Question

What should be the value of distance $d$, so that final image is formed on the object itself? $($Focal lengths of the lenses are written on the lenses$)$.

• A. $10\text{cm}$
• B. $20\text{cm}$
• C. $5\text{cm}$
• D. None of these

Answer 1

The correct option is A $10\text{cm}$

The first image will be formed by the first lens, convex lens. Let the image be formed at a distance $v_1$ when the object is at point $O$ at distance $u_1=-10\text{cm}$.
We know that the focal length of the convex lens is $f_1=10\text{cm}$.
So, we can use lens formula for the convex lens which is given as-
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
So, substituting the values for convex lens in the lens formula, we get,
$\frac{1}{v_1}-\frac{1}{-10}=\frac{1}{10}$

$\Rightarrow \frac{1}{v_1}+\frac{1}{10}=\frac{1}{10}$

$\Rightarrow \frac{1}{v_1}=0$

$\therefore v_1=\infty$
Now, the image from the convex lens will become the object for the second lens, which is a concave lens.
Now, according to the question, the focal length of the cancave lens, $f_2=-20\text{cm}$,
and from above calculation, we have the distance of object
$u_2=v_1=\infty$
Let, the distance of the image is $v_2$, then using the lens formula, we get
$\frac{1}{v_2}-\frac{1}{\infty }=\frac{1}{-20}$

$\therefore v_2=-20\text{cm}$
So, the final image formed by the concave lens is at $-20\text{cm}$
Negative sign gives the position of the final image which is formed at the left side of both lens.

Now, from the figure, we have
$v_2=d+10$
Taking the magnitude of $v_2=20\text{cm}$
$\Rightarrow 20=d+10$
$\therefore d=10\text{cm}$

Hence, distance between the both lens, $d=10\text{cm}$.

Alternative solution:
Let the distance of concave lens from the object be $x$.
It is evident from the ray diagram that, for the final image to form at object, the rays shoud hit the concave mirror parallel and the object $O$ should lie at the focus of concave mirror. Hence , $x=20\text{cm}$
$d=(20-10)=10\text{cm}$