Answer
given,
wavelength (λ)= 500 n m
thickness of film= 10⁻⁴ cm
refractive index = μ = 1.375
distance traveled is double which is equal to 2 x 10⁻⁴ cm
a) Number of wave


N = 2.91
N = 3
b) phase difference is equal to
Reflection from the first surface has a 180° (½λ) phase change.
There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.
net phase difference = 
= 270°
Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

Substituting the numerical values, we find

I would say A not 100℅ thou
Mechanical and Chemical. (Weathering and erosion)