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Ray Optics and Optical Instruments
Compound Microscope
1 question from 1 paper, each with its answer.
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Concepts tested
- Microscope Magnification
All Questions
2025 1 question
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A microscope has an objective of focal length $2\text{ cm}$, eyepiece of focal length $4\text{ cm}$ and the tube length of $40\text{ cm}$. If the distance of distinct vision of eye is $25\text{ cm}$, the magnification in the microscope is:
Magnification is $m = \frac{L}{f_o} \times \frac{D}{f_e} = \frac{40}{2} \times \frac{25}{4} = 125$.
The magnifying power of a compound microscope is given by $m = m_o \times m_e = \left(\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right)$. Substituting the given values: $m = \left(\frac{40}{2}\right) \times \left(\frac{25}{4}\right) = 20 \times 6.25 = 125$.