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) \( \boxed{0.04} \).
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: