List I contains four combinations of two lenses ( 1 and 2 ) whose...

I

$u = – 20cm$~${f}_{1} = + 10cm$

Using the lens formula,

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$⇒ \dfrac{1}{v1} + \dfrac{1}{20} = \dfrac{1}{10}⇒ \dfrac{1}{v1} = \dfrac{1}{10} - \dfrac{1}{20}$

$⇒ {v}_{1} = 20cm$

For second lens~$u = {v}_{1} - 5 = + 15cm$,~${f}_{2} = + 15cm$

$⇒ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f2}⇒ \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{15}⇒ \dfrac{1}{v} = \dfrac{2}{15}$

$v = + 7 . 5cm$ (from lens $2$)

Therefore,~$I \to P$

(II)~

Similarly,

$u = - 20cm, f = + 10cm$

$\dfrac{1}{v} + \dfrac{1}{20} = \dfrac{1}{10}⇒ \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20} ⇒ v = 20cm$

For second

$u = {v}_{1} - 5 = + 15cm$,~$f = - 10cm$

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} ⇒ \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{- 1}{10}$

$⇒ \dfrac{1}{v} = \dfrac{- 1}{10} + \dfrac{1}{15} = \dfrac{- 3 + 2}{30}⇒ \dfrac{1}{v} = - \dfrac{1}{30}$

$v = - 30cm$

$II \to R$

(III)~

$u = - 20$,~$f = + 10cm$

$\dfrac{1}{v} + \dfrac{1}{20} = \dfrac{1}{10}$

$⇒ \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20} = \dfrac{1}{20}$

$⇒ v = 20cm$

For second,

$⇒ u = 15cm$

$f = – 20cm$

$\dfrac{1}{v} - \dfrac{1}{15} = \dfrac{- 1}{20}$

$⇒ \dfrac{1}{v} = - \dfrac{1}{20} + \dfrac{1}{15} = \dfrac{- 3 + 4}{60} = \dfrac{1}{60}$

$v = 60cm$

$III \to Q$

(IV)~

$u = – 20cm$,~$f = – 20$

$⇒ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$⇒ \dfrac{1}{v} + \dfrac{1}{20} = \dfrac{- 1}{20}$

$v = - 10cm$

For second

$u = – 15cm$

$f = 10cm$

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$⇒ \dfrac{1}{v} + \dfrac{1}{15} = \dfrac{1}{10}$

$⇒ \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}$

$= \dfrac{3 - 2}{30} = \dfrac{1}{30}$

$v = 30cm$

$IV \to T$