Using the magnification formula for mirrors: \[ m = \frac{f}{u-f} \] For the concave mirror, the object distance is \( u = -18 \, \text{cm} \), and the focal length is \( f = \frac{R}{2} = 6 \, \text{cm} \), where \( R = 12 \, \text{cm} \): \[ m_1 = \frac{6}{18 - 6} = \frac{1}{2} \]

For the convex mirror, the object distance is the same, and the focal length is positive: \[ m_2 = \frac{6}{18 + 6} = \frac{1}{4} \] Hence, the ratio of the sizes of the images formed by the convex mirror and the concave mirror is: \[ \frac{m_2}{m_1} = \frac{1/4}{1/2} = \frac{1}{2} \]

Thus, the correct answer is: \[ \frac{1}{2} \]
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.