In a compound microscope the focal length of objective lens and eye lens are 4 cm and 10 cm respectively. An object is placed at 6 cm from the objective lens. If the final image is formed at the near point of eye, then the magnifying power of microscope is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 7
The magnifying power of a compound microscope is the product of the magnification by the objective and the angular magnification by the eyepiece. With the final image at the near point (D = 25 cm), \( m_e = 1 + D/f_e \).
Objective (lens formula, Cartesian signs, u = −6 cm, f = +4 cm):
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{4} - \frac{1}{6} = \frac{1}{12} \Rightarrow v = 12\,\mathrm{cm} \)
\( \displaystyle m_o = \frac{v}{u} = \frac{12}{-6} = -2 \)
Eyepiece:
\( \displaystyle m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{10} = 3.5 \)
\( |m| = |m_o|\,m_e = 2\times3.5 = 7 \)