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

2. 125

3. 150

4. 250

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Best answer

Answer: 2. 125

Key idea: for a compound microscope, the magnification is the product of the objective magnification and the eyepiece magnification.

Tube length $L=40$ cm, $f_o=2$ cm, $f_e=4$ cm, $D=25$ cm. Using the standard approximate formula:

$$m=\frac{L}{f_o}\cdot\frac{D}{f_e}=\frac{40}{2}\times\frac{25}{4}=20\times 6.25=125$$

The other values (100, 150, 250) do not follow from this formula.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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