Question

A compound microscope has an objective and an eyepiece of focal lengths \( f_0 \) and \( f_e \), respectively. To obtain a large magnification of a small object, the microscope should have:

• A. \( f_0 \) and \( f_e \) small, and \( f_e>f_0 \)
• B. \( f_0 \) and \( f_e \) small, and \( f_0>f_e \)
• C. \( f_0 \) and \( f_e \) large, and \( f_e>f_0 \)
• D. \( f_0 \) and \( f_e \) large, and \( f_0>f_e \)

Hint:

For a compound microscope, smaller focal lengths of the objective and eyepiece lead to greater magnification. A larger focal length for the eyepiece compared to the objective ensures an efficient formation of a magnified image.

Answer 1

The Correct Option is A

The magnification \( M \) of a compound microscope is given by the formula: \[ M = \frac{\text{angular magnification of the eyepiece} \times \text{magnification of the objective}}{1} \] The magnification of the objective is given by: \[ M_o = \frac{D}{f_0} \] Where:
- \( D \) is the least distance of distinct vision (usually taken as 25 cm),
- \( f_0 \) is the focal length of the objective lens. The magnification of the eyepiece is given by: \[ M_e = \frac{D}{f_e} \] Where \( f_e \) is the focal length of the eyepiece. To achieve large magnification, both \( f_0 \) and \( f_e \) should be small. The objective lens's focal length \( f_0 \) should be small to produce a larger magnification, and the eyepiece's focal length \( f_e \) should be slightly larger than \( f_0 \), since the eyepiece is used to magnify the image formed by the objective. Thus, to obtain a large magnification, the microscope should have both \( f_0 \) and \( f_e \) small, and \( f_e>f_0 \). Therefore, the correct answer is option (A).