Question

If in compound microscope objective with focal length 1.0 cm, eyepiece with focal length 2.0 cm, tube length of 20 cm and near point for an observer is 25 cm, the value of magnification will be ______.

• A. 2.5
• B. 250
• C. 25
• D. 2500

Hint:

The objective lens creates the \textbf{first} magnification (real image), and the eyepiece acts like a simple magnifying glass to enlarge that image further (virtual image).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The total magnification ($M$) of a compound microscope is the product of the magnification of the objective ($m_o$) and the eyepiece ($m_e$).
Step 2: Calculation:
Formula: $M \approx \frac{L}{f_o} \times \frac{D}{f_e}$ Given: $L$ (tube length) = 20 cm, $f_o$ = 1.0 cm, $f_e$ = 2.0 cm, $D$ (near point) = 25 cm. $$M = \left( \frac{20}{1.0} \right) \times \left( \frac{25}{2.0} \right)$$ $$M = 20 \times 12.5 = 250$$
Step 3: Final Answer:
The magnification is 250.