A camera fitted with a lens of focal length 50 mm is being used to photograph a flower that is 5 cm in diameter. The flower is placed 20 cm in front of the camera lens.

(a) At what distance from the film should the lens be adjusted to obtain a sharp image of the flower?

(b) What would be the diameter of the image of the flower on the film?

(c) What is the nature of camera lens?

Focal length, $f$ = 50 mm

Object distance, $u$ = $-$20 cm = $-$200 mm

To find: Image distance, $v$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-200)}=\frac {1}{50}$

$\frac {1}{v}+\frac {1}{200}=\frac {1}{50}$

$\frac {1}{v}=\frac {1}{50}-\frac {1}{200}$

$\frac {1}{v}=\frac {4-1}{200}$

$\frac {1}{v}=\frac {3}{200}$

$v=\frac {200}{3}$

$v=+66.6mm=6.66cm$

Thus, the film should be at a distance of 66.6 cm, and the positive $(+)$ sign implies that the image is formed behind the camera lens.

(b) Given:

Diameter of object, $h$ = 5 cm = 50 mm

Object distance, $u$ = $-$20 cm = $-$200 mm

Image distance, $v$ = 66.6 mm

To find: Diameter of image, $h'$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values, we get-

$\frac {66.6}{-200}=\frac {v}{50}$

$-\frac {666}{2000}=\frac {v}{500}$              

$-2000v=50\times {666}$              

$h'=-\frac {50\times {666}}{2000}$

$h'=-\frac {33,300}{2000}$              

$h'=-16.65mm=1.66cm$       

Thus, the diameter of the image $h'$ of the flower on the film is 1.66 cm 

(c) The nature of the lens is converging or convex, as the image is formed behind the lens.