Question 85 of 295

Two immiscible liquids of refractive indices $\dfrac{8}{5}$ and $\dfrac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is $6cm$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\dfrac{\alpha }{4}cm$. The value of $\alpha$ is _______.

Your answerCheck answer

Show answer and solution▾

Answer: 31

The formula to calculate the apparent height of the coin when seen through the combination of liquids can be written as
${h}_{app}=\dfrac{{h}_{1}}{{\mu }_{1}}+\dfrac{{h}_{2}}{{\mu }_{2}}...\left(1\right)$
From equation (1), it follows that
$\begin{matrix}{h}_{app} & = & \dfrac{6}{3/2}+\dfrac{6}{8/5}cm \\ & = & 4+\dfrac{15}{4}cm \\ & = & \dfrac{31}{4}cm\end{matrix}$
Hence, $\alpha =31$.