A compound microscope uses an objective lens of focal length $5, mm$ and an eyepiece lens of focal length $15, mm$ . The distance between the lenses (tube length) is $120, mm$ and the final image is formed at the near point of a relaxed eye ( $D = 25, cm$ ). Calculate the overall linear magnification of the microscope.

medium 2 marks Class 12 · Physics · Ray Optics and Optical Instruments

Answer

M_{total} = 20

The correct answer is $M_{total} = 20$ — Thus the overall magnification is approximately $400$ times.

Step-by-step solution

Let $u_{o}$ be the object distance from the objective. The intermediate image formed by the objective is at $v_{o} = f_{o} + (L - f_{o})$ because the final image is at the near point, so the intermediate image must lie $D$ from the eyepiece.
First, find $v_{o}$ using $f_{e} = 15 mm$ and final image at $D = 25 cm = 250 mm$ : $v_{e} = D = 250 mm$ . Thus the intermediate image must be $250 mm$ before the eyepiece.
Hence, from the objective, the intermediate image distance is $v_{o} = L - 250 = 120 - 250 = - 130 mm$ , which indicates the intermediate image is virtual and formed $130 mm$ in front of the eyepiece. For magnification calculation, use absolute values.
Linear magnification of the objective: $m_{o} = - \frac{v _{o}}{u _{o}}$ . We first find $u_{o}$ from the lens formula: $\frac{1}{f _{o}} = \frac{1}{u _{o}} + \frac{1}{v _{o}}$ . With $f_{o} = 5 mm$ , $v_{o} = - 130 mm$ , we get $\frac{1}{5} = \frac{1}{u _{o}} - \frac{1}{130}$ , leading to $\frac{1}{u _{o}} = 0.2 + 0.00769 = 0.20769$ , so $u_{o} \approx 4.81 mm$ .
$m_{o} = - \frac{- 130}{4.81} \approx + 27$ .
Angular magnification of the eyepiece: $M_{e} = \frac{D}{f _{e}} = \frac{250}{15} \approx 16.7$ .
Overall linear magnification: $M_{t o t a l} = m_{o} \times M_{e} \approx 27 \times 16.7 \approx 450$ ? Wait, the typical total magnification for such a microscope is large; check calculations.
Re-evaluate: Use the standard formula $M_{t o t a l} = \frac{- L}{f _{o}} \times \frac{D}{f _{e}}$ when the final image is at near point.
$M_{t o t a l} = \frac{- 120}{5} \times \frac{250}{15} = (- 24) \times 16.67 \approx - 400$ . Magnitude $400$ .
Thus the overall magnification is approximately $400$ times.

Common mistake: Ignoring the sign of intermediate image distance

Concept tested: Microscope magnification calculation

Published 20 September 2026 · How this solution was prepared and checked

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