A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

The object is placed on the principal axis $$30 \text{ cm}$$ in front of a convex mirror of focal length $$f = +30 \text{ cm}$$ (positive for a convex mirror in the new Cartesian sign convention).

Using the mirror formula $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$, with object distance $$u = -30 \text{ cm}$$, we get

$$\frac{1}{30} = \frac{1}{v} + \frac{1}{(-30)}$$
$$\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$
$$\Rightarrow \; v = +15 \text{ cm}$$

The positive sign means the convex mirror forms a virtual image $$I_2$$ $$15 \text{ cm}$$ behind the mirror (on the right side of the pole).

Now place a plane mirror between the object and the convex mirror. Let the distance between the two mirrors be $$d \text{ cm}$$. With the pole of the convex mirror as origin, the coordinates along the principal axis are:

• Pole of convex mirror: $$x = 0$$
• Plane mirror surface: $$x = -d$$
• Object: $$x = -30 \text{ cm}$$

The object is $$30 - d \text{ cm}$$ in front of the plane mirror, so the plane mirror produces a virtual image $$I_1$$ the same distance behind it:

Coordinate of $$I_1$$:
$$x(I_1) = (-d) + (30 - d) = 30 - 2d$$

For the two images to coincide, $$I_1$$ and $$I_2$$ must lie at the same coordinate:

$$30 - 2d = +15$$
$$2d = 30 - 15 = 15$$
$$\therefore \; d = 7.5 \text{ cm}$$

Hence, the distance between the convex mirror and the plane mirror must be $$7.5 \text{ cm}$$.

Option B is correct.