Question

An object is placed on the principal axis of a concave lens at a distance of 40 cm from it. If the focal length of the lens is also 40 cm, find the location of the image and the magnification. [3 MARKS]

Answer 1

Lens formula: 1 Mark
position: 1 Mark
magnification: 1 Mark

For a concave lens focal length f is negative, i.e. f = - 40 cm.

By sign convention, object is placed on the left of the lens, so u = - 40 cm.

Using the lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get

$\frac{1}{v}-\frac{1}{-40}=\frac{1}{-40}$

$v=-20cm$

The image is formed 20 cm from the lens. Negative sign shows that the image is formed on the same side of the lens as the object.

Now, magnification, $m=\frac{h^{{}^{\prime }}}{h}=\frac{v}{u}=\frac{-20}{-40}=\frac{1}{2}$

Positive sign shows that the image is erect and upwards.