\mu $ is the ratio of refractive index of the two mediums.
Complete step-by-step answer:
As it is given that a rectangular block of width $t$ can is placed at the medium with refractive index ${\mu _2}$ and the swimmer is at a refractive index of${\mu _1}$. The refractive index of rectangular block with respect to
Swimmer is $\mu = \dfrac{{{\mu _2}}}{{{\mu _1}}}$.
Now apply the formula for apparent distance,
$t' = t\left( {1 - \dfrac{1}{\mu }} \right)$ Where $t$ is the width of the rectangular block.
Replace the value of $\mu = \dfrac{{{\mu _2}}}{{{\mu _1}}}$ in the relation above,
\[
t' = t\left( {1 - \dfrac{1}{\mu }} \right) \\
t' = t\left( {1 - \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) \\
\]
The distance at which S will observe the P to be is,
$
d' = d - t' \\
d' = d - t\left( {1 - \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) \\
$
So the correct answer for this problem is C.
Note: The formula for calculating the apparent image for different refractive index of the medium should be remembered by students for solving such types of problems. We have subtracted the apparent distance from the distance $d$ while solving is because the distance between the observer and point P is given as $d$ and from S the apparent image will be formed before point P.
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