2025 Ray Optics and Optical InstrumentsCompound Microscope Easy Question 6 of 89 in Ray Optics and Optical Instruments

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:

A 100
B 125
C 150
D 250

The answer is B.

Solution

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: $m = \left(\frac{40}{2}\right) \times \left(\frac{25}{4}\right) = 20 \times 6.25 = 125$.

Concept Microscope Magnification

Ray Optics and Optical Instruments in past papers

89 questions from this chapter have appeared across 25 exam years.