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

Solution:

At 30 am distance an object is placed so that its image is formed 15 cm away from the lens and the size of the image is also 2.5 cm.
The distance of the object from the optical centre is equal to the distance of the image of the optical centre for a diverging lens. In the case of a diverging lens, both the object distance and the image distance from the optical centre are negative.

Given: Focal length f=-30 cm.
Distance of the image from the lens v=-15 cm.
The height of the object is h₁=5 cm. 

We will write the lens formula to find the distance of the object from the lens.

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

Here u is the distance of the image from the optical centre.
To find the value of u:

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

u = -30 cm

We know that magnification is given by:

$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}$
$\Rightarrow \frac{\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}$

Substitute, v = -15 cm, u = -30 cm and h₁ = 5 cm to find the value of h₂.

$\frac{-15 \mathrm{cm}}{-30 \mathrm{cm}}=\frac{{\mathrm{h}}_2}{5 \mathrm{cm}}$

h₂ = 2.5 cm.

As a result, the object should be located 30 cm from the optical centre, and the picture formed should be 2.5 cm in size.
Further information: The focal length is the distance between the focus point and the lens. For converging lenses, the focal length is always positive, but for diverging lenses, the focal length is always negative.
Hence, the correct answer is 30cm, 2.5 cm.