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.

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Updated on: 10-Oct-2022

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