Here, for the first lens, `mu_(1)=-10cm`
`f_(1)=20cm`
`(1)/(v_(1)) -(1)/(u_(1))=(1)/(f_(1))rArr (1)/(v_(1))=(1)/(20)-(1)/(10)`
`rArr v_(1)=-20cm`
The image is virtual and hence lies on the same side of the object . This wil behave as an object for the second lens.
For the second lens:
`(1)/(v_(2))-(1)/(u_(2))=(1)/(f_(2))`
Here, `u_(2)=-(20+5)`
`f_(2)=10cm `
`(1)/(v_(2))+(1)/(25)=(1)/(10)rArr v_(2)=(50)/(3)=16(2)/(3)cm`
The final image is at a distance of `16(2)/(3)` cm to the right of the second lens.
The magnification of the image is given by,
`m=(v_(1))/(u_(1))*(v_(2))/(u_(2))=(20)/(10)*(50)/(3xx25)=(4)/(3)=1.33`
