Your microscope has an objective lens of focal length 1.50 cm and an eyepiece of focal length 2.50 cm separated by 20.0 cm.What is the magnifying power if your near point is 25.0 cm?
A)150
B)-150
C)200
D)-75
E)None of these is correct.

Question

Quizplus · Accepted Answer

The magnifying power (M) of a compound microscope is given by the formula: $M = \frac{v_o \cdot v_e}{f_o \cdot f_e} \cdot \frac{D}{L}$, where $v_o$ is the image distance for the objective lens, $v_e$ is the image distance for the eyepiece, $f_o$ is the focal length of the objective lens, $f_e$ is the focal length of the eyepiece, $D$ is the near point distance, and $L$ is the length of the microscope (distance between the objective and eyepiece lenses). However, without additional information such as the image distance for the objective and eyepiece lenses, we cannot directly calculate the magnifying power using the given data alone. The formula for magnifying power in the question context typically involves the total magnification as the product of the magnifications of the objective and the eyepiece, and the near point correction factor. The correct formula for total magnification considering the near point is $M = (M_{objective} \times M_{eyepiece}) \times (1 + \frac{D}{f_e})$, where $M_{objective} = \frac{v_o}{f_o}$ and $M_{eyepiece} = \frac{D}{f_e}$. Given the lack of specific details on image distances or the exact method intended for calculation, and considering the options provided do not match a direct calculation from the given formula, "None of these is correct" is the appropriate choice.

Your Microscope Has an Objective Lens of Focal Length 1