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Q10PE

Expert-verifiedFound in: Page 929

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A scuba diver training in a pool looks at his instructor as shown in Figure 25.53. What angle does the ray from the instructor’s face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is ${\mathbf{25}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{\xb0}}}$.**

The $34.{3}^{\xb0}$angle does the ray from the instructor’s face make with the perpendicular to the water at the point where the ray enters.

**An angle is created by two rays, called the sides of the angle. They share a common end point, called the vertex of the angle, in Euclidean geometry. Angles generated by two rays lie in the plane in which the rays are contained.**

Using Snell's law of refraction, ${n}_{i}\mathrm{sin}{\theta}_{i}={n}_{f}\mathrm{sin}{\mathrm{\theta}}_{\mathrm{f}}$ we need to solve for the angle of refraction

${\theta}_{f}={\mathrm{sin}}^{-1}\left(\frac{{n}_{i}\mathrm{sin}{\theta}_{i}}{{n}_{f}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\frac{1.333\times \mathrm{sin}{25}^{\xb0}}{1.000}\right)\phantom{\rule{0ex}{0ex}}=34.{3}^{\xb0}$

Therefore, the required solution is $34.{3}^{\xb0}$.

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