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.
