A thin converging glass lens $\mu =1.5$ has a power of +5D. When this lens is immersed in a liquid of refractive index $\mu $ it acts as a diverging lens of focal length 100cm. The value of $\mu $ must be:
$\begin{align}
  & \left( A \right)\dfrac{5}{3} \\
 & \left( B \right)\dfrac{4}{3} \\
 & \left( C \right)\dfrac{5}{4} \\
 & \left( D \right)\dfrac{6}{5} \\
\end{align}$

Hint: Use mirror equation. A convex mirror is an example of a converging mirror and is a curved mirror in which the reflective surface bulges towards the light source. Convex mirrors generally are not used to focus light since they reflect light outwards. Virtual images are always formed by convex mirrors.

Complete answer:
Given that for a converging lens,
${{\mu }_{a}}=1$
where, ${{\mu }_{a}}$ is the refractive index of air
${{\mu }_{g}}=1.5$
where,
${{\mu }_{g}}$ is the refractive index of glass
${{P}_{C}}=+5D$ …………(1)
The power of lens is calculated using the lens maker’s formula
${{P}_{C}}=\dfrac{1}{f}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ …………….(2)
Equating equation (1) & (2) we get,
${{P}_{C}}=\dfrac{1}{f}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)=+5D$ …………..(3)
After that when the convergent lens is immersed in a liquid of refractive index $\mu $, then the lens becomes divergent.
Then f=-100cm
Thus again applying lens maker’s formula,
${{P}_{D}}=\dfrac{1}{f}=\left( \dfrac{{{\mu }_{g}}}{\mu }-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow {{P}_{D}}=\dfrac{-100}{100}=-1D$
${{P}_{D}}=\dfrac{1}{f}=\left( \dfrac{{{\mu }_{g}}}{\mu }-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)=-1D$……………(4)
Dividing equation (3)& (4) we get,
$\dfrac{{{P}_{C}}}{{{P}_{D}}}=\dfrac{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)}{\left( \dfrac{{{\mu }_{g}}}{\mu }-1 \right)}=-5$
$\Rightarrow \left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)=-5\left( \dfrac{{{\mu }_{g}}}{\mu }-1 \right)$
$\dfrac{1.5}{1}-1=-5\left( \dfrac{1.5}{\mu }-1 \right)$
$\dfrac{0.5}{5}=-\left( \dfrac{1.5}{\mu }-1 \right)$
$\dfrac{1}{10}=1-\dfrac{1.5}{\mu }$
$\dfrac{1.5}{\mu }=1-\dfrac{1}{10}$
$\dfrac{1.5}{\mu }=\dfrac{9}{10}$
$\dfrac{1}{\mu }=\dfrac{6}{10}$
$\therefore \mu =\dfrac{5}{3}$

Thus, option (A) is correct.

Additional information:
A convex mirror is an example of a converging mirror and is a curved mirror in which the reflective surface bulges towards the light source. Convex mirrors generally are not used to focus light since they reflect light outwards. Virtual images are always formed by convex mirrors. Plane mirrors and convex mirrors will always produce an upright image if the object is located in front of the focal point. The focal length of a convex mirror is the half of the radius of curvature. The focal length and radius of curvature of a convex mirror is positive.

Note:
Convex mirrors generally are not used to focus light since they reflect light outwards. The image is usually smaller than the object, but gets larger when the object is near the mirror. Virtual images are always formed by convex mirrors. Plane mirrors and convex mirrors will always produce an upright image if the object is located in front of the focal point.