NEET 2025 — Physics solution

NEET2025Physics

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
B
125
C
150
D
250

Solution

For a compound microscope in normal adjustment, the total magnification is $m = \frac{L}{f _{0}} \cdot \frac{D}{f _{e}}$ , where $L$ is the tube length, $f_{0}$ the objective focal length, $f_{e}$ the eyepiece focal length, and $D = 25$ cm.

Substituting: $m = \frac{40}{2} \times \frac{25}{4} = 20 \times 6.25 = 125$ .

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