(a) Define focal length of a pergent lens.(b) A pergent lens of focal length 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical center. Use lens formula to determine the distance of the object from the lens and the size of the image formed.(c) Draw a ray diagram to show the formation of image in the above situation.

(a) The focus $(F)$ of a divergent or concave lens is the point on the principal axis from which the light rays parallel to the axis is diverged after passing through the lens. The distance between the optical centre $(C)$ of the concave lens and principal focus $(F)$ is called focal length $(f)$.

(b) Given:

Focal length, $f=-30cm$

Size of the object, $h=6 cm$

Image distance, $v=-15 cm$

To find: Distance of the object from the lens, $(u)$ and the size of the image $(h')$.

Solution:

Applying lens formula:

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

Substituting the required values, we get-

$\frac {1}{(-30)}=\frac {1}{(-15)}-\frac {1}{u}$

$-\frac {1}{30}=-\frac {1}{15}-\frac {1}{u}$

$\frac {1}{u}=\frac {1}{30}-\frac {1}{15}$

$\frac {1}{u}=\frac {1-2}{30}$

$\frac {1}{u}=-\frac {1}{30}$

$u=-30cm$

Thus, the distance of the object from the lens, $(u)$ is 30 cm.

Now, 

Applying, magnification formula of the lens-

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

Putting the given values, we get-

$\frac {-15}{-30}=\frac {h'}{6}$

$\frac {1}{2}=\frac {h'}{6}$

$h'=\frac {6}{2}$

$h'=+3cm$

Thus, the size of the image is 3 cm.

(c) A ray diagram given below to show the formation of the image according to the above situation.

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Updated on: 10-Oct-2022

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