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
<em>A simple metallic band model is proposed for the transition metal mono antimonides, by analogy to the transition metals.</em>
Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
2 is c
3 is a
4 is b
5 is c
