Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm
I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J
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: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8
Explanation:
in the attachment