Question

A convex lens of focal length $40cm$ is held at a distance $12cm$ coaxially above a concave mirror of focal length $18cm$. A luminous point object placed $dcm$ above the lens on its axis gives rise to an image coincides with itself, then the value of $d$ is equal to:

• A. $15cm$
• B. $30cm$
• C. $24cm$
• D. $18cm$

Answer 1

The correct option is A $15cm$

The ray originating from point O at first gets refracted from the convex lens of focal length 40 cm and forms an image at point $I_1$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_2}$

Given: $u=-d{f}_2=+40cm$

$\frac{1}{v}-\frac{1}{-d}=\frac{1}{40}$

$⟹v=\frac{40d}{d-40}$

Now the image formed will act as source for the concave mirror, and if the final image is formed at point O, then the ray form point $I_1$ has to trace the same path after reflection from the mirror i-e, the ray from $I_1$ must fall normally on the mirror.Thus $I_1$ should be at the radius of curvature

$R$ of the mirror.

$R=2f_1=36cm$

Hence $12-\frac{40d}{d-40}=36$

$⟹d=15cm$