Question

A glass beaker is filled with water up to $5 \,cm$. It is kept on top of a $2 \,cm$ 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 $d \,cm$ . 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
$d=\frac{d_{1}}{\mu _{1}}+\frac{d_{2}}{\mu_{2}}+\dots$ $\,$ $\ldots\left(i\right)$
where, $d_{1}$ and $d_{2}$ are the thickness of slabs of medium with refractive index $\mu_{1}$ and $\mu_{2}$, respectively
Here, $d_{1}=5\,cm, \mu_{1}=133$
$d_{2}=2\,cm, \mu_{2}=15$
Substituting these values in Eq. (i), we get
Apparent depth, $d=\frac{5}{133}+\frac{2}{15}$
$=5.088\,cm$
$=5.1\,cm$