Question

A compound microscope consists of an objective lens of focal length 1 cm and eye piece of focal length 5 cm seperated by a distance 12.2 cm. At what distance from the objective lens should the object be placed so as to get the final image at least distance of distinct vision. Calculate the angular magnification also.

Answer 1

given, $f_0=1cm,f_e=5cm$ and L = 12.2 cm

so, final image seen at least division of distinct vision (25cm) , $v_e=-25cm$

now using formula, $\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$

or, 1/5 = 1/-25 - 1/$u_e$

or, 1/5 + 1/25 = -1/$u_e$

or, -6/25 = 1/$u_e$

or, $u_e$ = -25/6 = -4.2 cm

given length of tube , L = $|v_0|+|u_e|$ = 12.2 cm

or, $|v_0|+4.2=12.2\implies |v_0|=8cm$

again using lens maker formula,

$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$

or, 1/1 = 1/8 - 1/$u_0$

$u_0$ = -8/7 = -1.1 cm

now , angular magnification, m = $\frac{v_0}{-|u_0|}\left(1+\frac{D}{f_e}\right)$

= 8/-(-1.1)[ 1 + 25/5]

= (8/1.1) × 6

= 48/1.1

= 43.6

Answer 2

Given :

$\bullet \: \tt v_e = - 25 \: cm \\ \\ \bullet \: \tt f_e = + 5 \: cm$

To Find :

• Distance of the object
• Angular magnification in this case

Solution :

By using

$\tt \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \\ \\ \tt \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \\ \\ \tt \implies \frac{1}{u_e} = - \frac{1}{25} - \frac{1}{ 5} \\ \\ \tt \implies \frac{1}{u_e} = \frac{ - 1 - 5}{25} \\ \\ \tt \implies \frac{1}{u_e} = - \frac{6}{25} \\ \\ \tt \implies u_e = - \frac{25}{6} \\ \\\tt \implies u_e ≈ - 4.2 \: cm$

As the object is 12.2 cm away from the eyepiece, the image formed by the objective is 12.2 – 4.2 = 8.0 cm away from it. For the objective,

$\tt v = + 8.0 \: cm \\ \\ \tt f_o = + 1.0 \: cm$

By using

$\tt \frac{1}{v} - \frac{1}{u} = \frac{1}{f_o} \\ \\ \tt \frac{1}{u} = \frac{1}{v} - \frac{1}{f_o} \\ \\ \tt \implies \frac{1}{u} = \frac{1}{8.0} - \frac{1}{1.0} \\ \\\implies \tt \frac{1}{u} = \frac{1 - 8.0}{8.0} \\ \\ \implies\tt \frac{1}{u} = - \frac{7.0}{8.0} \\ \\\implies \tt u = - \frac{8.0}{7.0} \\ \\\large \boxed{ \tt u = - 1.1 \: cm}$

The angular magnification is

$\tt m = \dfrac{v}{u} \bigg(1 + \dfrac{d}{f_e} \bigg) \\ \\ \tt m = \frac{ + 8.0}{ - 1.1} \bigg(1 + \frac{25}{5} \bigg) \\ \\ \large \boxed{ \tt m ≈ - 44}$