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A tank is filled with water to a height of $12.5 \mathrm{~cm}$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4 \mathrm{~cm}$. What is the refractive index of water? If water is replaced by a liquid of refractive index $1.63$ up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
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Real depth $=12.5 \mathrm{~cm}$, Apparent depth $=9.4 \mathrm{~cm}$
Refractive index of water, $\mu=\frac{\text { real depth }}{\text { apparent depth }}=\frac{12.5}{9.4}$ $=1.33$
When the water is replaced by a liquid of refractive index, $(\mu=1.63)$ real depth $=12.5 \mathrm{~cm}$, apparent depth=?
$$
\begin{gathered}
\mu=\frac{\text { Real depth }}{\text { Apparent depth }} \\
\Rightarrow \text { apparent depth }=\frac{\text { Real depth }}{\mu}=\frac{12.5}{1.63}=7.67 \mathrm{~cm} .
\end{gathered}
$$
The microscope has to be moved to focus on the needle through a distance of $(9.4-7.67)=1.7 \mathrm{~cm}$.
Refractive index of water, $\mu=\frac{\text { real depth }}{\text { apparent depth }}=\frac{12.5}{9.4}$ $=1.33$
When the water is replaced by a liquid of refractive index, $(\mu=1.63)$ real depth $=12.5 \mathrm{~cm}$, apparent depth=?
$$
\begin{gathered}
\mu=\frac{\text { Real depth }}{\text { Apparent depth }} \\
\Rightarrow \text { apparent depth }=\frac{\text { Real depth }}{\mu}=\frac{12.5}{1.63}=7.67 \mathrm{~cm} .
\end{gathered}
$$
The microscope has to be moved to focus on the needle through a distance of $(9.4-7.67)=1.7 \mathrm{~cm}$.
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