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

Explanation

For 1st lens,
`u_1 = -30 cm, f_1 = + 10 cm, v_1 = ?`
As `(1)/(v_1)-(1)/(u_1)=(1)/(f_1) :. (1)/(v_1)-(1)/(-30) = (1)/(10)`
Image formed by first lens serves as object for second lens at a distance `=(15 - 5) = 10 cm` to the right of second lens, and the object is virtual Therefore, for second lens,
`u_2 = 10 cm, v_2 = ? f_2 = -10 cm`.
As `(1)/(f_2)=(1)/(v_2) - (1)/(u_2) :. -(1)/(10) = (1)/(v_2) - (1)/(10)`
`:. (1)/(v_2) = 0, v_2 = oo`
The virtual image is found at infinity to the right of `2nd` lens. This acts as an object for `3rd` lens. Therefore, for `3rd` lens, `u_3 = oo, v_3 = ?`
`f_3 = + 30 cm`
As `(1)/(v_3) - (1)/(u_3) = (1)/(f_3)`
`:. (1)/(v_3)-(1)/(oo) = (1)/(30)` or `v_3 = 30 cm`.
The final image is formed at `30 cm` to the right of third lens.