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Anna [14]
3 years ago
7

A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction

is 16.3◦ , find the critical angle for light traveling from the liquid back into the air. answer in units of ◦
Physics
1 answer:
lana66690 [7]3 years ago
3 0
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) (2)
where n_r is the refractive index of the second medium and n_i is the refractive index of the first medium.

We can find the ratio n_r / n_i by using Snell's law:
n_i \sin \theta_i = n_r \sin \theta_r (1)
where
\theta_i is the angle of incidence
\theta_r is the angle of refraction

By using the data of the problem and re-arranging (1), we find
\frac{n_r}{n_i} =  \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566

and if we use eq.(2) we can now find the value of the critical angle:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}
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