A jeweler examines a diamond with a magnifying glass. If the near-point distance

Question

A jeweler examines a diamond with a magnifying glass. If the near-point distance of the jeweler is 20.4 cm. and the focal length of the magnifying glass is 7.50 cm. find the angular magnification when the diamond is held 5.67 cm from the magnifying glass. Assume the magnifying glass is directly in front of the jeweler's eyes. A compound microscope has the objective and eyepiece mounted in a tube that is 17.6 cm long. The focal length of the eyepiece is 2.49 cm. and the near-point distance of the person using the microscope is 24.9 cm. If the person can view the image produced by the microscope with a completely relaxed eye. and the magnification is -4552. what is the distance between the objective lens and the object to be examined?

Explanation / Answer

A)Magnification is (image distance)/(object distance) i.e. di/do

The near point is the smallest distance at which the eye can focus on an object. In this example we can assume that the 'object' is actually the virtual image formed by the lens, and that the user adjusts the system so that this image is at his near point, so di = 20.4cm

With this assumption we get mag = 20.4/5.67 = 3.6

B) For eye piece, assume v = 24.9 and f = -2.49 and calculate u1 for eye piece as
u1 = (24.9*2.49)/[24.9+2.49] =2.26 cm
This gives image distance froms 17.6 - 2.26 = 15.34 cm

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