An astronomical telescope is being used to examine a relativelyclose object that

Question

An astronomical telescope is being used to examine a relativelyclose object that is only 139.00 m away from the objective of thetelescope. The objective and eyepiece have focal lengths of 1.600and 0.0640 m, respectively. Noting that the expression M =fo/fe is no longerapplicable because the object is so close, use the thin-lens andmagnification equations to find the angular magnification of thistelescope.

Then lens equation : 1/Do + 1/Di = 1/F
Magnification equations : (for telescope)-Fo/Fe(1-Fe/Die)        ˜-Fo/Fe

But because it is close Fo/Fe no longer applicable... Does thatmean that L (when you work backwards to obtain eq : M = -Fo(L+Doo)/Doo(L - Fo)-LFo     ) is = to zeroor L = Fo+Fe

I don't understand what happens when the object is so close and Idon't understand how to combine the thin lens eq with themagnification eq if the magnification eq for the telescope is just-Fo/Fe(1-Fe/Die)
I do understand that Doo is no longer =

Explanation / Answer

   from the theory we get the thin lens equation as    (1 / di) = (1 / fo) - (1 /do)               =(1 / 1.600 m) - (1 / 139.00 m)    di = ........ m    now the magnification will be    M = - (di / do)        = - (di /139.00 m)    now we use this first image as an object for thesecond lens    (1 / di') = (1 / fe) - (1 /do')               =(1 / 0.0640 m) - (1 / do')    di' = ........ m    in this case the magnification will be    M ' = - (di' / do')    the total magnification will be    M1 = M M'    the angular magnification is then    ' = - (hi' /di')    = ho / (do+ fo + fe)    Mang = ' /

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