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A microscope has an eyepiece with a 1.8 cm focal length and a 0.8 cm focal length objective lens. Assuming a relaxed eye, calculate a) the position of the object if the distance between the lenses is 16 cm, and b) the total magnification.

Sagot :

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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].

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