Question

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.

Answer 1

Answer:

$0.848\ \text{cm}$

$232.66$

Explanation:

N = Near point of eye = 25 cm

$f_o$ = Focal length of objective = 0.8 cm

$f_e$ = Focal length of eyepiece = 1.8 cm

l = Distance between the lenses = 16 cm

Object distance is given by

$v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}$

$u_o$ = Object distance for objective

From lens equation we have

$\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}\\\Rightarrow u_o=\dfrac{f_ov_o}{v_o-f_o}\\\Rightarrow u_o=\dfrac{0.8\times 14.2}{14.2-0.8}\\\Rightarrow u_o=0.848\ \text{cm}$

The position of the object is $0.848\ \text{cm}$.

Magnification of eyepiece is

$M_e=\dfrac{N}{f_e}\\\Rightarrow M_e=\dfrac{25}{1.8}\\\Rightarrow M_e=13.89$

Magnification of objective is

$M_o=\dfrac{v_o}{u_o}\\\Rightarrow M_o=\dfrac{14.2}{0.848}\\\Rightarrow M_o=16.75$

Total magnification is given by

$m=M_eM_o\\\Rightarrow m=13.89\times 16.75\\\Rightarrow m=232.66$

The total magnification is $232.66$.