An image formed on a screen is three times the size of the object. The object and screen are 80 cm apart when the image is sharply focussed.(a) State which type of lens is used.(b) Calculate focal length of the lens.

(a) Since the image is formed on a screen, it means that the image is real. And, we know that only a convex lens forms a real image, hence, a convex lens is used here.

(b) Given:

Magnification, $m$ = $-$3      $(\because image\ is\ real\ and\ inverted,\ 'm'\ will\ be\ negative)$

Distance between screen (image) and object $(-u+v)$ = 80 cm 

To find: Focal length, $f$.

Solution:

From the magnification formula, we know that-

$m=\frac {v}{u}$

Substituting the given values, we get-

$-3=\frac {80+u}{u}$        $[\because -u+v=80,\ then\ v=80+u]$

$-3u=80+u$   

$3u+u=-80$   

$4u=-80$   

$u=-\frac {80}{4}$   

$u=-20cm$ 

Thus, putting the value of $u$ in the equation given below-

 $v=80+u$

 $v=80+(-20)$

 $v=80-20$

 $v=60cm$

Thus, the object distance $u$ is 20 cm from the lens and image distance $v$ is 60 cm from the lens

Now,

From the lens formula, we know that-

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values, 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 of the lens is 15cm.

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Updated on: 10-Oct-2022

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