Answer:
[tex]0.848\ \text{cm}[/tex]
[tex]232.66[/tex]
Explanation:
N = Near point of eye = 25 cm
[tex]f_o[/tex] = Focal length of objective = 0.8 cm
[tex]f_e[/tex] = Focal length of eyepiece = 1.8 cm
l = Distance between the lenses = 16 cm
Object distance is given by
[tex]v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}[/tex]
[tex]u_o[/tex] = Object distance for objective
From lens equation we have
[tex]\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}\\\Rightarrow u_o=\dfrac{f_ov_o}{v_o-f_o}\\\Rightarrow u_o=\dfrac{0.8\times 14.2}{14.2-0.8}\\\Rightarrow u_o=0.848\ \text{cm}[/tex]
The position of the object is [tex]0.848\ \text{cm}[/tex].
Magnification of eyepiece is
[tex]M_e=\dfrac{N}{f_e}\\\Rightarrow M_e=\dfrac{25}{1.8}\\\Rightarrow M_e=13.89[/tex]
Magnification of objective is
[tex]M_o=\dfrac{v_o}{u_o}\\\Rightarrow M_o=\dfrac{14.2}{0.848}\\\Rightarrow M_o=16.75[/tex]
Total magnification is given by
[tex]m=M_eM_o\\\Rightarrow m=13.89\times 16.75\\\Rightarrow m=232.66[/tex]
The total magnification is [tex]232.66[/tex].
