Question

An object is placed in front of a diverging lens whose focal length is - 27.0 cm. A virtual image is formed whose height is 0.516 times the object height. How far is the object from the lens?

Answer 1

Given :

The lens is a diverging lens.

f = focal length of lens = - 27 cm

M = magnification = 0.516.

u = object distance = to be calculated.

Now, as the lens is diverging and the object is real, it means the image formed is virtual and erect and diminished.

We will be using cartesian sign convention, in which measurement in the direction of light is taken positive and opposite to it, it is taken negative. And measurements are taken from center of lens.

M = v/u [v = image distance]

=> 0.516 = v/u

=> v = 0.516u.

Using lens's euqation :

. [-ve sign denotes that the object is real. real object distance is always negative according to cartesian sign convention]