Answer :
Answer:
The magnification is [tex]m = -26.16[/tex]
Explanation:
From the question we are told that
The distance between the lenses is [tex]L =81.1 cm[/tex]
The focal length of the objective lens [tex]f_o = 78.0 \ cm[/tex]
Generally the length of the telescope that satisfies the condition for a relaxed eye can be mathematically represented as
[tex]L = f_0 + f_e[/tex]
Where [tex]f_e[/tex] is focal length of the eyepiece
Making [tex]f_e[/tex] the subject the of the formula above
[tex]f_e = L - f_o[/tex]
Now substituting value
[tex]f_e = 81.1 - 78.0[/tex]
[tex]=3.1\ cm[/tex]
Generally magnification can be mathematically represented as
[tex]m = - \frac{f_o}{f_e}[/tex]
This negative sign is a reminder that all real image are always inverted
[tex]=- \frac{81.1}{3.1}[/tex]
[tex]= - 26.16[/tex]