A Ray of light is incident at an angle of 45 degree Celsius at the interface of medium 1 with refractive index 1.0003 and medium 2 with refractive index 1.5 to the angle of refraction will be
Answers
Answer:
Given : i=45
0
r=30
0
(μ
air
)
To find : μ
water
Solution: From snell's law
μ
1
sini=μ
2
sinr
μ
air
sin45
0
=μ
w
sin30
0
2
1
=
2
μ
w
μ
w
2
Now let us take that for angle of incidence α the angle between reflected ray and refracted ray be 90
0
then
α+r+90
0
=180
0
/90
0
(since i = angle of reflection from laws of reflection)
Hence r=90−α
Now from snell's law
μ
1
sinα=μ
2
sinr
1sinα=
2
sin(90−α)
tanα=
2
α=tan
−1
2
Answer:
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Explanation:
A ray of light is incident at an angle of 45° at the interface of medium (1) and medium (2) as shown in the diagram. Redraw this diagram and complete it. If the angle of refraction is 30° find the refractive index of medium (2) with respect to medium (1). begin mathsize 12px style Given comma
open parentheses sin space 45 degree space equals space fraction numerator 1 over denominator square root of 2 end fraction space space and space sin space 30 degree space equals space 1 half close parentheses end style
If second medium is water in place of medium (2), will the angle of refraction increase or decrease? Why?
(refractive index of water =4/3)