If the focal length of an objective lens is decreased, then the magni…

The correct answer is Option (1) → the microscope will increase but that of the telescope will decrease

The magnifying power of an optical instrument is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object seen directly. The objective enters the two formulas in opposite ways.

For a compound microscope (final image at infinity), with tube length \( L \) and least distance of distinct vision \( D \):

    \( \displaystyle m = \frac{L}{f_{o}} \times \frac{D}{f_{e}} \;\Rightarrow\; m \propto \frac{1}{f_{o}} \)

For an astronomical telescope (normal adjustment):

    \( \displaystyle m = \frac{f_{o}}{f_{e}} \;\Rightarrow\; m \propto f_{o} \)

So a smaller \( f_{o} \) increases the magnifying power of the microscope and decreases that of the telescope.

NCERT (2026-27):
Points to Ponder – “… For a compound microscope, the magnifying power is given by m = m_e × m₀ … Approximately,

    \( \displaystyle m = \frac{L}{f_o} \times \frac{D}{f_e} \)

where f_o and f_e 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.

    \( \displaystyle m = \frac{\beta}{\alpha} = \frac{f_o}{f_e} \)

where f₀ and f_e are the focal lengths of the objective and eyepiece, respectively.”