Refraction of Light
3.0 Apparent shift of an object
3.0 Apparent shift of an object
Due to bending of light at the interface of different media, the image formation due to refraction creates an illusion by shifting the position of an object.
We will understand the refraction in details from the following cases,
1. When the object is in denser medium and observer is in the rarer medium.
2. When the object is in the rarer medium and the observer is in the denser medium.
3.0.1 When the object is in denser medium and observer is in the rarer medium
From Snell's law, $${\mu _2}\sin i = {\mu _1}\sin r$$ or $$\frac{{\sin i}}{{\sin r}} = \frac{{{\mu _1}}}{{{\mu _2}}}\quad ...(i)$$ $$\tan i = \frac{{AB}}{{AO}}$$
As $i$ is very small, So, $$\sin i \approx \tan i = \frac{{AB}}{{AO}}\quad ...(ii)$$
Similarly, $$\sin r \approx \tan = \frac{{AB}}{{AI}}\quad ...(ii)$$
From equation $(ii)$ and $(iii)$ we get,
$$\frac{{\sin i}}{{\sin r}} = \frac{{AI}}{{AO}}\quad ...(iv)$$
From equation $(i)$ and $(iv)$ we get,
$$\frac{{AI}}{{AO}} = \frac{{{\mu _1}}}{{{\mu _2}}}$$ or $$AI = AO\left( {\frac{{{\mu _1}}}{{{\mu _2}}}} \right)$$
As $\left( {{\mu _1} = 1,\;{\mu _2} = \mu } \right)$,
So, $$AI = \frac{{AO}}{\mu }$$
Therefore, apparent depth is $\mu $ times less than the actual depth.
Note:
So, if a beaker is filled with immissible transparent liquids of refractive indices ${\mu _1}$, ${\mu _2}$ & ${\mu _3}$ and individual depth $d_1$, $d_2$ & $d_3$ respectively. Then the apparent depth of the beaker is,
$$d = \frac{{{d_1}}}{{{\mu _1}}} + \frac{{{d_2}}}{{{\mu _2}}} + \frac{{{d_3}}}{{{\mu _3}}}$$
3.0.2 When the object is in rarer medium and observer is in the denser medium
From Snell's law,
$${\mu _1}\sin i = {\mu _2}\sin r$$ or $$\frac{{\sin i}}{{\sin r}} = \frac{{{\mu _2}}}{{{\mu _1}}}\quad ...(i)$$
$$\tan i = \frac{{AB}}{{AO}}$$
As (angle $i$ is very small). So,
$$\sin i \approx \tan i = \frac{{AB}}{{AO}}\quad ...(ii)$$
Similarly, $$\sin r = \tan r = \frac{{AB}}{{AI}}\quad ...(iii)$$
From equation $(ii)$ and $(iii)$ we get,
$$\frac{{\sin i}}{{\sin r}} = \frac{{AI}}{{AO}}\quad ...(iv)$$
From equation $(i)$ and $(iv)$ we get,
$$\begin{equation} \begin{aligned} \frac{{{\mu _2}}}{{{\mu _1}}} = \frac{{AI}}{{AO}} \\ AI = AO\left( {\frac{{{\mu _2}}}{{{\mu _1}}}} \right) \\\end{aligned} \end{equation} $$
As $\left( {{\mu _2} = \mu \quad \& \quad {\mu _1} = 1} \right)$. So,
$$AI = \mu (AO)$$
The apparent height is $\mu $ times more than the actual depth.
$${d_{{\text{apparent}}}} = \mu {d_{{\text{actual}}}}$$
