To determine the relative decrease in the focal length of a lens when the optical power changes, we need to understand the relationship between optical power and focal length.
The optical power \( P \) of a lens is given by the formula:
\(P = \frac{1}{f}\)
where:
Initially, the optical power of the lens is \( P_1 = 2.5 \, \text{D} \). Therefore, the focal length \( f_1 \) is:
\(f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}\)
The optical power is increased by \( 0.1 \, \text{D} \) to become \( P_2 = 2.6 \, \text{D} \). The new focal length \( f_2 \) is:
\(f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}\)
The decrease in focal length is:
\(\Delta f = f_1 - f_2 = 0.4 - 0.3846 = 0.0154 \, \text{m}\)
The relative decrease in focal length is calculated by the formula:
\(\text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} = 0.0385\)
Rounding to two decimal places, the relative decrease is approximately 0.04.
Therefore, the correct answer is 0.04.
The relationship between optical power \( P \) and focal length \( f \) is given by:
\[ P = \frac{1}{f}. \] Thus, the initial focal length is: \[ f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}. \]
After the optical power increases by 0.1 D, the new optical power becomes:
\[ P_2 = 2.5 + 0.1 = 2.6 \, \text{D}. \] The new focal length is: \[ f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}. \]
The relative decrease in focal length is: \[ \text{Relative decrease} = \frac{f_1 - f_2}{f_1} = \frac{0.4 - 0.3846}{0.4} = \frac{0.0154}{0.4} \approx 0.04. \]
The correct answer is option (1) \( \boxed{0.04} \).
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.