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) 477
B) 1070
C) 204
D) 520
E) 772

Question

Quizplus · Accepted Answer

The total magnification of a compound microscope is given by the product of the magnification of the objective lens and the magnification of the eyepiece. The magnification of the objective lens is given by the ratio of the distance between the object and the objective lens to the focal length of the objective lens. The magnification of the eyepiece is given by the ratio of the distance between the eyepiece and the image formed by the objective lens to the focal length of the eyepiece.For a person of normal eyesight, the distance between the eyepiece and the image formed by the objective lens is 25.0 cm. The focal length of the objective lens is 0.200 cm and the focal length of the eyepiece is 2.60 cm. Therefore, the magnification of the objective lens is 25.0 cm / 0.200 cm = 125. The magnification of the eyepiece is 25.0 cm / 2.60 cm = 9.62. Therefore, the total magnification of the microscope is 125 * 9.62 = 1202.5.The closest answer choice to 1202.5 is B) 1070.

The Distance Between the Object to Be Viewed and the Eyepiece