Question

A diverging lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens? Find the size of the image also.

Answer 1

Hint: For a diverging lens, the distance of the object from the optical centre is equal to the distance of the image of the optical centre. And in this case of diverging lens both the distance of object and the distance of image from the optical centre is negative.

Complete step-by-step answer:
Given: Focal length $f = - 30\,{\text{cm}}$.
Distance of the image from the lens $v = - 15\,{\text{cm}}$.
Height of the object is ${h_1} = 5\,{\text{cm}}$
We will write the lens formula to find the distance of the object from the lens.
\[\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}\]
Here u is the distance of the image from the optical centre.
Substitute $f = - 30\,{\text{cm}}$ and $v = - 15\,{\text{cm}}$ to find the value of $u$.
$
  \dfrac{1}{u} = \dfrac{1}{{ - 15}} - \dfrac{1}{{ - 30}} \\
  u = - 30\,{\text{cm}} \\
 $
We know that magnification is given by:
$m = \dfrac{v}{u} = \dfrac{{{h_2}}}{{{h_1}}}$
$\dfrac{v}{u} = \dfrac{{{h_2}}}{{{h_1}}}$
Substitute, $v = - 15\,{\text{cm}}$, $u = - 30\,{\text{cm}}$ and ${h_1} = 5\,{\text{cm}}$ to find the value of ${h_2}$.
$
  \dfrac{{ - 15\,{\text{cm}}}}{{ - 30\,{\text{cm}}}} = \dfrac{{{h_2}}}{{5\,{\text{cm}}}} \\
  {h_2} = 2.5\,{\text{cm}} \\
 $
Therefore, the object should be placed 30 cm from the optical centre and the size of the image formed is 2.5 cm.

Additional information: The distance between focal point and lens is called focal length. The focal length is always positive for converging lenses while the focal length is always negative for the diverging length.

Note:For diverging lenses, the focal length is negative and it is the distance from the point where the combination of beams appear to be diverging after passing through the lens. The focal length is positive for diverging lenses.