What is the position and nature of image formed by lens combination shown in figure? ($$f_1$$, $$f_2$$ are focal lengths)

$$\text{For the first lens (Convex Lens A): } u_1 = -20\text{ cm},\ f_1 = +5\text{ cm}$$

$$\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{5} \implies \frac{1}{v_1} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20} \implies v_1 = +\frac{20}{3}\text{ cm}$$

$$\text{The real intermediate image } I_1 \text{ forms at a distance of } \frac{20}{3}\text{ cm to the right of lens A.}$$

$$\text{For the second lens (Concave Lens B):}$$ $$\text{The separation between lens A and lens B is exactly } 2\text{ cm.}$$

$$\text{Therefore, } I_1 \text{ lies behind lens B at an object distance of: } u_2 = +\left(\frac{20}{3} - 2\right) = +\frac{14}{3}\text{ cm}\quad \text{(Virtual Object)}$$

$$\text{Focal length of the concave lens: } f_2 = -5\text{ cm}$$

$$\frac{1}{v_2} - \frac{1}{14/3} = \frac{1}{-5} \implies \frac{1}{v_2} = \frac{3}{14} - \frac{1}{5} = \frac{15 - 14}{70} = \frac{1}{70}$$

$$v_2 = +70\text{ cm}$$