\( M_{OA} = \dfrac{2}{5} \)
\( M_{OB} = \dfrac{a}{2} \)
\( = \tan \dfrac{\pi}{4} = \left| \dfrac{\dfrac{2}{5}}{1 + \dfrac{a}{2}} \right| \)
\( \Rightarrow 1 = \left| \dfrac{4 - 5a}{10 + 2a} \right| \Rightarrow a = \dfrac{-6}{7}, \dfrac{14}{3} \)
Product \( \Rightarrow -4, \text{Abs. value} = 4 \)

Consider the points \( A(5, 2) \) and \( B(2, a) \) joined by line segments from the origin \( O \). The given condition states that these segments subtend an angle \( \frac{\pi}{4} \) at the origin. Let the slopes of the lines \( OA \) and \( OB \) be \( M_{OA} \) and \( M_{OB} \) respectively.
The slope of line \( OA \) is: \[ M_{OA} = \frac{2}{5}. \]
The slope of line \( OB \) is: \[ M_{OB} = \frac{a}{2}. \]
Since the angle between the lines is \( \frac{\pi}{4} \),
we use the formula: \[ \tan \left( \frac{\pi}{4} \right) = \left| \frac{M_{OB} - M_{OA}}{1 + M_{OA} \cdot M_{OB}} \right|. \]
Given \( \tan \left( \frac{\pi}{4} \right) = 1 \),
we have: \[ 1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \frac{2}{5} \cdot \frac{a}{2}} \right|. \]
Simplifying the expression: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
Clearing the fractions: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
This gives two cases: \[ \frac{5a - 4}{10 + 2a} = 1 \quad \text{or} \quad \frac{5a - 4}{10 + 2a} = -1. \]
Case 1: \[ \frac{5a - 4}{10 + 2a} = 1. \]
Cross-multiplying: \[ 5a - 4 = 10 + 2a. \]
Rearranging terms: \[ 3a = 14 \implies a = \frac{14}{3}. \]
Case 2: \[ \frac{5a - 4}{10 + 2a} = -1. \]
Cross-multiplying: \[ 5a - 4 = -10 - 2a. \]
Rearranging terms: \[ 7a = -6 \implies a = -\frac{6}{7}. \]
The product of all possible values of \( a \) is: \[ a_1 \times a_2 = \left( \frac{14}{3} \right) \times \left( -\frac{6}{7} \right) = -4. \]
The absolute value of the product is: \[ |a_1 \times a_2| = 4. \]
Therefore: \[ 4. \]
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
