Question

What will be the depth of object 'O' for an observer just below the interface CD? $({\mu }_3>{\mu }_2>{\mu }_1)$

• A. $\frac{{\mu }_2{t}_1}{{\mu }_1}$
• B. $\frac{{\mu }_1{t}_2}{{\mu }_2}$
• C. $t_2+\frac{{\mu }_2{t}_1}{{\mu }_1}$
• D. $t_1+{\mu }_1{t}_2$

Answer 1

The correct option is C $t_2+\frac{{\mu }_2{t}_1}{{\mu }_1}$

For the observer just below CD, the position of 'O' is shifted due to refraction at interface AB.
$\Rightarrow$ Since ${\mu }_2>{\mu }_1,$ the incident ray gets refracted towards normal at interface AB and object will appear to situated at farther distance.
Thus, depth of object for observer,
$d=t_2+d_{app}$
Here, $\frac{{\mu }_1}{{\mu }_2}=\frac{d_{real}}{d_{app}}$
$\Rightarrow \frac{{\mu }_1}{{\mu }_2}=\frac{t_1}{d_{app}}$
or, $d_{app}=\frac{{\mu }_2}{{\mu }_1}{t}_1$
$\therefore d=t_2+\frac{{\mu }_2}{{\mu }_1}{t}_1$

Why this question? Tip: For observer just below interface CD, only one refraction at interface AB is responsible for image formation.