RE-NEET 2026 Physics Lens Formula MCQ Question

To find the position of the image formed by the lens combination, we first analyze the converging lens $L_1$ with a focal length of $+10 \, \text{cm}$. The object is placed $30 \, \text{cm}$ to the left of $L_1$. Using the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

where $f = 10 \, \text{cm}$ and $u = -30 \, \text{cm}$:

$\frac{1}{10} = \frac{1}{v} + \frac{1}{30}$

Solving gives $v = 15 \, \text{cm}$ (to the right of $L_1$). This image acts as the object for the concave lens $L_2$ (focal length $-10 \, \text{cm}$), located $3 \, \text{cm}$ to the right of $L_1$, making the object distance for $L_2$ $u = -12 \, \text{cm}$.

Applying the lens formula again:

$\frac{1}{-10} = \frac{1}{v} - \frac{1}{-12}$

Solving gives $v = -60 \, \text{cm}$, meaning the image is $60 \, \text{cm}$ to the left of $L_2$. Thus, the correct answer is option A.

The other options fail because they miscalculate the image distances or the sign conventions for the lenses.