For both the halves, position of object and image is same. Only difference is of
magnification. Magnification for one of the halves is given as `2(gt1)`. Thus can be for the first
one, because for this, `|v|gt|u|`. Therefore, magnification, `|m|=|v//u|gt1`.
So, for the first half
`|v//u|=2`
`|v|=2|u|`
or Let `u=-x`,
then `v=+2x`
and `|u|+|v|=1.8m`
i.e. `3x=1.8m`
or `x=0.6m`
Hence, `u=-0.6m`
and `v=+1.2m`
using `1/f=1/v-1/u=1/1.2-1/-0.6=1/0.4`
`:. f=0.4m`
For the second half, `1/f=1/(1.2-d)-1/-(0.6+d)`
or `1/0.4=1/(1.2-d)-1/-(0.6+d)`
Solving this, we get `d=0.6m`
Magnification for the second half will be
`m_2=v/u=0.6/(-(1.2))=-1/2`
and magnification for the first half is
`m_2=v/u=1.2/-(0.6)=-2`
the ray diagram is as follows:
