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

List I contains four combinations of two lenses ( $1$ and $2$ ) whose focal lengths (in $cm$ ) are indicated in the figures. In all cases, the object is placed $20 cm$ from the first lens on the left, and the distance between the two lenses is $5 cm$ . List II contains the positions of the final images.

	List-I		List-II
(I)		(P)	Final image is formed at $7.5 cm$ on the right side of lens $2$ .
(II)		(Q)	Final image is formed at $60.0 cm$ on the right side of lens $2$ .
(III)		(R)	Final image is formed at $30.0 cm$ on the left side of lens $2$ .
(IV)		(S)	Final image is formed at $6.0 cm$ on the right side of lens $2$ .
		(T)	Final image is formed at $30.0 cm$ on the right side of lens $2$ .

Which one of the following options is correct?

I \to P, II \to R, III \to Q, IV \to T

I \to Q, II \to P, III \to T, IV \to S

I \to P, II \to T, III \to R, IV \to Q

I \to T, II \to S, III \to Q, IV \to R

Solution:

I
$u = - 20 cm$ $f_{1} = + 10 cm$
Using the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v_{1}} + \frac{1}{20} = \frac{1}{10}$ $\Rightarrow \frac{1}{v_{1}} = \frac{1}{10} - \frac{1}{20}$
$\Rightarrow v_{1} = 20 cm$
For second lens $u = v_{1} - 5 = + 15 cm$ , $f_{2} = + 15 cm$
$\Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{1}{f_{2}}$ $\Rightarrow \frac{1}{v} - \frac{1}{15} = \frac{1}{15}$ $\Rightarrow \frac{1}{v} = \frac{2}{15}$
$v = + 7.5 cm$ (from lens $2$ )
Therefore, $I \to P$
(II)
Similarly,
$u = - 20 cm, f = + 10 cm$
$\frac{1}{v} + \frac{1}{20} = \frac{1}{10}$ $\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{20} \Rightarrow v = 20 cm$
For second
$u = v_{1} - 5 = + 15 cm$ , $f = - 10 cm$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{15} = \frac{- 1}{10}$
$\Rightarrow \frac{1}{v} = \frac{- 1}{10} + \frac{1}{15} = \frac{- 3 + 2}{30}$ $\Rightarrow \frac{1}{v} = - \frac{1}{30}$
$v = - 30 cm$
$II \to R$
(III)
$u = - 20$ , $f = + 10 cm$
$\frac{1}{v} + \frac{1}{20} = \frac{1}{10}$
$\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$
$\Rightarrow v = 20 cm$
For second,
$\Rightarrow u = 15 cm$
$f = - 20 cm$
$\frac{1}{v} - \frac{1}{15} = \frac{- 1}{20}$
$\Rightarrow \frac{1}{v} = - \frac{1}{20} + \frac{1}{15} = \frac{- 3 + 4}{60} = \frac{1}{60}$
$v = 60 cm$
$III \to Q$
(IV)
$u = - 20 cm$ , $f = - 20$
$\Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} + \frac{1}{20} = \frac{- 1}{20}$
$v = - 10 cm$
For second
$u = - 15 cm$
$f = 10 cm$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} + \frac{1}{15} = \frac{1}{10}$
$\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{15}$
$= \frac{3 - 2}{30} = \frac{1}{30}$
$v = 30 cm$
$IV \to T$

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

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

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