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The filament of a lamp is 80 cm from a screen and a converging lens forms an image of it on a screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.
Given:
Converging lens is a convex lens.
Magnification, $m$ = $-$3 (negative sign shows that the image is real since it forms on the screen)
Distance of the lens from filament = $u$
Distance of the lens from screen= $v$
To find: Distance of the lens from the filament and the focal length $f$ of the lens.
Solution:
According to the question-
$-u+v=80\ cm$ ............................. (i)
From the magnification formula, we know that-
$m=\frac {v}{u}$
Substituting the given values in the formula we get-
$-3=\frac {v}{u}$
$v=-3u$
Now, putting the value of $v$ in eq. (i), we get-
$-u+(-3u)=80\ cm$
$-u-3u=80\ cm$
$-4u=80\ cm$
$u=-\frac {80\ cm}{4}$
$u=-20cm$
Thus, the distance of the lens from the filament is 20cm.
Putting the value of $u$ in eq. (i) we get-
$-20\ cm+v=80\ cm$
$v=80\ cm-20\ cm$
$v=60\ cm$
Thus, the distance of the screen from the lens is 60cm.
For the focal length of the lens, we have the lens formula-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the value of $u$ and $v$ in the formula we get-
$\frac {1}{60}-\frac {1}{(-20)}=\frac {1}{f}$
$\frac {1}{60}+\frac {1}{20}=\frac {1}{f}$
$\frac {1}{f}=\frac {1+3}{60}$
$\frac {1}{f}=\frac {4}{60}$
$\frac {1}{f}=\frac {1}{15}$
$f=15cm$
Thus, the focal length $f$ of the lens is 15cm.