Find the size of the image formed in the situation shown in figure.

0.9 cm 0.6 cm 1.2 cm 1.4 cm

0.6 cm

Here u = –40 cm, R = –20 cm $μ = 1, μ_2 = 1.33$  We have,  $\frac{m_2}{v}-\frac{m_1}{u}=\frac{m_2-m_1}{R}$ $\frac{1.33}{v}=\frac{1}{40}-\frac{0.33}{20}$ v = -32 cm. The magnification is $m=\frac{h_2}{h_1}=\frac{m_1v}{m_1u}$ $\frac{h_2}{1}=-\frac{32}{1.33(-40)}$ $⇒ h_2 = 0.6 cm$ The image is erect.