When an object is placed 40 cm away from a spherical mirror an image of magnification $$\dfrac{1}{2}$$ is produced. To obtain an image with magnification of $$\dfrac{1}{3}$$, the object is to be moved:

The mirror formula and the lateral magnification formula for a spherical mirror are

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\qquad -(1)$$
$$m=-\frac{v}{u}\qquad\qquad\!\!\!\! -(2)$$

The object is kept in front of the mirror, so by the Cartesian sign convention
  • object distance $$u$$ is always negative,
  • for a convex mirror the focal length $$f$$ is positive and the image distance $$v$$ is positive (virtual, behind the mirror).
Since the given magnification $$\dfrac{1}{2}$$ is positive (erect image), the mirror must be convex.

Given data for the first position:
  $$u_1=-40\text{ cm},\qquad m_1=\frac{1}{2}$$

From $$-(2)$$ : $$-\frac{v_1}{u_1}=m_1\Longrightarrow v_1=-m_1u_1=-\frac{1}{2}\times(-40)=+20\text{ cm}$$

Substituting $$u_1$$ and $$v_1$$ in $$-(1)$$ to find the focal length:

$$\frac{1}{f}=\frac{1}{20}+\frac{1}{-40}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$$
$$\Rightarrow\; f=+40\text{ cm}$$

For the new position the required magnification is
  $$m_2=\frac{1}{3}\quad(\text{still positive, erect image})$$

Using $$-(2)$$ again:
$$-\frac{v_2}{u_2}=m_2\Longrightarrow v_2=-m_2u_2=-\frac{1}{3}u_2$$ $$-(3)$$

Insert $$v_2$$ from $$-(3)$$ into the mirror formula $$-(1)$$:

$$\frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{-\dfrac{u_2}{3}}+\frac{1}{u_2}=-\frac{3}{u_2}+\frac{1}{u_2}=-\frac{2}{u_2}$$

With $$f=+40\text{ cm}$$, we get

$$\frac{1}{40}=-\frac{2}{u_2}\;\;\Longrightarrow\;\;u_2=-80\text{ cm}$$

Comparison of object positions:
Initial distance $$|u_1|=40\text{ cm}$$
Final distance $$|u_2|=80\text{ cm}$$

The object must therefore be shifted farther from the mirror by

$$|u_2|-|u_1|=80-40=40\text{ cm}$$

Case 1: Moving the object 40 cm away from the mirror produces the required magnification $$\dfrac{1}{3}$$.

Hence, the correct option is Option A (40 cm away from the mirror).