As per the question the height of object [tex][h_{0} ][/tex] 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[tex][h_{i} ][/tex] 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
[tex]h_{0} =18 mm[/tex] [the height is positive as it is above the principal axis]
As per the magnification formula we know that
Magnification [tex]m= \frac{h_{i} }{h_{o} } = \frac{v}{u}[/tex]
⇒ [tex]\frac{h_{i} }{h_{o} } =\frac{v}{u}[/tex]
⇒[tex]\frac{h_{i} }{18} =\frac{-4}{-12}[/tex]
⇒[tex]h_{i} =\frac{18*[-4]}{[-12]}[/tex]
= 6 mm [ans]
Hence B is right.
