To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as

Where
Angle From vertical of the axis of the polarizing filter
Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by

Replacing we have that


Re-arrange the equation,

Re-arrange to find \theta





The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
Yes, our friend is right, because there is no contradiction to the law of conservation of mass in the above equation. It just the mass of the product is equal to the mass of reactants.. and that is shown in the equation you have presented earlier
Answer:
The phase difference is 
Explanation:
From the question we are told that
The distance between the slits is
The distance to the screen is 
The wavelength is 
The distance of the wave from the central maximum is 
Generally the path difference of this waves is mathematically represented as

Here
is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum
This implies that

=> 
![\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B%5Cfrac%7B5%2A10%5E%7B-3%7D%7D%7B1%7D%5D)

Substituting values into the formula for path difference
The phase difference is mathematically represented as

Substituting values

Converting to degree
the solution is subtracted by 360° in order to get the actual angle
<u>Answer:</u>
First, the thermometer is dipped into boiling water, and the mercury inside the thermometer rises to a high level, called the boiling point. This level is then marked as 100°C. The thermometer is then dipped into melting ice, which causes the mercury level to fall to a point called the ice point. This point is then marked as 0°C. The length of the thermometer from the 0°C mark to the 100°C point is then divided into 100 equal sections, and the rest of the levels are marked accordingly.