Q87 / 180 · NEET 2025 · Physics

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 :

  • A
    $100$
  • Correct answer
    B
    $125$
  • C
    $150$
  • D
    $250$

Solution

$\begin{array}{ll} \mathrm{f}_0=2 \mathrm{~cm} & \mathrm{f}_e=4 \mathrm{~cm} \\ \mathrm{~L}=40 \mathrm{~cm} & \mathrm{D}=25 \mathrm{~cm} \end{array}$ In the condition of normal adjustment. $\begin{aligned} \mathrm{m} & =\frac{\mathrm{L}}{\mathrm{f}_0} \cdot \frac{\mathrm{D}}{\mathrm{f}_e} \\ & =\frac{40}{2} \times \frac{25}{4} \\ & =125 \end{aligned}$

Hence, the correct answer is option: 'B'.

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