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A transparent glass cube of length $24 \mathrm{~cm}$ has a small air bubble trapped inside. When seen normally through one surface from air outside, its apparent distance is $10 \mathrm{~cm}$ from the surface. When seen normally from opposite surface, its apparent distance is $6 \mathrm{~cm}$. The distance of the air bubble from first surface is
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$15 \mathrm{~cm}$
Given: Length of cube $=12 \mathrm{~cm}$
$\begin{array}{ll} & \mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{l_1}{\mathrm{~h}_1}=\frac{24-l_1}{\mathrm{~h}_2} \\ & \text { putting } \mathrm{h}_1=10 \mathrm{~cm} \text { and } \mathrm{h}_2=6 \mathrm{~cm} \text { into (i), we get } \\ & \frac{l_1}{10}=\frac{24-l_1}{6} \\ & 6 l_1=240-10 l_1 \\ & 16 l_1=240 \\ \therefore \quad & l_1=15 \mathrm{~cm}\end{array}$
$\begin{array}{ll} & \mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{l_1}{\mathrm{~h}_1}=\frac{24-l_1}{\mathrm{~h}_2} \\ & \text { putting } \mathrm{h}_1=10 \mathrm{~cm} \text { and } \mathrm{h}_2=6 \mathrm{~cm} \text { into (i), we get } \\ & \frac{l_1}{10}=\frac{24-l_1}{6} \\ & 6 l_1=240-10 l_1 \\ & 16 l_1=240 \\ \therefore \quad & l_1=15 \mathrm{~cm}\end{array}$
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