Compound microscope: The focal lengths of the objective and eyepiece in a compound microscope are 0.80 cm and 2.5 cm, respectively. The image formed by the objective is 16 cm from it and the final image is 25 cm from the eye. What is the total overall magnification of the microscope?

Question

Quizplus · Accepted Answer

The total magnification of a compound microscope is the product of the magnification of the objective lens and the magnification of the eyepiece. The magnification of the objective lens (M_o) is given by the image distance (v_o) divided by the object distance (u_o), where the object distance is approximately equal to the focal length of the objective lens (f_o) because the object is very close to the focal point. Thus, M_o = v_o / f_o. The magnification of the eyepiece (M_e) is given by the near point of the eye (assumed to be 25 cm, the final image distance from the eye) divided by the focal length of the eyepiece (f_e), so M_e = 25 / f_e. Given:- f_o = 0.80 cm- f_e = 2.5 cm- v_o = 16 cm- Final image distance from the eye = 25 cmObjective lens magnification (M_o) = v_o / f_o = 16 cm / 0.80 cm = 20Eyepiece magnification (M_e) = 25 cm / f_e = 25 cm / 2.5 cm = 10Total magnification = M_o * M_e = 20 * 10 = 200Since the microscope inverts the image, the magnification is negative, so the total magnification is -200x.

Compound Microscope: the Focal Lengths of the Objective and Eyepiece