The objective lens in a laboratory microscope has a focal length 4.00 cm and provides an overall magnification of 250.What is the focal length of the eyepiece if the distance between the two lenses is 32.0 cm?
A)32.0 mm
B)25.0 mm
C)16.0 mm
D)8.00 mm
E)4.00 mm

Question

Quizplus · Accepted Answer

The overall magnification of a microscope is the product of the magnification of the objective lens ($M_o$) and the magnification of the eyepiece ($M_e$). The magnification of the objective lens is given by the tube length ($L$, the distance between the two lenses) divided by the focal length of the objective lens ($f_o$). The magnification of the eyepiece is approximately the near point distance (25 cm, a standard value for the closest distance at which the eye can focus comfortably) divided by the focal length of the eyepiece ($f_e$).Given:- Overall magnification ($M$) = 250- Focal length of the objective lens ($f_o$) = 4.00 cm = 40 mm- Distance between the two lenses ($L$) = 32.0 cm = 320 mmFirst, calculate the magnification of the objective lens ($M_o$):$$M_o = \frac{L}{f_o} = \frac{320 \, \text{mm}}{40 \, \text{mm}} = 8$$Since the overall magnification is the product of the magnifications of the objective and the eyepiece, we have:$$M = M_o \times M_e$$Rearrange to solve for the magnification of the eyepiece ($M_e$):$$M_e = \frac{M}{M_o} = \frac{250}{8} = 31.25$$Now, using the approximation for the eyepiece magnification:$$M_e = \frac{25 \, \text{cm}}{f_e} = 31.25$$Rearrange to solve for $f_e$:$$f_e = \frac{25 \, \text{cm}}{31.25} = \frac{250 \, \text{mm}}{31.25} = 8 \, \text{mm}$$Therefore, the focal length of the eyepiece is 8.00 mm.

The Objective Lens in a Laboratory Microscope Has a Focal