The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 5 cm respectively. If the magnifying power in normal adjustment is 45, the tube length of the microscope is (least distance of distinct vision = 25 cm)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 cm
In normal adjustment the final image of a compound microscope is at infinity, and the magnifying power is \( m=(L/f_o)(D/f_e) \). Here \( L \), the tube length, is the distance between the second focal point of the objective and the first focal point of the eyepiece.
\( f_o=1\,\mathrm{cm} \)
\( f_e=5\,\mathrm{cm} \)
\( D=25\,\mathrm{cm} \)
\( m=45 \)
\( \begin{aligned} L &= \frac{m\,f_o\,f_e}{D} \\ &= \frac{45\times1\,\mathrm{cm}\times5\,\mathrm{cm}}{25\,\mathrm{cm}} \\ &= 9\,\mathrm{cm} \end{aligned} \)
The separation of the two lenses is \( L+f_o+f_e \), here 15 cm; the tube length itself is 9 cm.
NCERT (2026-27):
The Microscope – “… The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.”
The Microscope – “… Thus, the total magnification … when the image is formed at infinity, is”
\( \displaystyle m=m_om_e=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right) \)\( \qquad (9.45) \)