Question

In the situation shown in the figure, three lenses of focal lengths $20\text{cm},30\text{cm}$ and $36\text{cm}$ are held coaxial such that separation between first and second lens is $12\text{cm}$ and between second and third lens is $10\text{cm}$. If the final image of an object at a distance of $30\text{cm}$ from first lens is at a distance $10x\text{cm}$ from the middle lens, then find the value of $x$.

Answer 1

For the lens
$u=-30\text{cm};f=20\text{cm}$
$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}\Rightarrow v=60\text{cm}$
For second lens
$u=+48\text{cm};f=-30\text{cm}$
$\frac{1}{v}=\frac{1}{-30}+\frac{1}{48}=\frac{-16+10}{480}=\frac{-6}{480}$
$\Rightarrow v=-80\text{cm}$
For third lens
$u=-90\text{cm};f=36\text{cm}$
$\frac{1}{v}=\frac{1}{36}-\frac{1}{90}=\frac{10-4}{360}=\frac{6}{360}$
$\Rightarrow v=60\text{cm}$
From middle lens, position of final image is $70\text{cm}$
$\Rightarrow 10x=70$
$\therefore x=7$