Question

How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are $1.25cm$ and $5cm$ respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of $30$ in normal adjustment.

Answer 1

(i)Telescope - It is an instrument used to see distant objects clearly by a magnified but apparent image . A telescope has an objective of large focal length and large aperature , whereas eyepiece is of comparatively smaller focal length and smaller aperature .In telescope size of image is smaller than actual size of object .By objective a smaller image is formed of the distant object ,which is adjusted between eyepiece and its focus so that we can get a larger final image of image formed by objective .

Microscope - It is an instrument used to see small objects clearly by a magnified but apparent image . A microscope has an objective of small focal length and small aperature , whereas eyepiece is of comparatively larger focal length and large aperature . In microscope size of image is larger than actual size of object . By objective a larger image is formed of the object (placed just outside the focus of objective) ,which is adjusted between eyepiece and its focus so that we can get a larger final image of image formed by objective .

(ii) Given - $f_o=1.25cm,f_e=5cm,M=30$ and we know $D=25cm$ ,

when the image is formed at infinity (normal adjustment) , angular magnification is given by ,

$M=-\frac{v_o}{u_o}.\frac{D}{f_e}=-\frac{v_o}{u_o}.\frac{25}{5}$ ,

$30=-5v_o/u_o$ ,

or $v_o=-6u_o$

now by lens formula , for objective lens

$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$ ,

$\frac{1}{1.25}=\frac{1}{-6u_o}-\frac{1}{u_o}$ ,

or $\frac{1}{1.25}=\frac{-7}{6u_o}$ ,

or $u_o=(-7\times 1.25)/6=-1.458cm$