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An object is fixed at the bottom of a vessel and water is filled in the vessel upto a height of $10 \mathrm{~cm}$. A plane mirror is placed at a height of $7 \mathrm{~cm}$ from the surface of water in such a way that its reflecting surface faces the water. The distance of the image from the mirror is (Refractive index of water, $n=1.33$ )
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Verified Answer
The correct answer is:
14.5 cm
According to the question,
the apparent distance of object from the mirror, distance of image from the mirror, $d$
$$
=7 \mathrm{~cm}+\text { apparent depth }
$$
$$
\begin{aligned}
\because \text { Apparent depth } & =\frac{\text { Real depth }}{\mu} \\
& =\frac{10}{1.33} \quad(\because \mu=1.33 \text {, given })
\end{aligned}
$$
Hence, $d=7 \mathrm{~cm}+\frac{10}{1.33}=14.5 \mathrm{~cm}$
the apparent distance of object from the mirror, distance of image from the mirror, $d$
$$
=7 \mathrm{~cm}+\text { apparent depth }
$$
$$
\begin{aligned}
\because \text { Apparent depth } & =\frac{\text { Real depth }}{\mu} \\
& =\frac{10}{1.33} \quad(\because \mu=1.33 \text {, given })
\end{aligned}
$$
Hence, $d=7 \mathrm{~cm}+\frac{10}{1.33}=14.5 \mathrm{~cm}$
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