$v=+10\, \mathrm{cm}$
$\Rightarrow \mathrm{f}=5\,\mathrm{cm}$
Glass plate shift $=\mathrm{t}\left(1-\frac{1}{\mu}\right)=1.5\left(1-\frac{2}{3}\right)=0.5\,\mathrm{cm}$
So, new $u=10-0.5=9.5\,\mathrm{cm}$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{-9.5}=\frac{1}{5}$
After solving we get,
$v=\frac{47.5}{4.5}$. Hence, shift $\frac{47.5}{4.5}-10=\left(\frac{2.5}{4.5}\right)=0.55\, \mathrm{cm}$
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The ketone. $(A)$ is
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