The distance between the object to be viewed and the eyepiece of a compound microscope is 25.0 cm.The focal length of its objective lens is 0.200 cm and the eyepiece has a focal length of 2.60 cm.What is the magnitude of the total magnification of the microscope when used by the person of normal eyesight if it is designed for minimum eyestrain?
A)1070
B)204
C)520
D)477
E)772

Question

Quizplus · Accepted Answer

The total magnification of a compound microscope for minimum eyestrain is given by the formula $M = \frac{v_o}{u_o} \times \frac{D}{f_e}$, where $v_o$ is the image distance for the objective lens, $u_o$ is the object distance for the objective lens (which is approximately equal to its focal length for small objects), $D$ is the least distance of distinct vision (usually taken as 25 cm for a person with normal eyesight), and $f_e$ is the focal length of the eyepiece. The image distance $v_o$ can be found using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ for the objective lens, where $f$ is the focal length of the objective lens, $v$ is the image distance, and $u$ is the object distance. Since the total length of the microscope is the distance between the object and the eyepiece, and the image formed by the objective lens acts as the object for the eyepiece, $v_o = 25.0 cm - 2.60 cm = 22.4 cm$. The magnification by the objective lens, $M_o = \frac{v_o}{u_o} = \frac{22.4 cm}{0.200 cm} = 112$, and the magnification by the eyepiece, $M_e = \frac{D}{f_e} = \frac{25.0 cm}{2.60 cm} \approx 9.62$. Therefore, the total magnification is $M = M_o \times M_e = 112 \times 9.62 \approx 1077$, which is closest to the given option A) 1070.

The Distance Between the Object to Be Viewed and the Eyepiece