1. Use magnification formula for a spherical mirror

For mirrors, $m=-\frac{v}{u}$ where $u$ is object distance and $v$ is image distance.

Also, mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

2. First condition: object at 40 cm gives magnification $\frac{1}{2}$

Given object is placed $40\text{ cm}$ from the mirror. Using Cartesian convention, $u=-40\text{ cm}$

Given magnification magnitude is $\frac{1}{2}$. For a real diminished image formed by a concave mirror, $m=-\frac{1}{2}$

So, $-\frac{v}{u}=-\frac{1}{2}$ $\Rightarrow \frac{v}{u}=\frac{1}{2}$ $v=\frac{u}{2}=\frac{-40}{2}=-20\text{ cm}$

Now apply mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ $\frac{1}{f}=\frac{1}{-20}+\frac{1}{-40}=-\frac{1}{20}-\frac{1}{40}=-\frac{3}{40}$ $f=-\frac{40}{3}\text{ cm}$

3. Second condition: required magnification $\frac{1}{3}$

Again for a real diminished image, $m=-\frac{1}{3}$

Let new object distance be $u'$ and new image distance be $v'$. Then, $-\frac{v'}{u'}=-\frac{1}{3}$ $\Rightarrow v'=\frac{u'}{3}$

Using mirror formula with same focal length: $\frac{1}{f}=\frac{1}{v'}+\frac{1}{u'}$ $-\frac{3}{40}=\frac{1}{u'/3}+\frac{1}{u'}$ $-\frac{3}{40}=\frac{3}{u'}+\frac{1}{u'}=\frac{4}{u'}$ $u'=-\frac{160}{3}\text{ cm}$

So the new object distance from mirror is $\left|u'\right|=\frac{160}{3}\approx 53.3\text{ cm}$

4. How much should the object be moved?

Initial distance $=40\text{ cm}$

Final distance $\approx 53.3\text{ cm}$

Required shift: $53.3-40=13.3\text{ cm}$

So the object must be moved about $13.3$ cm away from the mirror.

5. Check options

• A: 20 cm away $\Rightarrow$ final distance $60$ cm, not correct.
• B: 20 cm towards $\Rightarrow$ final distance $20$ cm, not correct.
• C: 80 cm away $\Rightarrow$ final distance $120$ cm, not correct.
• D: 40 cm away $\Rightarrow$ final distance $80$ cm, not correct.

None of the given options matches the correct result.

6. Conclusion

The correct movement should be: $\boxed{13.3\text{ cm away from the mirror}}$

Hence, the stored answer D is incorrect.