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Geometrical Optics question

2025 · Q148
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Geometrical Optics question

2025 · Q148

NEETPhysicsGeometrical OpticsMCQ+4 / −1

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. A
    150
  2. B
    250
  3. C
    100
  4. D
    125
View written solutionFree

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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