Answer:
angle at which it will leave glass is 34.33 degree with normal
Explanation:
As per Snell's law we know that
[tex]n_1sin i = n_2 sin r[/tex]
here we have
[tex]n_1 = 1.5 [/tex]
[tex]i = 30^o[/tex] = Angle of incidence
[tex]n_2 = 1.33[/tex]
now from above formula we will have
[tex]1.5 sin30 = 1.33 sin r[/tex]
[tex]r = 34.33[/tex]
so angle at which it will leave glass is 34.33 degree with normal
The light leaves the glass at about 42° relative to its normal
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Let's recall Snell's Law about Refraction as follows:
[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]
where:
n₁ = incident index
θ₁ = incident angle
n₂ = refracted index
θ₂ = refracted angle
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Given:
incident angle = θ₁ = 30°
index of refraction of air = n₃ = 1.0
index of refraction of glass = n₂ = 1.5
index of refraction of water = n₁ = 1.33
Asked:
refracted angle = θ₃ = ?
Solution:
Surface between Glass - Water:
[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]
[tex]\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1[/tex] → Equation A
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Surface between Air - Glass:
[tex]n_2 \sin \theta_2 = n_3 \sin \theta_3[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)[/tex] ← Equation A
[tex]\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}[/tex]
[tex]\sin \theta_3 \approx 0.665[/tex]
[tex]\boxed{\theta_3 \approx 42^o}[/tex]
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Grade: High School
Subject: Physics
Chapter: Light
