The lens combination as shown consists of two thin lenses \(L_1\) and \(L_2\) of focal lengths \(+10\ \text{cm}\) and \(-10\ \text{cm}\), respectively.
The object is placed \(30\ \text{cm}\) to the left of \(L_1\), and the distance between the two lenses is \(3\ \text{cm}\).
The position of the image formed is:
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In multi-lens systems, always solve one lens at a time.
The image formed by the first lens becomes the object for the second lens.
Step 1: Plan the two-step approach. Light passes through $L_1$ first, then $L_2$. The image made by $L_1$ becomes the object for $L_2$. We use the lens formula $\dfrac1f = \dfrac1v - \dfrac1u$ each time with proper signs. Step 2: Apply the formula to the convex lens. For $L_1$: $f_1 = +10$ cm, $u_1 = -30$ cm. So $\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{1}{15}$, giving $v_1 = 15$ cm to the right of $L_1$. Step 3: Shift to the second lens. The lenses are $3$ cm apart, so this image is $15 - 3 = 12$ cm to the right of $L_2$. It lies beyond $L_2$, making it a virtual object: $u_2 = +12$ cm. Step 4: Apply the formula to the concave lens. For $L_2$: $f_2 = -10$ cm. So $\dfrac{1}{v_2} = -\dfrac{1}{10} + \dfrac{1}{12} = -\dfrac{1}{60}$. Step 5: Solve for the image distance. \[ v_2 = -60 \text{ cm} \] Step 6: Read the sign. The negative sign means the final image forms $60$ cm to the left of the concave lens, which is option C. \[ \boxed{60 \text{ cm to the left of the concave lens}} \]
