The magnification produced when an object is placed at a distance of 20 cm from a spherical mirror is [tex]+\frac{1}{2}[/tex]. Where should the object be placed to reduce the magnification to [tex]+\frac{1}{3}[/tex]?



Answer :

Certainly! To determine where the object should be placed to reduce the magnification from [tex]\( \frac{1}{2} \)[/tex] to [tex]\( \frac{1}{3} \)[/tex], we can follow these steps:

### Step-by-Step Solution:

1. Understand the initial setup:
- The object distance from the mirror is [tex]\( u_1 = 20 \)[/tex] cm.
- The initial magnification, [tex]\( m_1 \)[/tex], is [tex]\( \frac{1}{2} \)[/tex].

2. Use the magnification formula:
[tex]\[ m = -\frac{v}{u} \][/tex]
where [tex]\( v \)[/tex] is the image distance and [tex]\( u \)[/tex] is the object distance. For the initial condition:
[tex]\[ m_1 = -\frac{v_1}{u_1} \][/tex]

3. Calculate the initial image distance ([tex]\( v_1 \)[/tex]):
We know:
[tex]\[ \frac{1}{2} = -\frac{v_1}{20} \][/tex]
By multiplying both sides by [tex]\(-20\)[/tex], we get:
[tex]\[ v_1 = -10 \text{ cm} \][/tex]

4. Determine the final configuration:
- The final magnification, [tex]\( m_2 \)[/tex], is [tex]\( \frac{1}{3} \)[/tex].

We again use the magnification formula for the new setup:
[tex]\[ m_2 = -\frac{v_1}{u_2} \][/tex]
Here, we already know [tex]\( v_1 = -10 \text{ cm} \)[/tex].

5. Calculate the new object distance ([tex]\( u_2 \)[/tex]):
Applying the values we know:
[tex]\[ \frac{1}{3} = -\frac{-10}{u_2} \][/tex]
Simplifying this, we have:
[tex]\[ \frac{1}{3} = \frac{10}{u_2} \][/tex]
By multiplying both sides by [tex]\( u_2 \)[/tex] and then multiplying both sides by 3, we get:
[tex]\[ u_2 = 30 \text{ cm} \][/tex]

### Conclusion:
To reduce the magnification to [tex]\( \frac{1}{3} \)[/tex], the object should be placed at a distance of 30 cm from the spherical mirror.