Figure illustrates an aligned system consisting of three thin lenses. The system is located in air. Determine:

(a) the position of the point of convergence of a parallel incoming from the left after passing through the system,
(b) the distance between teh first lens and a point lying on the axis to the left of the system, at which that point and its image are located symmetrically with respect to the lens system.

Answer

(a)3.3 cm (b) 50/3 cm

Explanation

When a parallel beam comes from `'oo'` from left , after refraction from `L_(1)` it forms image at 10 cm from `L_(1)`.
This image will behave as a object for `L_(2)`
So object distance from `'L_(2)' = +5 cm`
So image formed at v : `(1)/(v) - (1)/(5) = (1)/(-10)`
`implies (1)/(v) = (1)/(5) - (1)/(10) =(1)/(10) implies v = 10 cm`
Therefore `L_(2)` will form a image at +10 cm from `L_(2)` on the right side . This image will act as object for `L_(3)`
Object distance for `L_(3) = +5 cm`
final image at v = ?
`(1)/(v) - (1)/(u) = (1)/(f) implies (1)/(v) - (1)/((+5)) = (1)/((10)) implies (1)/(v) = (1)/(5) + (1)/(10)`
`implies (1)/(v) = (3)/(10) implies v = (10)/(3) ` cm from `L_(3)`
Ray converges at d =`(10)/(3)` cm from `L_(3)`
(b)
`(1)/(v) + (1)/(x) = (1)/(f) implies v = (fx)/(x-f) implies 5 - (fx)/(x-f) implies (1)/(v_(t)) - (1)/(v) = (1)/(f)`
`(1)/(v_(1)) + ((1)/(5 - (fx)/(x-f))) = (1)/(-f) implies (1)/(v_(1)) = (1)/(f) - ((1)/(5-(fx)/(x-f)))`
`5 - v_(1) = u implies ( 5 - (fu)/(u-f)) = v_(1) implies (5u - 5f -fu)/(u-f)`
`(1)/(v_(1)) = (u-f)/(5u-5f -fu)` + (1)/(f) + (f)/(5 - (fx)/(x-f)) = (-u+f)/(5u-5f-fx)`
`(1)/(f) + (x-f)/((5x -5f -fx) = (f-u)/(5x -5f -fu)`
Putting the value after solving x = `(50)/(3)` cm