Find the position of the image formed by the lens combination given in the Fig.

Explanation

`rArr` For the image formed by first lens,
`1/(v_1)- (1)/(u_1) = 1/f`
`therefore (1)/(v_1) - (1)/(-30) = 1/10`
`therefore v_1 = 15 cm `
`rArr` Thus image formed by the first lens is formed at 15 cm distance on the right-hand side. This image is on the right-hand side of the second lens at 15 - 5 = 10 cm distance and so it acts as a vlrlual object for the second lens.
`rArr` Now for the second lens ,
`(1)/(v_2) - (1)/(u_2) = (1)/(f_2)`
`therefore 1/(v_2) - (1)/(10) = (1)/(10) " " [ therefore` According to sign convention ]
`therefore v_2 = oo`
This distance `v_2 ( = oo)` is the object distance fOl the third lens. So, the third image formed due to it should be on the principal focus of the third lens. Thus, as the focal length ofthe third lens is 30 cm the final image is formed at 30 cm distance on the right side of the third lens.
For the third lens,
`(1)/(v_3) - (1)/(u_3) = (1)/(f_3)`
`therefore (1)/(v_3) - (1)/(oo) = (1)/(30)`
`therefore (1)/(v_3) = (1)/(30)`
`therefore v_3 = 30 cm `
`therefore` Final image is formed at 30 cm distance on the right side of the third lens.