Question

A glass beaker is filled with water up to $5cm$. It is kept on top of a $2cm$ thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, its apparent depth from the water surface is $dcm$ . Value of $d$ is close to (the refractive indices of water and glass are $1.33$ and $1.5$, respectively)

• A. 2.5 cm
• B. 5.1 cm
• C. 3.7 cm
• D. 6.0 cm

Answer 1

Answer: (B)

Apparent depth $d$ in case of more than one medium is

$d=\frac{d_1}{{\mu }_1}+\frac{d_2}{{\mu }_2}+\dots$ $\dots \left(i\right)$

where, $d_1$ and $d_2$ are the thickness of slabs of medium with refractive index ${\mu }_1$ and ${\mu }_2$, respectively

Here, $d_1=5cm,{\mu }_1=133$

$d_2=2cm,{\mu }_2=15$

Substituting these values in Eq. (i), we get

Apparent depth, $d=\frac{5}{133}+\frac{2}{15}$

$=5.088cm$

$=5.1cm$