Q35. A compound microscope has fₒ = 1 cm and fₑ = 5 cm. An object is placed 1.2 cm in front of the objective. For final image at 25 cm from the eyepiece, the required tube length is:
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Q35. A compound microscope has fₒ = 1 cm and fₑ = 5 cm. An object is placed 1.2 cm in front of the objective. For final image at 25 cm from the eyepiece, the required tube length is:

G1 Physics · Accepted Answer

GIVEN: Compound microscope fₒ = 1 cm, fₑ = 5 cm; object 1.2 cm before objective; final virtual image at D = 25 cm. Step 1: Objective object distance uₒ = −1.2 cm = −6/5 cm. Step 2: Objective lens formula: 1/fₒ = 1/vₒ − 1/uₒ ⇒ 1 = 1/vₒ + 5/6. Step 3: 1/vₒ = 1 − 5/6 = 1/6 ⇒ vₒ = 6 cm. Step 4: Eyepiece requires intermediate image at distance s = fₑD/(fₑ + D) = (5 × 25)/(5 + 25) = 125/30 = 25/6 cm. Step 5: Tube length L = vₒ + s = 6 + 25/6 = (36 + 25)/6 = 61/6 cm = 10.1667 cm. FINAL ANSWER: 61/6 cm…