List I contains four combinations of two lenses ( 1 and 2 ) whose focal lengths (in cm ) are indicated in…

I

$u=–20 \mathrm{cm}$ $f_1=+10 \mathrm{cm}$

Using the lens formula,

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

$\Rightarrow \frac{1}{v_1}+\frac{1}{20}=\frac{1}{10}$$\Rightarrow \frac{1}{v_1}=\frac{1}{10}-\frac{1}{20}$

$\Rightarrow v_1=20 \mathrm{cm}$

For second lens $u=v_1-5=+15 \mathrm{cm}$, $f_2=+15 \mathrm{cm}$

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f_2}$$\Rightarrow \frac{1}{v}-\frac{1}{15}=\frac{1}{15}$$\Rightarrow \frac{1}{v}=\frac{2}{15}$

$v=+7.5 \mathrm{cm}$ (from lens $2$)

Therefore, $\mathrm{I}\to P$

(II) 

Similarly,

$u=-20 \mathrm{cm},f=+10 \mathrm{cm}$

$\frac{1}{v}+\frac{1}{20}=\frac{1}{10}$$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{20}\Rightarrow v=20 \mathrm{cm}$

For second

$u=v_1-5=+15\mathrm{cm}$, $f=-10\mathrm{cm}$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}-\frac{1}{15}=\frac{-1}{10}$

$\Rightarrow \frac{1}{v}=\frac{-1}{10}+\frac{1}{15}=\frac{-3+2}{30}$$\Rightarrow \frac{1}{v}=-\frac{1}{30}$

$v=-30 \mathrm{cm}$

$\mathrm{II}\to R$

(III) 

$u=-20$, $f=+10\mathrm{cm}$

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

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$

$\Rightarrow v=20\mathrm{cm}$

For second,

$\Rightarrow u=15\mathrm{cm}$

$f=– 20\mathrm{cm}$

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

$\Rightarrow \frac{1}{v}=-\frac{1}{20}+\frac{1}{15}=\frac{-3+4}{60}=\frac{1}{60}$

$v=60 \mathrm{cm}$

$\mathrm{III}\to Q$

(IV) 

$u=–20\mathrm{cm}$, $f=–20$

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{-1}{20}$

$v=-10\mathrm{cm}$

For second

$u=–15\mathrm{cm}$

$f=10\mathrm{cm}$

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

$\Rightarrow \frac{1}{v}+\frac{1}{15}=\frac{1}{10}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}$

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

$v=30\mathrm{cm}$

$\mathrm{IV}\to T$