Question

The apparent depth of a needle lying in a water beaker is found to be $9 \,cm$. 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. $10\, cm$
• B. $9\, cm$
• C. $12 \,cm$
• D. $7\, cm$
• E. $8\, cm$

Answer 1

Answer: (E)

Given, apparent depth $\left(d_{\text {app }}\right)_{\text {water }}=9\, cm$
$ \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} $
$ =8\, cm$