A microscope has an objective of focal length
2 cm, eyepiece of focal length 4 cm and the tube
length of 40 cm. If the distance of distinct vision
of eye is 25 cm, the magnification in the
microscope is
(1) 100 125
(3) 150 (4) 250
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We have a compound microscope problem. The microscope has an objective lens with focal length 2 centimeters, an eyepiece with focal length 4 centimeters, a tube length of 40 centimeters, and the distance of distinct vision is 25 centimeters. We need to find the total magnification of this microscope.
The total magnification of a compound microscope is the product of the objective magnification and the eyepiece magnification. The objective magnification is the tube length divided by the objective focal length. The eyepiece magnification is the distance of distinct vision divided by the eyepiece focal length. So the formula becomes M equals L over f naught times D over f e.
Now let's substitute the given values into our formula. The objective magnification is 40 divided by 2, which equals 20. The eyepiece magnification is 25 divided by 4, which equals 6.25. Multiplying these together: 20 times 6.25 equals 125.
The magnification of the compound microscope is 125. This matches option 2 in the multiple choice question. We used the formula M equals L over f naught times D over f e, where L is the tube length, f naught is the objective focal length, D is the distance of distinct vision, and f e is the eyepiece focal length.