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

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 $m= \frac{h_{i} }{h_{o} } = \frac{v}{u}$

                        ⇒ $\frac{h_{i} }{h_{o} } =\frac{v}{u}$

                        ⇒$\frac{h_{i} }{18} =\frac{-4}{-12}$

                        ⇒$h_{i} =\frac{18*[-4]}{[-12]}$

                                      = 6 mm [ans]

Hence B is right.

Answer 2

The answer is

18 / x = 12 / 4 
12x = 72
x = 6mm