Answer to Question #113220 in Optics for Gautam pradhan

Question

(THE FOCAL LENGTH OF THE EYEPIECE OF A CERTAIN MICROSCOPE IS 18.0MMTHE FOCAL LENGTH OF 19.7CMTHE FINAL IMAGE FORMED BY THE EYEPIECE IS AT INFINITYTREAT ALL LENAES THIN A WHAT IS A DISTANCE FROM THE OBJECTIVE TO THE OBJECT BEGIN VIEWED B WHAT IS THE MAGNITUDE OF LINEAR MAGNIFICATION PRODUCED BY THE OBJECTIVE C WHAT IS THE OVERALL ANGULAR MAGNIFICATION OF MICROSCOPE)

Answer 1

As per the given question,

The focal length of the eyepiece $(f_e)= 18mm=1.8cm$

Focal length of the objective lens $(f_o)=19.7cm$

Let the length of the microscope =L

let the object position of objective lens is $u_o$ , image position of the objective lens $v_o$ similarly object position for the eyepiece lens is $u_e$ and image position is $v_e=\infty$

$L=u_e+v_o$

we know that distance of distinct vision D=25cm

But on the eye piece, image is formed at infinity So, $u_e=f_e$

$v_o=L-f_e$

Now, $\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}$

$\Rightarrow \dfrac{1}{v_o}=\dfrac{1}{1.8}+\dfrac{1}{\infty}$

$\Rightarrow L-f_e=f_o$

$\Rightarrow L=f_o+f_e =1.8+19.7=21.5cm$

magnification produced by objective lens $m_o=\dfrac{L}{f_o}=\dfrac{21.5}{19.7}=1.091$

Overall angular magnification$=m_oM_e=1.091\times \dfrac{D}{f_e}=1.09\times\dfrac{25}{1.8}=15.12$