Why does light behave as shown in the image when it passes from air to glass?
2 answers:
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Ir=Initial Intensity/Area of spread=Io4πr2
Ir∝1r2
It is seen from this expression that intensity is inversely proportional to the square of the distance. As we move away from the light source the intensity decreases at the rate of square of the distance from the source.
Brightness being the perception of intensity. more the intensity more bright the object is perceived and vice versa.
Ir∝1r2
It is seen from this expression that intensity is inversely proportional to the square of the distance. As we move away from the light source the intensity decreases at the rate of square of the distance from the source.
Brightness being the perception of intensity. more the intensity more bright the object is perceived and vice versa.
Answer:
a) The angle of refraction is approximately 34.7
b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°
Explanation:
According to Snell's law, we have;
The refractive index of the glass, n₁ = 1.66
The angle of incident of the light as it moves into water, θ₁ = 27.2°
a) The refractive index of water, n₂ = 1.333
Let θ₂ represent the angle of refraction of the light in water
By plugging in the values of the variables in Snell's Law equation gives;
θ₂ = arcsin(0.5692292265) ≈ 34.7°
The angle of refraction of the light in water, θ₂ ≈ 34.7°
b) When the angle of refraction, θ₂ = 90°, we have;
θ₁ ≈ arcsin(0.803) ≈ 53.42°
The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°