Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be = 1.9000000000000001 J
and the final kinetic energy from point B be = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:
Where,
Indicates the intensity of the light before passing through the polarizer
I = Resulting intensity
= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light
From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore
In the case of the second polarizer the angle is directly 60 degrees therefore
In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be
Then,
Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.