Question

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

• A. 20 cm away from the mirror.
• B. 80 cm away from the mirror.
• C. 20 cm towards the mirror.
• D. 40 cm away from the mirror.

Answer 1

Answer: (B)

80 cm away from the mirror.

Answer 2

For a convex mirror, we have:

$$m = -\frac{v}{u}\quad \text{and}\quad \frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$

For a convex mirror, object distance $u$ is positive, image distance $v$ is negative, and focal length $f$ is negative.

1. Initial arrangement:
   Given $u=40$ cm and $|m|=\frac{1}{2}$, using

   $$\frac{1}{2} = \frac{|v|}{40} \quad \Rightarrow \quad |v|=20\, \text{cm}.$$

   Since the image is virtual, $v=-20$ cm.
   Mirror formula gives:

   $$\frac{1}{f} = \frac{1}{40} + \frac{1}{-20} = \frac{1}{40} - \frac{1}{20} = -\frac{1}{40} \quad \Rightarrow \quad f=-40\, \text{cm}.$$

2. For the new magnification $\frac{1}{3}$:
   Let the new object distance be $u'$ and the new image distance be $v'$. Then:

   $$m = -\frac{v'}{u'} = \frac{1}{3}\quad \Rightarrow \quad v'=-\frac{u'}{3}.$$

   Applying the mirror formula:

   $$\frac{1}{f}=\frac{1}{u'}+\frac{1}{v'} = \frac{1}{u'} +\left(-\frac{3}{u'}\right) = -\frac{2}{u'}.$$

   With $f=-40$ cm:

   $$\frac{1}{-40}=-\frac{2}{u'} \quad \Rightarrow \quad u'=\frac{2 \times 40}{1}=80\, \text{cm}.$$

Therefore, the object should be moved to 80 cm away from the mirror.