In a microscope the objective is having focal length f_0 = 2text cm$f_{0} = 2\text{ cm}$ and eye-piece is having focal length f_e = 4text cm$f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.

### Related Formula m approx fracLf_0 times fracDf_e$$m \approx \frac{L}{f_0} \times \frac{D}{f_e}$$ ### Core Logic For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation: m simeq fracl Df_0 f_e$$m \simeq \frac{l D}{f_{0} f_{e}}$$ where l$l$ is the tube length, D = 25text cm$D = 25\text{ cm}$ is the least distance of distinct vision. ### Step 1: Substitute Values Given values: l = 32text cm$l = 32\text{ cm}$ f_0 = 2text cm$f_0 = 2\text{ cm}$ f_e = 4text cm$f_e = 4\text{ cm}$ D = 25text cm$D = 25\text{ cm}$ (standard assumption when not given) m = frac322 times frac254$$m = \frac{32}{2} \times \frac{25}{4}$$ m = 16 times frac254 = 4 times 25 = 100$$m = 16 \times \frac{25}{4} = 4 \times 25 = 100$$

### Pattern Recognition Compound microscope formulas: Normal adjustment to$\to$ image at infty$\infty$, m = (L/f_0)(D/f_e)$m = (L/f_0)(D/f_e)$. Image at near point D$D$, m = (L/f_0)(1 + D/f_e)$m = (L/f_0)(1 + D/f_e)$. Default to standard approximation when given tube length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments