Question

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?

Answer 1

The light leaves the glass at about 42° relative to its normal

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Further explanation

Let's recall Snell's Law about Refraction as follows:

$\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}$

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:

$n_1 \sin \theta_1 = n_2 \sin \theta_2$

$\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1$ → Equation A

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Surface between Air - Glass:

$n_2 \sin \theta_2 = n_3 \sin \theta_3$

$\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2$

$\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)$ ← Equation A

$\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1$

$\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o$

$\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}$

$\sin \theta_3 \approx 0.665$

$\boxed{\theta_3 \approx 42^o}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Light

Answer 2

Answer:

angle at which it will leave glass is 34.33 degree with normal

Explanation:

As per Snell's law we know that

$n_1sin i = n_2 sin r$

here we have

$n_1 = 1.5$

$i = 30^o$ = Angle of incidence

$n_2 = 1.33$

now from above formula we will have

$1.5 sin30 = 1.33 sin r$

$r = 34.33$

so angle at which it will leave glass is 34.33 degree with normal