Arrange the magnifying power in the following cases of optical instruments in increasing order of their values.
(A) A compound microscope of objective of focal length f0 = 1.0 cm, an eyepiece of focal length 2.0 cm and tube length 20 cm, when the final image is at infinity.
(B) A small telescope with an objective lens of focal length 150 cm and an eyepiece of focal length 5.0 cm in normal adjustment.
(C) A simple microscope with a converging lens of focal length 5.0 cm.
(D) A compound microscope of objective of focal length f0 = 1.0 cm, an eyepiece of focal length 2.0 cm and tube length 20 cm, when the final image is formed at the near point.
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (B), (A), (D)
Work out each magnifying power, using D = 25 cm for the near point and the tube length L = 20 cm.
(A) Compound microscope, final image at infinity:
\( \displaystyle m_A = \frac{L}{f_o} \times \frac{D}{f_e} = \frac{20}{1.0} \times \frac{25}{2.0} = 250 \)
(B) Telescope in normal adjustment:
\( \displaystyle m_B = \frac{f_o}{f_e} = \frac{150}{5.0} = 30 \)
(C) Simple microscope:
\( \displaystyle m_C = \frac{D}{f} = \frac{25}{5.0} = 5 \)
With the image at the near point this becomes \( 1 + D/f = 6 \); the question does not say which, but either value is the smallest of the four.
(D) Compound microscope, final image at the near point:
\( \displaystyle m_D = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) = 20 \times 13.5 = 270 \)
So the increasing order is (C), (B), (A), (D).
Shortcut: A simple microscope (about 5) is always weaker than this telescope (30), and the compound microscopes (hundreds) are the strongest; the near-point setting (D) is always larger than the infinity setting (A). This gives (C), (B), (A), (D) with no calculation, and only Option (4) has it.