A convex lens made of glass $\left({\mu }_0=3/2\right)$ has focal length $f$ in air. The image of an object placed in front of it is inverted real and magnified. Now the whole arrangement is immersed in water $\left({\mu }_w=4/3\right)$ without changing the distance between object and lens. Then 

$\begin{array}{l}\frac{1}{f_{\mathrm{ait}}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ \frac{1}{f_{\text{water }}}=\left(\frac{3/2}{4/3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\end{array}$

From these two equations we get, 

$f_{\text{water }}=4f_{\text{air }}=4f$

In air image was inverted, real and magnified. Therefore, object was lying between $f$ and 2$f$. Now the focal length has changed to 4$f$. Therefore, the object now lies between pole and focus. 

Hence, the new image will be virtual and magnified.