If the focal length of objective lens is increased, the magnifying power of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Microscope will decrease & telescope will increase
The objective focal length enters the two magnifying powers in opposite ways.
Compound microscope (final image at the near point, tube length \( L \)):
\( \displaystyle m \approx \frac{L}{f_o}\cdot\frac{D}{f_e} \;\Rightarrow\; m \propto \frac{1}{f_o} \)
Telescope:
\( \displaystyle m = \frac{f_o}{f_e} \;\Rightarrow\; m \propto f_o \)
So increasing \( f_o \) decreases the magnifying power of the microscope and increases that of the telescope.
NCERT (2026-27):
Points to Ponder – “… For a compound microscope, the magnifying power is given by \( m = m_e \times m_o \) where \( m_e = 1 + (D/f_e) \), is the magnification due to the eyepiece and \( m_o \) is the magnification produced by the objective. Approximately,
\( \displaystyle m = \frac{L}{f_o} \times \frac{D}{f_e} \)
where fo and fe are the focal lengths of the objective and eyepiece, respectively, and L is the distance between their focal points.”
Points to Ponder – “… 9. Magnifying power m of a telescope is the ratio of the angle β subtended at the eye by the image to the angle α subtended at the eye by the object.
\( \displaystyle m = \frac{\beta}{\alpha} = \frac{f_o}{f_e} \)”