Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.
Mathematically the expression is given as
Where,
= Initial Intensity
I = Final Intensity after pass through the polarizer
= Angle between the polarizer and the light
Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as
Using the Malus Law we have,
Angle with respect to maximum is