Rohit has been given two converging lenses of focal lengths 1.25 cm a…

The correct answer is Option (4) → \( 11.67\,\mathrm{cm} \)

In a compound microscope the total magnification is the product of the magnifications of the objective and the eyepiece, \( m = m_o\,m_e \). The lens with the smaller focal length is the objective: \( f_o = 1.25\,\mathrm{cm} \), \( f_e = 5\,\mathrm{cm} \), and \( D = 25\,\mathrm{cm} \). Distances are measured from the lens, with the Cartesian sign convention (light travels from left to right); magnifications are taken as magnitudes.

With the final image at the least distance of distinct vision, the eyepiece magnification is

    \( \displaystyle m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6 \)

    \( \displaystyle m_o = \frac{m}{m_e} = \frac{30}{6} = 5 \)

For the objective, the real image is 5 times as far from the lens as the object: \( v_o = 5x \), \( u_o = -x \). The lens formula gives

    \( \displaystyle \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \;\Rightarrow\; \frac{1}{5x} + \frac{1}{x} = \frac{1}{1.25} \)

    \( \displaystyle \frac{6}{5x} = 0.8 \;\Rightarrow\; x = 1.5\,\mathrm{cm} \)

    \( u_o = -1.5\,\mathrm{cm} \)
    \( v_o = 5 \times 1.5 = 7.5\,\mathrm{cm} \)

For the eyepiece, the final virtual image is at \( v_e = -25\,\mathrm{cm} \):

    \( \displaystyle \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \)

    \( \displaystyle u_e = -\frac{25}{6} \approx -4.17\,\mathrm{cm} \)

The separation between the lenses is the objective image distance plus the distance of that image from the eyepiece:

    \( \displaystyle L = v_o + |u_e| = 7.5 + 4.17 = 11.67\,\mathrm{cm} \)

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):
Exercises – …