Refer the figure given below. $$\mu_1$$ and $$\mu_2$$ are refractive indices of air and lens material. The height of image will be __________ cm.

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

$$m = \frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u}$$

$$\frac{1.54}{v} - \frac{1}{-40} = \frac{1.54 - 1}{-20}$$

$$\frac{1.54}{v} + \frac{1}{40} = \frac{0.54}{-20}$$

$$v = \frac{1.54 \times 40}{-2.08} = -\frac{61.6}{2.08} \approx -29.615\text{ cm}$$

$$m = \frac{\mu_1 v}{\mu_2 u} = \frac{1 \times \left(-\frac{61.6}{2.08}\right)}{1.54 \times (-40)}$$

$$m = \frac{-61.6}{2.08 \times (-61.6)} = \frac{1}{2.08}$$

$$h_i = \frac{1}{2.08} \times 2 = \frac{2}{2.08} \approx 0.96\text{ cm} \approx 1\text{cm}$$