A point object O is placed in front of two thin symmetrical coaxial...

For ${L}_{1}$:

$\dfrac{1}{{v}_{1}} + \dfrac{1}{6} = \dfrac{1}{24}$

$⇒ \dfrac{1}{{v}_{1}} = \dfrac{1}{24} - \dfrac{1}{6}$

$= \dfrac{1 - 4}{24}$

$⇒ \dfrac{1}{{v}_{1}} = \left(\dfrac{- 3}{24}\right)$

$⇒ {v}_{1} = – 8cm$

For ${L}_{2}$:

$u = – 18cm$

$⇒ \dfrac{1}{v} + \dfrac{1}{18} = \dfrac{1}{9}$

$⇒ \dfrac{1}{v} = \dfrac{1}{9} - \dfrac{1}{18}$

$= \dfrac{2 - 1}{18}$

$⇒ \dfrac{1}{v} = \dfrac{1}{18}$

$⇒ v = 18cm$

Distance between object and distance $= 16 + 18 = 34cm$.