Explanation

(b) If object in a denser medium is seen from a rarer medium then image of object will appear at a leaser distance. The distance between object and its image, called as normal shift is given by
$x = t\left\lbrack 1 - \frac{1}{\mu} \right\rbrack$
Here, $t = 6\ cm,\ \mu = 1.5$
$\therefore\ \ \ \ \ \ \ x = 6\left\lbrack 1 - \frac{1}{1.5} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6\left\lbrack \frac{0.5}{1.5} \right\rbrack = 2\ cm\ $