# Step-by-Step Solution ## Given Data - **Focal length of objective lens:** \( f_{\text{objective}} = 3.75 \, \text{mm} \) - **Overall magnification:** \( M = -130 \) - **Near-point distance:** \( N = 25 \, \text{cm} =250 \, \text{mm} \) Let the focal length of the eyepiece be \( f_e \) (in mm). --- ## Step 1: Formula for Microscope Magnification The total magnification of a compound microscope is given by: \[ M = M_{\text{objective}} \times M_{\text{eyepiece}} \] Where: - \( M_{\text{objective}} \) is the objective magnification, - \( M_{\text{eyepiece}} \) is the eyepiece magnification. ### Objective Magnification: \[ M_{\text{objective}} \approx \frac{L}{f_{\text{objective}}} \] Where \( L \) is the tube length (distance between the objective and eyepiece focal points), typically \( 160 \, \text{mm} \) for standard microscopes. ### Eyepiece Magnification (for relaxed eye at near point \( N \)): \[ M_{\text{eyepiece}} = 1 + \frac{N}{f_e} \] --- ## Step 2: Substitute Values Assume standard tube length \( L = 160 \, \text{mm} \): \[ M = \frac{L}{f_{\text{objective}}} \left( 1 + \frac{N}{f_e} \right) \] \[ -130 = \frac{160}{3.75} \left( 1 + \frac{250}{f_e} \right) \] --- ## Step 3: Solve for \( f_e \) First, calculate \( \frac{160}{3.75} \): \[ \frac{160}{3.75} = 42.67 \] So, \[ -130 = 42.67 \left( 1 + \frac{250}{f_e} \right) \] Divide both sides by 42.67: \[ \frac{-130}{42.67} = 1 + \frac{250}{f_e} \] \[ -3.047 = 1 + \frac{250}{f_e} \] \[ -3.047 - 1 = \frac{250}{f_e} \] \[ -4.047 = \frac{250}{f_e} \] Multiply both sides by \( f_e \) and divide by \(-4.047 \): \[ f_e = \frac{250}{-4.047} \] \[ f_e = -61.8 \, \text{mm} \] --- ## Step 4: Round to Two Significant Figures \[ f_e \approx -62 \, \text{mm} \] --- ## Final Answer **The focal length of the eyepiece should be:** \[ \boxed{-62 \ \text{mm}} \] --- ### Explanation - The negative sign indicates the orientation of the final image (inverted), consistent with the negative overall magnification. - The calculation assumes the standard microscope tube length and that the final image is at the near point.