Answer to Question #197302 in Optics for Charles

Question

A microscope has an objective of focal length 1.0cm and an eyepiece of focal length 8.0cm.Both lenses are separated by a distance of 12.0 cm. A tiny object 1mm long is placed 1.2cm from the objective. Find the position and size of the final image

Answer 1

$u =$ distance of object from objective lens.

$v =$ image formed by objective lens.

$f_0 =$ objective focal length.

$f_e =$ eye piece focal length.

We know,

$\dfrac{1}{f_0} = \dfrac{1}{u}+\dfrac{1}{v}$

$1 = \dfrac{1}{v}-\dfrac{1}{1.2}$

$\dfrac{1}{v} = \dfrac{2.2}{1.2}$

$v = \dfrac{1.2}{2.2}$

$v= 0.54cm$

Distance of object from eye lens $= 12-0.54 = 11.46cm$

Now,

$u_1=$ distance of object from eye piece = -11.46cm

$v_1 =$ distance of image formed by eye piece

Hence,

$\dfrac{1}{f_e} = \dfrac{1}{u_1}+ \dfrac{1}{v_1}$

$\dfrac{1}{8} = -\dfrac{1}{11.46} + \dfrac{1}{v_1}$

$v_1 = 4.76cm$

We know,

Magnification $= -\dfrac{h_i}{h_0} = \dfrac{f_0}{f_e}$

$-\dfrac{h_i}{1} = \dfrac{1}{8}$

$h_i = 0.125mm$