Shown in the figure is a vertically erect object placed on the optic axis at a distance (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 (4/3)f from the pole, facing concave mirror, The position of the final image formed

f

Two rays come from the tip of the object. They converge after reflection to form a real image at I₁.

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 I₂. Infact I₂ is the real image of the image I₁ of the object O, which serves as a virtual object to the mirror,

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}⇒\frac{1}{-2.5f}+\frac{1}{v}=\frac{1}{-f}$

$⇒\frac{1}{v}=-\frac{1}{f}+\frac{2}{5f}=-\frac{3}{5f}$

$⇒v=-\frac{5}{3}f$

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

⇒ The image I₁ is formed which behaves as a virtual object for the plane mirror & therefore finally a real erect image I₂ of same size, will be formed at equal distance x infront of the plane mirror as shown in the figure.

⇒ I₂ will be formed at a distance (5/3)f – 2x, from concave mirror where x = 5/3f − 4/3f = 1/3f

⇒ The final image distance v’ = (5/3)f - 2(1/3f) = f

⇒ Final image is formed at the focus.