An angular magnification of 30 X is desired using an objective lens of focal length 1.25 cm and an eyepiece of focal length 5 cm. What should be the separation between objective and eyepiece when the final image is formed at the near point?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 11.67 cm
In a compound microscope the total magnification is the product of the objective and eyepiece magnifications, \( M = m_o\,m_e \). The separation of the lenses is the image distance of the objective plus the object distance of the eyepiece. Magnitudes are used throughout, with the near point \( D = 25\,\mathrm{cm} \).
Eyepiece magnification (final image at the near point)
\( \displaystyle m_e = 1+\frac{D}{f_e} = 1+\frac{25}{5} = 6 \)
Objective magnification
\( \displaystyle m_o = \frac{M}{m_e} = \frac{30}{6} = 5 \)
Image distance of the objective: the object distance is \( |u_o| = v_o/5 \). With real-is-positive distances the lens formula for a convex lens is \( \frac{1}{v_o}+\frac{1}{|u_o|} = \frac{1}{f_o} \):
\( \displaystyle \frac{1}{v_o}+\frac{5}{v_o} = \frac{1}{1.25} \Rightarrow v_o = 6\times1.25\,\mathrm{cm} = 7.5\,\mathrm{cm} \)
Object distance of the eyepiece: the final image is virtual at \( 25\,\mathrm{cm} \) from the eyepiece, \( v_e = -25\,\mathrm{cm} \), and \( \frac{1}{v_e}-\frac{1}{u_e} = \frac{1}{f_e} \):
\( \displaystyle \frac{1}{u_e} = \frac{1}{v_e}-\frac{1}{f_e} = -\frac{1}{25}-\frac{1}{5} = -\frac{6}{25}\Rightarrow |u_e| = \frac{25}{6} = 4.17\,\mathrm{cm} \)
Separation of the lenses
\( L = v_o+|u_e| = 7.5+4.17 = 11.67\,\mathrm{cm} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
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