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) \( \boxed{0.04} \).
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Choose the correct nuclear process from the below options:
\( [ p : \text{proton}, n : \text{neutron}, e^- : \text{electron}, e^+ : \text{positron}, \nu : \text{neutrino}, \bar{\nu} : \text{antineutrino} ] \)
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: