Bošković, Ruđer Josip, Abhandlung von den verbesserten dioptrischen Fernröhren aus den Sammlungen des Instituts zu Bologna sammt einem Anhange des Uebersetzers

Table of Notes

< >
[Note]
Page: 128
[Note]
Page: 129
[Note]
Page: 129
[Note]
Page: 136
[Note]
Page: 137
[Note]
Page: 142
[Note]
Page: 149
[Note]
Page: 149
[Note]
Page: 149
[Note]
Page: 149
[Note]
Page: 150
[Note]
Page: 150
[Note]
Page: 150
[Note]
Page: 151
[Note]
Page: 151
[Note]
Page: 186
[Note]
Page: 186
[Note]
Page: 188
< >
page |< < (34) of 199 > >|
    <echo version="1.0RC">
      <text xml:lang="de" type="free">
        <div xml:id="echoid-div10" type="section" level="1" n="7">
          <p>
            <s xml:id="echoid-s366" xml:space="preserve">
              <pb o="34" file="0038" n="38" rhead="Abhandlung"/>
            ben werde; </s>
            <s xml:id="echoid-s367" xml:space="preserve">ſuchet man den Abſtand D L des
              <lb/>
              <note position="left" xlink:label="note-0038-01" xlink:href="note-0038-01a" xml:space="preserve">Fig. 3.
                <lb/>
              Tab. I.</note>
            Brennpunkts L von der nächſten Fläche. </s>
            <s xml:id="echoid-s368" xml:space="preserve">Fig. </s>
            <s xml:id="echoid-s369" xml:space="preserve">3
              <lb/>
            Tab. </s>
            <s xml:id="echoid-s370" xml:space="preserve">I.</s>
            <s xml:id="echoid-s371" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s372" xml:space="preserve">51. </s>
            <s xml:id="echoid-s373" xml:space="preserve">Der Abſtand B C der zwey Gläſer ſey
              <lb/>
            = β; </s>
            <s xml:id="echoid-s374" xml:space="preserve">die Dicke des zweyten Glaſes = γ;
              <lb/>
            </s>
            <s xml:id="echoid-s375" xml:space="preserve">der halbe Durchmeſſer der Vörderfläche des
              <lb/>
            zweyten Glaſes = c, der halbe Durchmeſ-
              <lb/>
            ſer der Hinterfläche D P = d: </s>
            <s xml:id="echoid-s376" xml:space="preserve">das Verhält,
              <lb/>
            niß der Sinus des Einfalls- und Brechungswin-
              <lb/>
            kels M : </s>
            <s xml:id="echoid-s377" xml:space="preserve">1, da die Straalen aus der Luft in
              <lb/>
            das Glas kommen, die Brennweite des zwey-
              <lb/>
            ten Glaſes für die mit der Achſe parallel, und
              <lb/>
            unendlich nahe einfallenden Straalen = H. </s>
            <s xml:id="echoid-s378" xml:space="preserve">
              <lb/>
            Nun muß man aus den halben Durchmeſſern
              <lb/>
            c, d, aus M, H, der Oeffnung e, und den
              <lb/>
            Abſtande C I des Punktes I (nach welchem die
              <lb/>
            Straalen gerichtet ſind) von der erſten Fläche CO,
              <lb/>
            eben ſo die Länge D L beſtimmen, wie wir
              <lb/>
            oben B I aus a, b, m, h, e und A G oder p
              <lb/>
            gefunden haben.</s>
            <s xml:id="echoid-s379" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s380" xml:space="preserve">52. </s>
            <s xml:id="echoid-s381" xml:space="preserve">Der Werth des B I iſt r - {r
              <emph style="super">2</emph>
            m a/q
              <emph style="super">2</emph>
            } -
              <lb/>
            r
              <emph style="super">2</emph>
            ρ. </s>
            <s xml:id="echoid-s382" xml:space="preserve">Wäre dieſer nur allein r, könnten wir
              <lb/>
            auf eben die Weiſe die Größen Q, R, σ aus
              <lb/>
            c, d, M, g, r beſtimmen, auf welche wir die
              <lb/>
            ähnlichen q, r, ρ aus a, b, m, f, p erhalten
              <lb/>
            haben, und alsdenn gälte D L = R -
              <lb/>
            {R
              <emph style="super">2</emph>
            M γ/Q
              <emph style="super">2</emph>
            } - R
              <emph style="super">2</emph>
            σ. </s>
            <s xml:id="echoid-s383" xml:space="preserve">Weil aber C I um {r
              <emph style="super">2</emph>
            m a/q
              <emph style="super">2</emph>
            } +
              <lb/>
            β + r
              <emph style="super">2</emph>
            ρ kleiner iſt, als r, muß man (48
              <lb/>
            gemäß) noch dieſen Unterſchied mit {DL
              <emph style="super">2</emph>
            /B I
              <emph style="super">2</emph>
            }, </s>
          </p>
        </div>
      </text>
    </echo>
Impressum