Question #31746

Mar 24, 2017

We know the conjugate focci relation for lens as follows

$\textcolor{b l u e}{\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \ldots \ldots . \left[1\right]}$

Where

$u \to \text{object distance}$

$v \to \text{image distance}$

$f \to \text{focal length}$

Here

$u \to \text{real object distance} = - 25 c m$

$f \to \text{focal length for convex lens} = + 20 c m$

Inserting these in equation [1] we get

$\textcolor{b l u e}{\frac{1}{v} - \frac{1}{u} = \frac{1}{f}}$

$\textcolor{b l u e}{\implies \frac{1}{v} - \frac{1}{- 25} = \frac{1}{20}}$

$\textcolor{b l u e}{\implies \frac{1}{v} = \frac{1}{20} - \frac{1}{25}}$

$\textcolor{b l u e}{\implies \frac{1}{v} = \frac{5 - 4}{100} = \frac{1}{100}}$

$\textcolor{b l u e}{\implies v = 100 c m}$

So the lens forms a real image at a distance of $100 c m$from it.

It has been said that on placing a convex mirror 40 cm behind the lens, the final image coincides with object itself. This is possible only if retracing of rays is brought about by the reflection on convex mirror i.e the rays emerging from lens are incident on the convex mirror normally and it is possible if they are directed to the center of curvature of the mirror as shown in the ray diagram above.

So distance of radius of curvature from lens is $v = 100 c m$

Hence the radius of curvature of mirror is $r = \left(100 - 40\right) c m = 60 c m$

Again we know that the focal length of mirror $f = \frac{r}{2}$

So the focal length of the mirror will be $f = \frac{60}{2} c m = 30 c m$