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(D) The image formed by the first lens acts as an object for the second lens.For the first lens $(L_1)$:Given $u_1 = -0.40 \,m$ and $f_1 = +0.20 \,m$.

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infinity

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(D) The image formed by the first lens acts as an object for the second lens.For the first lens $(L_1)$:Given $u_1 = -0.40 \,m$ and $f_1 = +0.20 \,m$.Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{0.20} + \frac{1}{-0.40} = 5 - 2.5 = 2.5 \,m^{-1}$.So,$v_1 = \frac{1}{2.5} = 0.40 \,m$.This image is formed $0.40 \,m$ to the right of $L_1$.For the second lens $(L_2)$:The distance between the lenses is $0.30 \,m$. The image from $L_1$ is $0.40 \,m$ to the right of $L_1$,which means it is $0.40 - 0.30 = 0.10 \,m$ to the right of $L_2$.Since the light rays are converging towards a point $0.10 \,m$ behind $L_2$,this acts as a virtual object for $L_2$.Thus,$u_2 = +0.10 \,m$ and $f_2 = -0.10 \,m$ (concave lens).Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-0.10} + \frac{1}{0.10} = -10 + 10 = 0$.Therefore,$v_2 = \infty$.The final image is formed at infinity.

An object is placed $0.40 \,m$ from one of the two lenses $L_1$ and $L_2$ of focal lengths $0.20 \,m$ and $0.10 \,m$ respectively,as depicted in the figure. The separation between the lenses is $0.30 \,m$. The final image formed by this two-lens system is at

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