Q28. An astronomical telescope has fₒ = 120 cm, fₑ = 6 cm and is in normal adjustment. Its eyepiece is moved 1 cm away from the objective without moving the objective or object. The final image is:
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Question

Q28. An astronomical telescope has fₒ = 120 cm, fₑ = 6 cm and is in normal adjustment. Its eyepiece is moved 1 cm away from the objective without moving the objective or object. The final image is:

Options

G1 Physics · Accepted Answer

GIVEN: Astronomical telescope fₒ = 120 cm, fₑ = 6 cm; initial normal length 126 cm; eyepiece moved 1 cm away. Step 1: Objective image of distant object remains 120 cm from objective. Step 2: New eyepiece position = 120 + 6 + 1 = 127 cm from objective. Step 3: Thus eyepiece object distance uₑ = 120 − 127 = −7 cm (real intermediate image on its left). Step 4: Eyepiece lens formula: 1/vₑ = 1/fₑ + 1/uₑ = 1/6 − 1/7 = (7 − 6)/42 = 1/42. Step 5: vₑ = +42 cm; positive vₑ means a real final image to the…