If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, what is the height of the image? a. 14 millimeters b. 6 millimeters c. 22 millimeters d. 8 millimeters

As per the question the height of object  18 mm

The object distance [u] is given as 12 mm

The image distance[v] is given as 4 mm

The lens taken here is a diverging nature.hence the lens is a concave lens.

The image formed in a concave lens is always virtual,erect and diminished.

As per the question the image is formed in front of the lens.

we have to calculate the image height  which is calculated as follows-

putting the sign convention on the above data we get

u= -12 mm              v=-- -4 mm     [ - sign is due to the fact that the measurement is opposite to the direction of light

               [the height is positive as it is above the principal axis]

As per the magnification formula we know that

Magnification

                        ⇒

                        ⇒

                        ⇒

                                      = 6 mm [ans]

Hence B is right.

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary