Answer:
The magnification would be "103.55". A further explanation is given below.
Explanation:
The given values are:
Distance between lens and eyepiece,
L = 85 cm
Eyepiece is,
= 123 D
Now,
The refractive power of eye piece will be:
⇒ [tex]\frac{1}{f_e}=123D[/tex]
[tex]f_e=\frac{1}{123D}[/tex]
[tex]f_e=0.813 \ cm[/tex]
The length of the telescope will be:
⇒ [tex]L=f_0+f_e[/tex]
⇒ [tex]f_0=L-f_e[/tex]
On substituting the values, we get
⇒ [tex]=85-0.813[/tex]
⇒ [tex]=84.187 \ cm[/tex]
Now,
The magnification of the telescope will be:
⇒ [tex]M=\frac{f_0}{f_e}[/tex]
⇒ [tex]=\frac{84.187}{0.813}[/tex]
⇒ [tex]=103.55[/tex]
