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

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.