Question

If an observer sees the bottom of the vessel shown in Fig. at 8 cm, find the refractive index of the medium in which the observer is present.

Answer 1

Calculate the apparent depth using the formula $\mu = \dfrac{{real}}{{apparent}}$ and now this becomes the real depth for the new case and 8 cm is the apparent depth and using the same formula now calculate required refractive index.

Complete step- by-step solution
When the light is travelling from medium 1 to 2 the refractive index can be written as,
$_1{\mu _2} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$
If the object is placed in a different medium then due to refraction, object appears to be displaced from its real position so, when object is in denser medium and observer is in rarer medium,
$\mu = \dfrac{{real}}{{apparent}}$
It is given that real depth is 10 cm and μ can be written as ${}_g{\mu _w} = \dfrac{{{\mu _w}}}{{{\mu _g}}}$ where, ${\mu _g} = \dfrac{3}{2};{\mu _w} = \dfrac{4}{3}$
Substitute in the formula and we get apparent depth.

$$_g{\mu _w} = \dfrac{{real}}{{apparent}} \\\ \dfrac{{{\mu _w}}}{{{\mu _g}}} = \dfrac{{real}}{{apparent}} \\\ \dfrac{{\dfrac{4}{3}}}{{\dfrac{3}{2}}} = \dfrac{{10}}{{apparent}} \\\ apparent = \dfrac{{45}}{4} \\\$$

Using the same format the refractive index at 8 cm (μr) which is the apparent depth now
$\dfrac{{\dfrac{3}{2}}}{{{\mu _r}}} = \dfrac{{45}}{{4 \times 8}} \\\ {\mu _r} = \dfrac{{16}}{{15}} \\\$

Hence the refractive index at 8 cm is$\dfrac{{16}}{{15}}$ .

Note In case of more immiscible liquids as layers present then refractive index of the combination is

${\mu _c} = \dfrac{{real(d)}}{{app(d)}} = \dfrac{{{d_1} + {d_2}...}}{{\dfrac{{{d_1}}}{{{\mu _1}}} + \dfrac{{{d_2}}}{{{\mu _2}}}...}}$