Question

A lens forms an image of an object which is placed at a distance 15 cm in front of it. if the image forms at a distance 60 cm on the same side Find : (i) the focal length, (ii) the magnification, and (iii) the nature of image.

Answer 1

Step 1: Given that:

For a lens;

Object distance,u= $\text{-15 cm}$
Image distance,v= $\text{-60 cm}$

(Note:

1. The object is placed left to the optical centre of a lens therefore the sign is taken as negative.
2. As in the given question, the image is also formed on the same side of the object then the image distance is also taken as negative.)​

Step 2: i) Calculation of focal length of the lens:

Lens formula is
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Where; f = focal length of the lens, u = Object distance and v= image distance

$\frac{1}{f}=-\frac{1}{60}+\frac{1}{15}$

$\frac{1}{f}=\frac{-1+4}{60}$

$\frac{1}{f}=\frac{3}{60}$

$f=+20cm$

Step 3: ii) Calculation of magnification of the lens:

For a lens, magnification is; $m=\frac{h_i}{h_o}=\frac{v}{u}$

Where; $h_i$ is the height of the image and $h_o$ is the height of the object.

$m=\frac{-60}{-15}$

$m=4$

Step 4:iii) Determination of the nature of the image:

As. m= +4

Then, the image is virtual and erect.

Thus,

Focal length of the lens = +20cm

Magnification = +4

Nature of image = Virtual and erect