Correct Answer - a.
Each half lens will form an image in the same plane. The optic axes of the lenses are displaced,
`(1)/(upsilon)-(1)/((-30))=(1)/(20),upsilon=60cm`
From similar triangle `OI_(1)I_(2)` and `OP_(1)P_(2)` , we have
`(I_(1)I_(2))/(P_(1)P_(2))=(u+upsilon)/(u)I_(1)I_(2)=(90)/(30)(2xx0.05)=0.3cm`
Thus, the two images are 0.3cm apart. Alternatively, imagine two arrows(see figure) that act as objects for the lens.
Magnification, `m=(upsilon)/(u)=((+60))/((-30))=-2`
Image of height of arrow is `y=2xx(0.005)=0.10cm`
Thus, two inverted images are formed whose tips are at `I_(1)` and `I_(2)` respectively.
Thus, `I_(1)I_(2)=2y+Delta=(2xx0.01)+0.1=0.3cm`
