A convex lens of refractive index 1.5 and focal length f=18 ~cm is...

Using the lensmaker's equation, in air:
$\frac{1}{f_{air}} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ where $(n-1) = 0.5$ for glass with $n = 1.5$.
In water:
$\frac{1}{f_{water}} = \left(\frac{n}{n_{water}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$= \left(\frac{1.5}{4/3} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$= \frac{1}{8}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Since \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{2}{f_{air}} = \frac{2}{18}$,<br>We get$\frac{1}{f_{water}} = \frac{1}{8} \cdot \frac{2}{18} = \frac{1}{72}$, so$f_{water} = 72$cm.<br>The difference is$f_{water} - f_{air} = 72 - 18 = 54$cm.<br>Therefore$\alpha = \frac{54}{18} = 3.