Q41. An astronomical telescope has eyepiece focal length 5 cm. With final image at D = 25 cm, its tube length is 104 1/6 cm. The normal-adjustment tube length and near-point angular magnification magnitude are respectively:
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Question

Q41. An astronomical telescope has eyepiece focal length 5 cm. With final image at D = 25 cm, its tube length is 104 1/6 cm. The normal-adjustment tube length and near-point angular magnification magnitude are respectively:

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G1 Physics · Accepted Answer

GIVEN: Telescope eyepiece fₑ = 5 cm; near point D = 25 cm; near-point length Lᴅ = 104 1/6 = 625/6 cm. Step 1: Eyepiece object distance for final image at near point: s = fₑD/(fₑ + D) = (5 × 25)/(5 + 25) = 25/6 cm. Step 2: Objective focal length fₒ = Lᴅ − s = 625/6 − 25/6 = 600/6 = 100 cm. Step 3: Normal-adjustment length L∞ = fₒ + fₑ = 100 + 5 = 105 cm. Step 4: Near-point angular magnification |Mᴅ| = (fₒ/fₑ)(1 + fₑ/D). Step 5: |Mᴅ| = (100/5)(1 + 5/25) = 20 × (6/5) = 24. FINAL ANSWER: 105 cm and…