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標題:
F.5 Physics Homework
發問:
The focal lengths of the objective and the eye-piece of a compound microscope are 10 mm and 20 mm respectively, their separation being 60 mm. Where must a small object be placed in order that the image seen by the observer may be 0.12 m from the eye-piece? What is the overall magnification?
Apply the thin lens formula to the eye-piece v = -0.12 m(=-120 mm), f = 20 mm hence, 1/u - 1/120 = 1/20 u = 17.14 mm The object of the eye-piece is indeed the image produced by the objective. Apply the thin lens formula to the objective, v = (60-17.14) mm = 42.86 mm, f = 10 mm 1/u + 1/42.86 = 1/10 u = 13 mm The object should be placed at distance of 13 mm from the objective. Mangitfication = (42.86/13)x(120/17.14) = 23
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F.5 Physics Homework
發問:
The focal lengths of the objective and the eye-piece of a compound microscope are 10 mm and 20 mm respectively, their separation being 60 mm. Where must a small object be placed in order that the image seen by the observer may be 0.12 m from the eye-piece? What is the overall magnification?
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最佳解答:Apply the thin lens formula to the eye-piece v = -0.12 m(=-120 mm), f = 20 mm hence, 1/u - 1/120 = 1/20 u = 17.14 mm The object of the eye-piece is indeed the image produced by the objective. Apply the thin lens formula to the objective, v = (60-17.14) mm = 42.86 mm, f = 10 mm 1/u + 1/42.86 = 1/10 u = 13 mm The object should be placed at distance of 13 mm from the objective. Mangitfication = (42.86/13)x(120/17.14) = 23
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