Q39. An astronomical telescope has normal-adjustment angular magnification magnitude 10 and lens separation 110 cm. It is readjusted for a final image at D = 25 cm. Its new tube length is:
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Q39. An astronomical telescope has normal-adjustment angular magnification magnitude 10 and lens separation 110 cm. It is readjusted for a final image at D = 25 cm. Its new tube length is:

G1 Physics · Accepted Answer

GIVEN: Telescope in normal adjustment has |M| = 10 and length L∞ = 110 cm. Near point D = 25 cm. Step 1: From |M| = fₒ/fₑ = 10, obtain fₒ = 10 fₑ. Step 2: Normal length fₒ + fₑ = 110 ⇒ 11 fₑ = 110 ⇒ fₑ = 10 cm, fₒ = 100 cm. Step 3: For final virtual image at D, eyepiece object distance s = fₑD/(fₑ + D). Step 4: s = (10 × 25)/(10 + 25) = 250/35 = 50/7 cm = 7.142857 cm. Step 5: New tube length L = fₒ + s = 100 + 50/7 = 750/7 = 107 1/7 cm. FINAL ANSWER: 107 1/7 cm. Option B.