Answer:
- the coating’s index of refraction is 1.25
- the required thickness is 104.1667 nm
Explanation:
Given the data in the question;
Thickness of coating t = 100 nm
wavelength λ = 500nm
we know that refractive index is;
t = λ/4n
make n, the subject of formula
t4n = λ
n = λ / 4t
we substitute
n = 500 / ( 4 × 100 )
n = 500 / 400
n = 1.25
Therefore, the coating’s index of refraction is 1.25
2)
given that;
Index of refraction of the coating; n = 1.20
λ = 500 nm
thickness of coating t = ?
t = λ / 4n
we substitute
t = 500 / ( 4 × 1.2 )
t = 500 / 4.8
t = 104.1667 nm
Therefore, the required thickness is 104.1667 nm
Answer:
When one system vibrating at its natural frequency is put closer to a stationary system, the stationary system receives impulses.
At resonant frequency, the system vibrating at its own natural frequency suddenly goes on decreasing in order to cope with neighboring system.
These decrease in frequency is known as damping.
gas molecules having at least one oxygen atom
E is the vapourising state
Solo dígales cómo se siente, dígales que no está contento porque está triste y no sabe qué hacer al respecto, que está deprimido, y que solo quiere que las cosas mejoren.
