AP EAPCET 06-Jul-2022 Shift 1 Paper

Concave lens forms image at a distance $v$ then using lens equation,$\;\frac{1}{f_1}=\;\frac{1}{v_1}-\;\frac{1}{u_1}$$\;\Rightarrow \;\; \;\frac{1}{-20}=\;\frac{1}{v_1}-\;\frac{1}{-30}$$\;\Rightarrow \;\; \;\frac{1}{v_1}=-\;\frac{1}{30}-\;\frac{1}{20}=\;\frac{-50}{600}$$\Rightarrow v_1=-12 \text{cm}$ so, 1st image is formed at $12 \text{cm}$ left of concave lens. Now, first image will serve as an object for convex lens.So, object distance will be $17 \text{cm}$ to left of convex lens.$\;u_2=-17 \text{cm}$$\;f_2=+10 \text{cm}$using lens equation again,$\;\frac{1}{f_2}=\;\frac{1}{v_2}-\;\frac{1}{u_2}$$\Rightarrow \;\; v_2=\;\frac{170}{7}=24.2 \text{cm}$from convex lens. But distance between concave and convex lens is $5 \text{cm}$. Hence, image will be formed at a distance of $24.2+5=29.2 \text{cm}$ to the right of concave lens.