Bošković, Ruđer Josip
,
Abhandlung von den verbesserten dioptrischen Fernröhren aus den Sammlungen des Instituts zu Bologna sammt einem Anhange des Uebersetzers
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{(m - 1) d r′/f′}. </
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">Soiſt demnach d r = - {f′ d m/(m - 1)
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}.
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">Setzet man dieſes mit RR ({d m/f} + {d M/g}) gleich,
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ſo wird - {f′ d m/(m - 1)
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RR} = {d m/f} + {d M/g},
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oder {f′/(m - 1)
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RR} + {1/f} = - {d M/d m} X {1/g}.</
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<
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">Nimmt man anſtatt {1/RR} ſeinen Werth,
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ſo bekommt man verſchiedene Theile, derer eini-
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ge {1/g
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}, andre {1/g} enthalten, aber auch einige
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ohne {1/g}: </
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">und aus dieſen entſtehet eine Glei-
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chung des zweyten Grades für {1/g}, das man aus
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f′, f, M, m und {d M/d m} findet.</
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<
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">Jedoch weil die Größe y nicht gar ſo klein
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iſt, geziemet es ſich, daß wir für {1/R} jenen
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Werth annehmen, den die gänzliche Verbeſſe-
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rung erfodert, und wir (81) = {d m/f′} X ({m - 1/d m}
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- {M - 1/d M}) gefunden haben. </
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<
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">Auf dieſe Weiſe
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wird {1/(m - 1) R} = {1/f} X (1 - {M - 1/m - 1} X
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{d m/d M}), und die vorige Gleichung {f′/(m - 1)
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