A compound microscope has an objective of focal length 1.5 cm, an eyepiece of focal length 5 cm and a tube length of 15 cm. The final image is moved from infinity to the near point (25 cm from the eye). The increase in its magnifying power is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10
The objective gives a magnification \( m_o=L/f_o \). The eyepiece works like a simple microscope. It gives \( D/f_e \) with the final image at infinity and \( 1+D/f_e \) with the final image at the near point. The total magnifying power is \( m_o \) times the eyepiece value.
\( \begin{aligned} m_o &= \frac{L}{f_o} \\ &= \frac{15}{1.5} \\ &= 10 \end{aligned} \)
\( \begin{aligned} m_{\infty} &= m_o\frac{D}{f_e} \\ &= 10\times\frac{25}{5} \\ &= 50 \end{aligned} \)
\( \begin{aligned} m_{D} &= m_o\left(1+\frac{D}{f_e}\right) \\ &= 10\times6 \\ &= 60 \end{aligned} \)
\( \begin{aligned} m_{D}-m_{\infty} &= 60-50 \\ &= 10 \end{aligned} \)
The eyepiece value rises by 1, and this rise is multiplied by \( m_o \). Hence, the magnifying power increases by 10.
NCERT (2026-27):
The microscope – “… the (linear) magnification due to the objective, namely h′/h, equals”
\( \displaystyle m_O=\frac{h^{\prime}}{h}=\frac{L}{f_o} \)\( \qquad (9.43) \)
Summary – “8. … For a compound microscope, the magnifying power is given by m = me × m0 where me = 1 + (D/fe), is the magnification due to the eyepiece and mo is the magnification produced by the objective. …”
The microscope – “When the final image is formed at infinity, the angular magnification due to the eyepiece … is”
\( \displaystyle m_e=\frac{D}{f_e} \)\( \qquad [9.44(b)] \)