1. **Problem Statement:** We need to understand how a compound microscope forms a magnified image of a tiny object placed at the least distance of distinct vision (usually denoted as $D$, typically 25 cm). 2. **Concept and Setup:** A compound microscope consists of two lenses: the objective lens (close to the object) and the eyepiece lens (close to the eye). The object is placed just beyond the focal length of the objective lens to form a real, inverted, and magnified image. 3. **Ray Diagram Explanation:** - The objective lens forms a real, inverted, and magnified image of the object at point $I_1$. - This image acts as the object for the eyepiece lens. - The eyepiece lens acts as a simple magnifier, producing a virtual, magnified image at the least distance of distinct vision $D$. 4. **Deriving the Magnification Expression:** - Let $f_o$ and $f_e$ be the focal lengths of the objective and eyepiece lenses respectively. - Let $v$ be the image distance from the objective lens, and $u$ the object distance from the objective lens. - The linear magnification by the objective lens is: $$m_o = \frac{v}{u}$$ - The angular magnification by the eyepiece (acting as a magnifier) when the final image is at distance $D$ is: $$m_e = 1 + \frac{D}{f_e}$$ 5. **Total Magnification:** The total magnification $M$ of the compound microscope is the product of the objective magnification and the eyepiece magnification: $$ M = m_o \times m_e = \frac{v}{u} \times \left(1 + \frac{D}{f_e}\right) $$ 6. **Additional Notes:** - The image distance $v$ for the objective is approximately equal to the tube length $L$ (distance between the objective and eyepiece lenses). - The object distance $u$ is slightly greater than $f_o$. Hence, the compound microscope produces a highly magnified virtual image of a tiny object placed near the least distance of distinct vision.