Question

If a symmetrical bi-concave thin lens is cut into two identical halves, and they are placed in different arrangements as shown, then

• A. Three images will be formed in case $\left(i\right)$
• B. Two images will be formed in case $\left(ii\right)$
• C. The ratio of focal lengths in $\left(ii\right)$ and $\left(iii\right)$ is $1$
• D. The ratio of focal lengths in $\left(ii\right)$ and $\left(iii\right)$ is $2$

Answer 1

The correct option is C The ratio of focal lengths in $\left(ii\right)$ and $\left(iii\right)$ is $1$

Case- (i):

From the figure it is evident that the principle axis of each lens is not coinciding. Therefore, two images are formed.

Case- (ii):

The principle axis of one lens coincides with the other . so, one image is formed.

Let, $f_2$ and $f_3$ be the combined focal lengths in case $\left(ii\right)$ and case ($iii)$ respectively and the radius of curvature is $R$ for each curved surface.

Now, applying lens makers formula for case $\left(ii\right)$,

$\frac{1}{f_2}=(\mu -1)(\frac{1}{-R}-\frac{1}{R})$

$\Rightarrow \frac{1}{f_2}=-\frac{2}{R}(\mu -1).....\left(1\right)$

Now , for case $\left(iii\right)$.


we can write that,
$$\frac{1}{f_3}=\frac{1}{F_1}+\frac{1}{F_2}$$
$\Rightarrow \frac{1}{f_3}=(\mu -1)(\frac{1}{\infty }-\frac{1}{R})+(\mu -1)(\frac{1}{-R}-\frac{1}{\infty })$

$\Rightarrow \frac{1}{f_3}=-\frac{2}{R}(\mu -1)....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we get, $\frac{1}{f_2}=\frac{1}{f_3}$

$\therefore \frac{f_2}{f_3}=1$

Hence, option (c) is the correct answer.