Question

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
A		P	Final image is farmed at $7.5 cm$ on the right side of lens 2 .
B		Q	Final image is formed at $60.0 \,cm$ on the right side of lens 2.
C		R	Final image is formed at $30.0 \,cm$ on the left side of lens 2.
D		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?

• A. $I \rightarrow$ P, II $\rightarrow$ R, III $\rightarrow$ Q, IV $\rightarrow$ T
• B. I $\rightarrow$ Q, II $\rightarrow$ P, III $\rightarrow$ T, IV $\rightarrow$ S
• C. I $\rightarrow$ P, II $\rightarrow$ T, III $\rightarrow$ R, IV $\rightarrow$ Q
• D. $I \rightarrow T$, II $\rightarrow$ S, III $\rightarrow$ Q, IV $\rightarrow$ R

Answer 1

Answer: (A)

(I) $v _1=\frac{ uf }{ u + f } $
$=\frac{(-20)(10)}{(-20)+(10)}=+20$
$ u _2=+15 $
$ v _2= \frac{(15)(15)}{(15)+(15)}=+7.5$
(II)$ v _1=+20 $
$ u _2=+15$
$v _2=\frac{(15)(-10)}{(15)+(-10)}=-30$
(III) $v _1=+20$
$u _2=+15$
$v_2=\frac{(15)(-20)}{(15)+(-20)}=60$
(IV) $ v _1=\frac{(-20)(-20)}{(-20)+(-20)}=-10 $
$ u _2=-15 $
$ v _2=\frac{(-15)(10)}{(-15)+(10)}=30$