Answer:
[tex]30.4^{\circ}[/tex]
Explanation:
Refraction occurs when a ray of light passes from a medium into another medium. In this situation, the ray of light bends and changes speed, according to Snell's Law:
[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]
where
[tex]n_1[/tex] is the index of refraction of the 1st medium
[tex]n_2[/tex] is the index of refraction of the 2nd medium
[tex]\theta_1[/tex] is the angle of incidence (the angle between the incident ray and the normal to the boundary)
[tex]\theta_2[/tex] is the angle of refraction (the angle between the refracted ray and the normal to the boundary)
In this problem, we have:
[tex]n_1=1.00[/tex] is the index of refraction of air
[tex]n_2=1.33[/tex] is the index of refraction of water
[tex]\theta_1=42.3^{\circ}[/tex] is the angle of incidence in air
Solving for [tex]\theta_2[/tex], we find the angle of refraction in water:
[tex]\theta_2=sin ^{-1}(\frac{n_1}{n_2}sin \theta_1)=sin^{-1}(\frac{1.00}{1.33}sin42.3^{\circ})=30.4^{\circ}[/tex]
In the water, the angle of refraction will be "30.4°". To understand the calculation, check below.
According to the question,
Air's index of refraction, n₁ = 1.00
Water's index of refraction, n₂ = 1.33
In air, angle of incidence, θ₁ = 42.3°
Let,
Angle of refraction be "θ₂".
We know the relation,
→ θ₂ = Sin⁻¹([tex]\frac{n_1}{n_2}[/tex] Sinθ₁)
By substituting the values, we get
= Sin⁻¹([tex]\frac{1.00}{1.33}[/tex] Sin 42.3°)
= 30.4°
Thus the approach above is correct.
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