Physics, 16.03.2020 19:26 chellybean

An incident beam of light moves through liquid water ( 1.33) and impinges upon an interface with glass ( 1.52) with an angle of incidence of 47.0o. What is the angle of the refracted beam (i. e., in the glass) with respect to the surface normal

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Answer from: tatia65

The angle of the refracted beam with respect to the surface normal is $\theta_{2} =39.79^{o}$ .

Explanation:

The direction of a beam of light changes when it crosses the interface between two media with different index of refraction. We will consider that n₁ and n₂ are the index of refraction of the first and the second medium and θ₁ and θ₂ are the angles that the beam of light makes with the surface normal in the first and the second medium respectively.

Snell's Law relates the angle of incidence θ₁ and the angle of refraction θ₂ of a beam of light with the index of refraction in each medium:

$n_{1} sin\theta_{1} =n_{2} sin\theta_{2}$

We are told that:

$n_{1} =1.33$ $n_{2} =1.52$ $\theta_{1} =47^{o}$

if we rearrange the equation:

$sin\theta_{2} =\frac{n_{1}}{n_{2}} sin\theta_{1}$

we get an expression for the angle of the refracted beam:

$\theta_{2}=arcsin(\frac{n_{1}}{n_{2}} sin\theta_{1})$

replacing the values we get that:

$\theta_{2}=arcsin(\frac{1.33}{1.52}\ sin47^{o})$

$\theta_{2} =arcsin(0.875\ .\ 0.731)$

$\theta_{2}=arcsin(0.640)$

finally we get that:

$\theta_{2} =39.79^{o}$

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