Question

The length of the compound microscope is $15\text{ cm}$ . The magnifying power for relaxed eye is 25 . If the focal length of eye lens is $6\text{ cm}$ then the object distance for objective lens will be}

• A. $1.3\text{ cm}$
• B. $1.5\text{ cm}$
• C. $1.7\text{ cm}$
• D. $1.9\text{ cm}$

Hint:

Relaxed eye means the intermediate image falls exactly on the focal point of the eye lens ($f_e$).

Answer 1

The Correct Option is B

Step 1: Concept
For a relaxed eye, the final image is at infinity. Length of microscope $L = v_o + f_e$. Magnifying power $M = \frac{v_o}{u_o} \cdot \frac{D}{f_e}$ (taking $D=25 \text{ cm}$).

Step 2: Meaning
$L = 15$, $f_e = 6 \implies v_o = 15 - 6 = 9 \text{ cm}$.

Step 3: Analysis
$M = 25 = \frac{9}{u_o} \cdot \frac{25}{6}$ $25 = \frac{9 \times 25}{6 \times u_o} \implies 1 = \frac{1.5}{u_o} \implies u_o = 1.5 \text{ cm}$.

Step 4: Conclusion
The object distance for the objective lens is $1.5 \text{ cm}$. Final Answer: (B)