Question

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

Answer 1

The answer is

18 / x = 12 / 4
12x = 72
x = 6mm

Answer 2

As per the question the height of object
`[h_(0) ]` 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
`[h_(i) ]` 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

`h_(0) =18 mm` [the height is positive as it is above the principal axis]

As per the magnification formula we know that

Magnification
`m= (h_(i) )/(h_(o) ) = (v)/(u)`

⇒
`(h_(i) )/(h_(o) ) =(v)/(u)`

⇒
`(h_(i) )/(18) =(-4)/(-12)`

⇒
`h_(i) =(18*[-4])/([-12])`

= 6 mm [ans]

Hence B is right.