question_answer

Three immiscible liquids of densities \[{{d}_{1}}>{{d}_{2}}>{{d}_{3}}\] and refractive indices \[{{\mu }_{1}}>{{\mu }_{2}}>{{\mu }_{3}}\] are put in a beaker. The height of each liquid column is h/3. A dot is made at bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Answer:

The image formed by first medium acts as an object for second medium.

Let the apparent depth be \[{{O}_{1}}\]for the object seen from \[{{m}_{2}},\]then \[{{O}_{1}}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}.\frac{h}{3}\] Since, apparent depth = real depth/refractive index (\[\mu \]). Since, the image formed by medium 1 acts as an object for medium 2. If seen from \[{{\mu }_{3}}\] the apparent depth is\[{{O}_{2}}\]. Similarly, the image formed by medium 2 acts as an object for medium 3. \[{{O}_{2}}=\frac{{{\mu }_{3}}}{{{\mu }_{2}}}\,\,\left( \frac{h}{3}+{{O}_{1}} \right)=\frac{{{\mu }_{3}}}{{{\mu }_{2}}}\,\,\left( \frac{h}{3}+\frac{{{\mu }_{2}}h}{{{\mu }_{1}}3} \right)=\frac{h}{3}\,\,\left( \frac{{{\mu }_{3}}}{{{\mu }_{2}}}+\frac{{{\mu }_{3}}}{{{\mu }_{1}}} \right)\] As, seen from outside, the apparent height is \[{{O}_{3}}=\frac{1}{{{\mu }_{3}}}\,\left( \frac{h}{3}+{{O}_{2}} \right)=\frac{1}{{{\mu }_{3}}}\left[ \frac{h}{3}+\frac{h}{3}\,\left( \frac{{{\mu }_{3}}}{{{\mu }_{2}}}+\frac{{{\mu }_{3}}}{{{\mu }_{1}}} \right) \right]\] This is the required expression of apparent depth.