(4)

Focal length of objective lens, $f_o=2\text{cm}$

Focal length of eyepiece lens, $f_e=4\text{cm}$

Distance of distinct vision, $D=25\text{cm}$

         $m=\frac{L}{f_0}\times \frac{D}{f_e}=\frac{40}{2}\times \frac{25}{4}$

         $m=125$

(2)

Given, $f_0=140\text{cm}$  and  $f_e=5.0\text{cm}$

Magnifying power of a telescope, $M=\frac{f_0}{f_e} \therefore M=\frac{140}{5.0}=28$

(4)

As the focal length is large, so it enhances the magnifying power of telescope. The large aperture or diameter of lens helps in collecting large amount of light from the object so that the image formed is bright.

The resolving power of telescope is

            $\text{R.P.}=\frac{D}{1.22\lambda }$

where, D is diameter or aperture of lens and $\lambda$ is wavelength of light used.

So, all options are correct.

(2)

Here $f_o=40\text{cm,}$ $f_e=4\text{cm}$

Tube length $\left(l\right)$ = Distance between lenses = $v_o+f_e$

For objective lens,

       $u_o=-200\text{cm}, v_o= ?$

       $\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o} \text{or} \frac{1}{v_o}-\frac{1}{-200}=\frac{1}{40}$

or    $\frac{1}{v_o}=\frac{1}{40}-\frac{1}{200}=\frac{4}{200} \therefore v_o=50\text{cm}$

$\therefore l=50+4=54$$\text{cm}$

(2)

For eye-piece lens,

$m=\frac{f}{f+u}=\frac{h_1}{h_0}\Rightarrow \frac{f_e}{f_e-(f_o+f_e)}=\frac{I}{L}$

$\Rightarrow \frac{f_o}{f_e}=-\frac{L}{I}=\text{Magnification of the telescope}$

(4)

Magnifying power of a microscope,

         $m=\left(\frac{{\displaystyle L}}{{\displaystyle f_o}}\right)\left(\frac{{\displaystyle D}}{{\displaystyle f_e}}\right)$

where $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece respectively, and L is the distance between their focal points and D is the least distance of distinct vision.

If $f_o$ increases, then $m$ will decrease.

Magnifying power of a telescope, $m=\frac{f_o}{f_e}$

where $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece respectively.

If $f_o$ increases, then $m$ will increase.