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
- A150
- B250
- C100
- D125
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Correct answer: D
To determine the magnification of the microscope, we use the formula:
$ m = \frac{L}{f_o} \times \frac{D}{f_e} $
Where:
$ m $ is the total magnification.
$ L $ is the tube length of the microscope.
$ f_o $ is the focal length of the objective lens.
$ D $ is the distance of distinct vision of the eye.
$ f_e $ is the focal length of the eyepiece.
Given:
$ L = 40 $ cm (tube length)
$ f_o = 2 $ cm (focal length of the objective)
$ D = 25 $ cm (distance of distinct vision)
$ f_e = 4 $ cm (focal length of the eyepiece)
Substitute these values into the formula:
$ m = \frac{40}{2} \times \frac{25}{4} $
Calculate each part:
$ \frac{40}{2} = 20 $
$ \frac{25}{4} = 6.25 $
Multiply these results:
$ m = 20 \times 6.25 = 125 $
Thus, the magnification of the microscope is 125.
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