Explanation

(b) According to Cartesian sign convention
$u = - 40\ cm,R = - 20\ cm$
$\mu_{1} = 1,{\ \mu}_{2} = 1.33$
Applying equation for refraction through spherical surface, we get
$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R}$
$\frac{1.33}{v} - \frac{1}{- 40} = \frac{1.33}{- 20}$
After solving, $v = - 32\ $cm
The magnification is $m = \frac{h_{2}}{h_{1}} = \frac{\mu_{1}v}{\mu_{2}u}$
$\therefore\ \frac{h_{2}}{1} = - \frac{1(32)}{1.33( - 40)}$
Or $\ h_{2} = 0.6\ $cm
The positive sign shows that the image is erect