Shown in the figure is a vertically erect object placed on the optic axis at a distance<img alt="\mathrm{ (5/2)f}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%285/2%29f%7D"

Shown in the figure is a vertically erect object placed on the optic axis at a distance $\mathrm{ (5/2)f}$ from a concave mirror of focal length f. If a plane mirror is placed perpendicular to the optic axis at a distance $\mathrm{(4/3)f}$ from the pole, facing concave mirror, The position of the final image formed

Option: 1
$\mathrm{f}$

Option: 2
$\mathrm{2f}$

Option: 3
$\mathrm{3f}$

Option: 4
$\mathrm{4f}$

Answers (1)

Two rays come from the tip of the object. They converge after reflection to form a real image at $\mathrm{I}_{1}$ .

Since the plane mirror is introduced in the mid, the reflected rays from the concave mirror get reflected by the plane mirror to form an image $\mathrm{I}_{2}$ .
Infact $\mathrm{I}_{2}$ is the real image of the image $\mathrm{I}_{1}$ of the object $\mathrm{O}$ ,which serves as a virtual object to the mirror,

$\mathrm{\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \Rightarrow \frac{1}{-2.5 f}+\frac{1}{v}=\frac{1}{-f}}$
$\mathrm{ \Rightarrow \frac{1}{v}=-\frac{1}{f}+\frac{2}{5 f}=-\frac{3}{5 f}}$
$\mathrm{ \Rightarrow v=-\frac{5}{3} f}$

(Negative sign indicates the image is left the mirror).

$\mathrm{ \Rightarrow }$ The image $\mathrm{ I_{1}}$ is formed which behaves as a virtual object for the plane mirror & therefore finally a real erect image $\mathrm{ I_{2}}$ of same size, will be formed at equal distance $\mathrm{ x}$ infront of the plane mirror as shown in the figure.

$\mathrm{ \Rightarrow \mathrm{I}_{2}}$ will be formed at a distance $\mathrm{(5 / 3) \mathrm{f}-2 \mathrm{x}}$ , from concave mirror
where $\mathrm{x=5 / 3 f-4 / 3 f=1 / 3 f}$

$\mathrm{\Rightarrow}$ The final image distance $\mathrm{v^{\prime}=(5 / 3) f-2(1 / 3 f)=f}$
$\mathrm{\Rightarrow}$ Final image is formed at the focus.