1. **Problem Statement:** Derive the magnification produced by a compound microscope in two cases: (A) when the final image is formed at infinity, and (B) when the final image is formed at the near point. 2. **Background:** A compound microscope consists of two lenses: the objective lens and the eyepiece lens. The total magnification is the product of the magnifications produced by each lens. 3. **Key Formulas:** - Magnification by objective lens: $M_o = \frac{v}{u}$ where $v$ is the image distance and $u$ is the object distance for the objective. - Angular magnification by eyepiece: $M_e = \frac{D}{f_e}$ when image is at infinity, and $M_e = 1 + \frac{D}{f_e}$ when image is at near point. - Here, $D$ is the least distance of distinct vision (usually 25 cm), $f_e$ is the focal length of the eyepiece. --- ### (A) Image formed at infinity 4. **Step 1:** The objective forms a real, inverted, and magnified image of the object at distance $v$ from the objective. 5. **Step 2:** The eyepiece acts as a simple magnifier to view this image. For the final image at infinity, the eyepiece is adjusted so that the image formed by the objective lies at the focal point of the eyepiece. 6. **Step 3:** The magnification by the objective is approximately $M_o = \frac{L}{f_o}$ where $L$ is the tube length (distance between objective and eyepiece) and $f_o$ is the focal length of the objective. 7. **Step 4:** The angular magnification by the eyepiece when the final image is at infinity is $M_e = \frac{D}{f_e}$. 8. **Step 5:** Therefore, the total magnification is: $$ M = M_o \times M_e = \frac{L}{f_o} \times \frac{D}{f_e} $$ --- ### (B) Image formed at near point 9. **Step 1:** When the final image is formed at the near point, the eyepiece forms a virtual image at the least distance of distinct vision $D$. 10. **Step 2:** The angular magnification of the eyepiece in this case is: $$ M_e = 1 + \frac{D}{f_e} $$ 11. **Step 3:** The objective magnification remains approximately the same: $$ M_o = \frac{L}{f_o} $$ 12. **Step 4:** Hence, the total magnification is: $$ M = M_o \times M_e = \frac{L}{f_o} \times \left(1 + \frac{D}{f_e}\right) $$ --- **Summary:** - For image at infinity: $M = \frac{L}{f_o} \times \frac{D}{f_e}$ - For image at near point: $M = \frac{L}{f_o} \times \left(1 + \frac{D}{f_e}\right)$ This derivation shows how the tube length, focal lengths, and least distance of distinct vision affect the magnification of a compound microscope.