AIIMS 2018 Physics Lens Systems MCQ Question

To find the magnification of the two-lens system, we first calculate the image formed by the convex lens using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. For the convex lens with $f_1 = 100 \, \text{cm}$ and object distance $u = -50 \, \text{cm}$, we find the image distance $v_1$ to be 100 cm, indicating a real image. This image acts as the object for the concave lens, which is 90 cm away, resulting in an object distance $u_2 = -10 \, \text{cm}$. The concave lens has a focal length $f_2 = -8 \, \text{cm}$, giving an image distance $v_2$ of -8 cm. The total magnification is the product of the magnifications of both lenses, calculated as $m_1 \times m_2 = \left( \frac{v_1}{u_1} \right) \times \left( \frac{v_2}{u_2} \right) = \left( \frac{100}{-50} \right) \times \left( \frac{-8}{-10} \right) = 0.08$. Other options are incorrect as they do not