A point object ‘O’ is placed midway between two concave mirrors placed at a distance ‘d’ apart. Then the value of ‘d’ for which object & image coincide is [Each mirror has focal length f]

Given that two concave mirrors each having focal length f are kept facing each other separated by distance d as shown.

AB = d

${\text{AF}}_{\text{1}}\text{=}{\text{BF}}_{\text{2}}\text{=}\text{f}$

$\text{OA}\text{=}\text{OB}\text{=}\frac{d}{2}$

An object O is kept exactly midway between them. We are asked to get image. I also taking place coinciding with O.

It is possible in the following ways

(i) After first reflection from each mirror.

We know from image properties

When u = 2f then v also 2f.

(i.e., object at centre of curvature)

i.e., OA = 2f = OB.

We get images due to both mirrors coincide with O.

$\frac{\text{d}}{\text{2}}=2\text{f}$

$\Rightarrow \text{d}\text{=}\text{4f}$

(ii) We can get the reflected image from 1^st mirror coincide with object after under going reflection from second mirror.

1^st  Image I₁ from M₁

$\text{u}=-\frac{\text{d}}{\text{2}}$  , v = ?

$\text{f}=-\text{f}$

We have $\frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{u}}\text{+}\frac{\text{1}}{\text{v}}$

$\frac{-1}{f}=\frac{-2}{\text{d}}+\frac{1}{\text{v}}$

$\Rightarrow \frac{1}{\text{v}}=\frac{2}{\text{d}}-\frac{1}{\text{f}}=\frac{2\text{f}-\text{d}}{\text{f}-\text{d}}$

$\therefore \text{v}=\frac{\text{fd}}{2\text{f}-\text{d}}$ .

Now I₁ is formed from second mirror at $\text{u}=-\left(\text{d}-\text{v}\right)$ .

We should get

$\text{f}=-\text{ve}$  , $\text{v}=\frac{-\text{d}}{2}$

$\frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{u}}\text{+}\frac{\text{1}}{\text{v}}$

$\Rightarrow$ $\frac{-1}{\text{f}}=\frac{-1}{\text{d}-\text{v}}-\frac{2}{\text{d}}$

$\Rightarrow$ $\frac{1}{\text{f}}=\frac{1}{\text{d}-\text{v}}+\frac{2}{\text{d}}$

But $\text{d}-\text{v}=\text{d}-\frac{\text{fd}}{2\text{f}-\text{d}}$

$=\frac{2\text{fd}-{\text{d}}^2-\text{fd}}{2\text{f}-\text{d}}$

$\because \text{d}-\text{v}=\frac{f\text{d}-{\text{d}}^2}{\left(2\text{f}-\text{d}\right)}$

$\Rightarrow$ $\frac{1}{\text{f}}-\frac{2}{\text{d}}=\frac{1}{\text{d}-\text{v}}$

$\Rightarrow$ $\frac{\text{d}-2\text{f}}{\text{fd}}=\frac{(2\text{f}-\text{d})}{\text{fd}-{\text{d}}^2}$

$\Rightarrow$ $-\left(\text{fd}-{\text{d}}^2\right)=\text{fd}$

$\Rightarrow$ $-\text{d}\left(\text{f}-\text{d}\right)=\text{fd}$

$\Rightarrow$ $-\text{f}+\text{d}=\text{f}$

$\Rightarrow$ d = 2f

So d can became 4f or 2f

ray diagram (i) for d = 4f

(ii) d = 2f