A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the final image of the tower if it is formed at 25 cm?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30 cm
The distant tower forms a real image in the focal plane of the objective. The eyepiece then magnifies this image, and with the final image at the near point \( D = 25\,\mathrm{cm} \) its magnification is \( 1 + D/f_e \).
Angle subtended by the tower at the objective:
\( \displaystyle \alpha = \frac{100\,\mathrm{m}}{3000\,\mathrm{m}} = \frac{1}{30}\,\mathrm{rad} \)
Height of the image formed by the objective:
\( \displaystyle h_1 = f_o\,\alpha = 150\,\mathrm{cm} \times \frac{1}{30} = 5\,\mathrm{cm} \)
Magnification by the eyepiece:
\( \displaystyle m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6 \)
Height of the final image:
\( h = m_e\,h_1 = 6 \times 5\,\mathrm{cm} = 30\,\mathrm{cm} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
