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(A) For the plano-concave lens $(L_1)$: Using the lens maker's formula, $\frac{1}{f_1} = (\mu - 1) \left( -\frac{1}{R} \right) = (1.75 - 1) \left( -\f

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$120$

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(A) For the plano-concave lens $(L_1)$: Using the lens maker's formula, $\frac{1}{f_1} = (\mu - 1) \left( -\frac{1}{R} ight) = (1.75 - 1) \left( -\frac{1}{30} ight) = 0.75 	imes \left( -\frac{1}{30} ight) = -\frac{0.75}{30} = -\frac{1}{40}$. So, $f_1 = -40\,cm$. Since the object is at infinity, the rays incident on $L_1$ are parallel. After passing through $L_1$, they appear to diverge from a point $40\,cm$ to the left of $L_1$. For the plano-convex lens $(L_2)$: The distance between the lenses is $d = 40\,cm$. The virtual image formed by $L_1$ acts as a virtual object for $L_2$. The distance of this virtual object from $L_2$ is $u_2 = -(40 + 40) = -80\,cm$. Using the lens maker's formula for $L_2$, $\frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R} ight) = (1.75 - 1) \left( \frac{1}{30} ight) = \frac{0.75}{30} = \frac{1}{40}$. So, $f_2 = 40\,cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$: $\frac{1}{v_2} - \frac{1}{-80} = \frac{1}{40} \Rightarrow \frac{1}{v_2} = \frac{1}{40} - \frac{1}{80} = \frac{2-1}{80} = \frac{1}{80}$. So, $v_2 = 80\,cm$ to the right of $L_2$. The distance $x$ from the concave lens $(L_1)$ to the final image is $x = d + v_2 = 40 + 80 = 120\,cm$.

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