Q22. An astronomical telescope has fₒ = 100 cm and fₑ = 5 cm. The final image is at the near point D = 25 cm. Angular magnification magnitude and tube length are approximately:
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Q22. An astronomical telescope has fₒ = 100 cm and fₑ = 5 cm. The final image is at the near point D = 25 cm. Angular magnification magnitude and tube length are approximately:

G1 Physics · Accepted Answer

GIVEN: Telescope fₒ = 100 cm; fₑ = 5 cm; eye near point D = 25 cm; final image virtual at −25 cm. Step 1: Eyepiece imaging: uₑ = −s, vₑ = −D. Hence 1/fₑ = −1/D + 1/s. Step 2: 1/s = 1/5 + 1/25 = 6/25 ⇒ s = 25/6 = 4.1667 cm. Step 3: Tube length L = fₒ + s = 100 + 25/6 = 104.1667 cm ≈ 104.17 cm. Step 4: Angular magnification |M| = fₒ/s = 100/(25/6) = 600/25 = 24. Step 5: Equivalently |M| = (fₒ/fₑ)(1 + fₑ/D) = 20 × 1.2 = 24. FINAL ANSWER: 24 and 104.17 cm. Option B.