Question

The apparent depth of a needle lying in a water beaker is found to be $9cm$. If water is replaced by a liquid of refractive index $1.5$, then the apparent depth of needle will be ( mu of water is $\frac{4}{3}$ )

• A. $10cm$
• B. $9cm$
• C. $12cm$
• D. $7cm$
• E. $8cm$

Answer 1

Answer: (E)

Given, apparent depth ${\left(d_{\text{app }}\right)}_{\text{water }}=9cm$
${\mu }_{\text{water }}=4/3$
${\mu }_{\text{liquid }}=1.5$
As we know,
$d_{\text{app }}=\frac{\text{ actual depth }}{\text{ refractive index }(\mu )}$
$\Rightarrow d_{\text{app }}\propto \frac{1}{\mu }$
$\Rightarrow \frac{{\left(d_{\text{app }}\right)}_{\text{liquid }}}{{\left(d_{\text{app }}\right)}_{\text{water }}}=\frac{{\mu }_{\text{water }}}{{\mu }_{\text{liguid }}}$
Substituting the given values, we get
$\frac{{\left(d_{\text{app }}\right)}_{\text{liquid }}}{9}=\frac{4/3}{1.5}$
$\Rightarrow {\left(d_{\text{app }}\right)}_{\text{liquid }}=\frac{4\times 9\times 2}{3\times 3}$
$=8cm$