A compound microscope has an objective of focal length 2 cm, an eyepiece of focal length 2.5 cm and a tube length of 16 cm. Take \(D = 25\) cm. Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Magnification due to the objective | (I) 11 |
| (B) Angular magnification due to the eyepiece, final image at infinity | (II) 80 |
| (C) Angular magnification due to the eyepiece, final image at the near point | (III) 10 |
| (D) Total magnifying power, final image at infinity | (IV) 8 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) Magnification due to the objective — This matches (IV) 8. Here \( m_o=L/f_o \), which gives \( 16/2=8 \).
(B) Angular magnification due to the eyepiece, final image at infinity — This matches (III) 10. Here \( m_e=D/f_e \), which gives \( 25/2.5=10 \).
(C) Angular magnification due to the eyepiece, final image at the near point — This matches (I) 11. Here \( m_e=1+D/f_e \), one more than the value at infinity.
(D) Total magnifying power, final image at infinity — This matches (II) 80. The total is the product \( m_o m_e \), which gives \( 8\times10=80 \).
NCERT (2026-27):
The microscope – “… the (linear) magnification due to the objective, namely h′/h, equals”
\( \displaystyle m_O=\frac{h^{\prime}}{h}=\frac{L}{f_o} \)\( \qquad (9.43) \)
The microscope – “… the (angular) magnification me due to it … when the final image is formed at the near point, is”
\( \displaystyle m_e=1+\frac{D}{f_e} \)\( \qquad [9.44(a)] \)
“When the final image is formed at infinity, the angular magnification due to the eyepiece … is”
\( \displaystyle m_e=\frac{D}{f_e} \)\( \qquad [9.44(b)] \)
“Thus, the total magnification … when the image is formed at infinity, is”
\( \displaystyle m=m_o m_e=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right) \)\( \qquad (9.45) \)