Question

The total magnification produced by a compound microscope is 24 when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is 5 cm, the magnification produced by the objective is

• A. 4
• B. 4.8
• C. 120
• D. 6

Hint:

Always assume \( D = 25 \, \text{cm} \) for optical instrument problems unless specified otherwise.

Answer 1

The Correct Option is A

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 magnification of the eyepiece (\( m_e \)).
Step 2: Key Formula or Approach:
1. \( M = m_o \times m_e \) 2. Magnification of eyepiece (image at near point D): \( m_e = 1 + \frac{D}{f_e} \) (Standard value for \( D \) is 25 cm).
Step 3: Detailed Explanation:
Given: Total Magnification \( M = 24 \) Focal length of eyepiece \( f_e = 5 \, \text{cm} \) Least distance of distinct vision \( D = 25 \, \text{cm} \) Calculate \( m_e \): \[ m_e = 1 + \frac{25}{5} = 1 + 5 = 6 \] Calculate \( m_o \): \[ M = m_o \times m_e \] \[ 24 = m_o \times 6 \] \[ m_o = \frac{24}{6} = 4 \]
Step 4: Final Answer:
The magnification produced by the objective is 4.