Q.48 A compound microscope has its objective with linear magnification of 10. In order to achieve
a final magnification of 100, the angular magnification of the eyepiece should be ________.
Magnification Formula
In a compound microscope, total magnification \( m \) equals the product of the objective’s linear magnification \( m_o \) and the eyepiece’s angular magnification \( m_e \): \( m = m_o \times m_e \).
Here, \( m_o = 10 \) and desired \( m = 100 \), so \( m_e = \frac{100}{10} = 10 \).
The objective creates a real, enlarged intermediate image with linear magnification \( m_o = \frac{L}{f_o} \) (where \( L \) is tube length, \( f_o \) is objective focal length), while the eyepiece acts as a simple magnifier.
Detailed Solution
Substitute values directly: \( 100 = 10 \times m_e \), yielding \( m_e = 10 \).
For final image at infinity (relaxed eye), \( m_e = \frac{D}{f_e} \) (\( D = 25 \) cm least distance of distinct vision); at near point, \( m_e = 1 + \frac{D}{f_e} \).
The problem specifies “angular magnification,” standardly \( m_e = 10 \) without further conditions.
Magnification Components
| Component | Magnification Type | Value | Formula |
|---|---|---|---|
| Objective | Linear (\( m_o \)) | 10 | \( \frac{L}{f_o} \) |
| Eyepiece | Angular (\( m_e \)) | 10 | \( \frac{D}{f_e} \) |
| Total | Combined | 100 | \( m_o \times m_e \) |
Exam Tips for CSIR NET
- Practice variations: image at infinity vs. near point changes \( m_e \).
- Common trap: confusing linear/angular—objective always linear.
- Similar questions yield integers like 10, 20; verify via \( m / m_o \).
