Question

The position of final image formed by the given lens combination from the third lens will be at a distance of $\left(f_{1}=+10, f_{2}=-10\, cm\right.$ and $\left.f_{3}=+30\, cm \right)$

• A. 15 cm
• B. Infinity
• C. 45 cm
• D. 30 cm

Answer 1

Answer: (D)

Given, lens combination and position of object is as shown below
For first lens, $u=-30\, cm , f=10\, cm$ By lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} $
$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30} $
$\Rightarrow \frac{1}{v}=\frac{1}{15} $ or $ v=15\, cm$
Now, image of first lens acts like object for second lens.
For second lens.

$u=+10\, cm , f=-10\, cm$
Again, using lens formula,
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-10}+\frac{1}{10}$
$\Rightarrow \frac{1}{v}=0 \text { or } v=\infty$
This means light rays becomes parallel after refraction from the second lens.
When these parallel light rays passes through third lens, their image is formed at focus. Hence, distance of image from third lens is $30\, cm$.